in general, denotes a continual succession of things in the same order, and having the same relation or connection with each other. In this sense we say, a series of emperors, kings, bishops, and the like.

In natural history, a series is used for an order or subdivision of some class of natural bodies; comprehending all such as are distinguished from the other bodies of that class, by certain characters which they possess in common, and which the rest of the bodies of that cast have not.

(1.) Series, in Arithmetic or Algebra, a rank or progression of quantities which succeed one another according to some determinate law. For example, the numbers

3, 5, 7, 9, 11, 13, 15, &c.

constitute a series, the law of which is, that each term exceeds that before it by a given number, viz. 2. Again, the numbers

3, 6, 12, 24, 48, 96, 192, &c.

constitute a series of a different kind, each term being the product of the term before it, and the given number 2.

(2.) As the law according to which the terms of a series are formed may be infinitely varied, there may be innumerable kinds of series. We shall enumerate a few of the most common.

1. Arithmetical Series. The general form of a series of this kind is

\[a, a+d, a+2d, a+3d, a+4d, \ldots\]

and its law is, that the difference between any two adjacent terms is the same quantity, viz. \(d\). The first of the two preceding examples is a series of this nature.

2. Geometrical Series. Its general form is

\[a, ar, ar^2, ar^3, ar^4, \ldots\]

In this kind of series each term is the product of that which precedes it and a constant number \(r\), which is called the common ratio of the terms. The second of the above examples is a particular case of a geometrical series.

3. Harmonic Series is that in which the first of any three of its consecutive terms is to the third, as the difference between the first and second to the difference between the second and third; hence we readily find, that putting \(a\) and \(b\) for its first two terms, its general form will be

\[\frac{ab}{2a-b}, \frac{ab}{3a-2b}, \frac{ab}{4a-3b}, \ldots\]

If we suppose \(a = 1\) and \(b = \frac{1}{2}\), we get

\[1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \ldots\]

as a particular example of a harmonic series.

4. Recurring Series. Let its terms be denoted by

\[A, B, C, D, E, F, \ldots\]

Then we shall form a recurring series if, \(m\) and \(n\) being put for given quantities, we take

\[C = mA + nB, \quad E = mC + nD,\] \[D = mB + nC, \quad F = mD + nE.\]

For example, let us suppose \(A = 1\), \(B = 2x\), \(m = 4x^2\), \(n = 3x\); then \(C = 10x^2\), \(D = 38x^2\), \(E = 154x^2\), \(F = 614x^2\), so that the first six terms of the series are

\[1, 2x, 10x^2, 38x^2, 154x^2, 614x^2.\]

We have here supposed each term to be formed from the two which come immediately before it; but the name recurring series is given to every one in which the terms are formed in like manner from some assigned number of the terms which precede that sought. Thus, putting, as before, \(A, B, C, D, E, F, \ldots\) for the terms of the series, and \(m, n, p, q\), for given quantities, we shall have another recurring series, if we suppose them so related that

\[mA + nB + pC + qD = 0,\] \[mB + nC + pD + qE = 0,\] \[mC + nD + pE + qF = 0.\]

The two series of quantities \(sin.\), \(a, sin.\), \(2a, sin.\), \(3a, \ldots\) and \(cos.\), \(a, cos.\), \(2a, cos.\), \(3a, \ldots\) are both recurring, as is manifest from the law which connects the quantities one with another. (See Algebra, § 251.)

(3.) As in general it is the sum of the terms of a series which is the object of investigation, it is usual to connect them by the sign \(+\) or \(-\), and to apply the name series to the expression thus formed. Accordingly,

\[1 + 3 + 5 + 7 + 9 + \cdots + \{1 + 2(n - 1)\}\]

(where \(n\) denotes the number of terms) is called an arithmetical series; and in like manner

\[1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots + \frac{1}{2^{n-1}}\]

is a geometrical series.

(4.) A series may either consist of a definite number of terms, or their number may be supposed greater than any that can be assigned, and in this case the series is said to be infinite. The number of terms of a series may be infinite, and yet their sum finite. This is true, for example, of the series

\[\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots\]

which is equivalent to unity, or 1.

(5.) We have already treated of several branches of the doctrine of series in the articles ALGEBRA, FLUXIONS, and LOGARITHMS; and, in particular, we have given four different methods for expanding a quantity into a series, viz.

1. By Division or Evolution. 2. By the Method of Indeterminate Co-efficients. 3. By the Binomial Theorem. (ALGEBRA, sect. xvii.) 4. By Taylor's Theorem. (FLUXIONS, § 28.)

We shall here treat briefly of another branch of the theory, namely, how to find the sum of any proposed number of terms of certain series, or the sum of their terms continued ad infinitum, when that sum is finite.

(6.) There is a great analogy between the terms of a series and the ordinates of a curve which are supposed to stand upon the axis at equal distances from one another, the first ordinate reckoned from the extremity of the axis being analogous to the first term of the series, the second ordinate to the second term, and so on. From this analogy it follows immediately, that like as the nature of a curve is indicated by an equation expressing the value of an indefinite ordinate in terms of its corresponding abscissa, so also the nature of a series may be shown by an equation which shall express the relation between any term, and the num- Series.

ber that denotes the place or order of that term in the series. In conformity to this method, putting the symbols $T(1), T(2), T(3), \ldots$ to denote the terms of any series whatever, we may express it generally thus—

$$T(1) + T(2) + T(3) + \cdots + T(n),$$

where the characters (1), (2), are meant to denote the place or order of the terms to which they are joined (the first term being supposed to have the place 1, the second term the place 2, and so on), and (n) is put for any indefinite number.

The nature of the arithmetical series

$$a + (a + d) + (a + 2d) + (a + 3d) + \cdots,$$

will be defined by the equation

$$T(n) = a + (n - 1)d;$$

and, in like manner, the nature of the geometrical series

$$a + ar + ar^2 + ar^3 + \cdots,$$

will be expressed by the equation

$$T(n) = ar^{n-1}.$$

(7.) As the expression for the value of the indefinite term $T(n)$ becomes identical with all the terms of the series in succession, by substituting the numbers 1, 2, 3, &c., one after another, for $n$, that expression is called the general term of the series. In the series

$$a + b + \frac{ab}{2a-b} + \frac{ab}{3a-2b} + \frac{ab}{4a-3b} + \cdots,$$

the general term is evidently

$$\frac{(v-1)a-(v-2)b}{v},$$

(8.) We shall investigate the sum of any number of terms of such series as have their general terms expressed by any one of the following algebraic functions:

$$\frac{v(v+1)}{1 \cdot 2}, \frac{v(v+1)(v+2)}{1 \cdot 2 \cdot 3}, \frac{v(v+1)(v+2)(v+3)}{1 \cdot 2 \cdot 3}, \ldots,$$

Problem I. It is proposed to find the sum of $n$ terms of the series of which the general term is the first function.

By putting 1, 2, 3, &c. to $n$ successively for $v$, it appears that the series to be summed is

$$1 + 2 + 3 + \cdots + n.$$

Now, as $v = \frac{v(v+1)}{2} - \frac{(v-1)v}{2}$, we have, by putting in this formula 1, 2, 3, $\cdots$ to $n$ successively, for $v$,

$$1 = \frac{1 \cdot 2}{2} - 0,$$ $$2 = \frac{2 \cdot 3}{2} - \frac{1 \cdot 2}{2},$$ $$3 = \frac{3 \cdot 4}{2} - \frac{2 \cdot 3}{2},$$ $$4 = \frac{4 \cdot 5}{2} - \frac{3 \cdot 4}{2},$$ $$\vdots$$ $$n-1 = \frac{(n-1)n}{2} - \frac{(n-2)(n-1)}{2},$$ $$n = \frac{n(n+1)}{2} - \frac{(n-1)n}{2}.$$

Let the sum of the quantities on each side of the sign $=$ be now taken; then, observing that each of the fractions on the right-hand side, with the exception of $\frac{n(n+1)}{1 \cdot 2}$, occurs twice, once with the sign $+$, and again with the sign $-$, by which it happens that their aggregate is $= 0$, it is evident that we have

$$1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{1 \cdot 2}.$$

Prob. II. It is proposed to sum $n$ terms of the series having for its general term the second function

$$\frac{v(v+1)}{1 \cdot 2}.$$

This series, by substituting 1, 2, 3, &c. successively for $v$, is found to be

$$\frac{1 \cdot 2}{1 \cdot 2} + \frac{2 \cdot 3}{1 \cdot 2} + \frac{3 \cdot 4}{1 \cdot 2} + \cdots + \frac{n(n+1)}{1 \cdot 2}.$$

We now, following the mode of proceeding employed in last problem, put the expression $\frac{v(v+1)}{1 \cdot 2}$ under this form,

$$\frac{v(v+1)(v+2)}{1 \cdot 2 \cdot 3} - \frac{(v-1)v(v+1)}{1 \cdot 2 \cdot 3},$$

to which it is evidently equivalent, and, substituting 1, 2, 3, &c. successively for $v$, find

$$\frac{1 \cdot 2}{1 \cdot 2} = \frac{1 \cdot 2 \cdot 3}{1 \cdot 2 \cdot 3} - 0,$$ $$\frac{2 \cdot 3}{1 \cdot 2} = \frac{2 \cdot 3 \cdot 4}{1 \cdot 2 \cdot 3} - \frac{1 \cdot 2 \cdot 3}{1 \cdot 2 \cdot 3},$$ $$\frac{3 \cdot 4}{1 \cdot 2} = \frac{3 \cdot 4 \cdot 5}{1 \cdot 2 \cdot 3} - \frac{2 \cdot 3 \cdot 4}{1 \cdot 2 \cdot 3},$$ $$\frac{4 \cdot 5}{1 \cdot 2} = \frac{4 \cdot 5 \cdot 6}{1 \cdot 2 \cdot 3} - \frac{3 \cdot 4 \cdot 5}{1 \cdot 2 \cdot 3},$$ $$\vdots$$ $$\frac{n(n+1)}{1 \cdot 2} = \frac{n(n+1)(n+2)}{1 \cdot 2 \cdot 3} - \frac{(n-1)n(n+1)}{1 \cdot 2 \cdot 3}.$$

In this problem, as in the former, it appears that each quantity on the right side of the equations, except $\frac{n(n+1)(n+2)}{1 \cdot 2 \cdot 3}$, occurs twice, and with contrary signs; therefore, taking the aggregate of the terms on each side, we have

$$\frac{1 \cdot 2}{1 \cdot 2} + \frac{2 \cdot 3}{1 \cdot 2} + \frac{3 \cdot 4}{1 \cdot 2} + \frac{4 \cdot 5}{1 \cdot 2} + \cdots + \frac{n(n+1)}{1 \cdot 2} = \frac{n(n+1)(n+2)}{1 \cdot 2 \cdot 3}.$$

(9.) It will be obvious, by a little attention to the solutions of these two problems, that in each, the terms of the series to be summed are the differences betwixt the adjacent terms of another series, namely, that which has for its general term the function next in order to the general term of the series under consideration; that is, the terms of the series whose general term is $v$ are the differences betwixt those of the series having $\frac{v(v+1)}{1 \cdot 2}$ for its general term; and again, the terms of this last are the differences of the terms of the series having $\frac{v(v+1)(v+2)}{1 \cdot 2 \cdot 3}$ for its general term. Now, as the sum of the differences of any series of quantities whatever which begins with 0 must necessarily be the last term of that series, it follows that the sum of all the terms of each of the series we have considered must be equal to the last term of the next following series; and this term is necessarily the expression formed by substituting $n$ for $v$ in its general term, that is, the sum of the series $1 + 2 + 3 + \cdots + n$, which has $v$ for its general term, is $\frac{n(n+1)}{1 \cdot 2}$; and the sum of the series

---

1 For example, let the quantities be $\theta, a, b, c, d$; then it is manifest that $(a - 0) + (b - a) + (c - b) + (d - c) = d$. The next series, which has \( \frac{v(v+1)(v+2)}{1 \cdot 2 \cdot 3} \) for its general term, as well as all that succeed, will be found to have the very same property, as may be proved as follows. Let \( p \) denote any term of the series of natural numbers 1, 2, 3, &c. Then, because

\[ 1 = \frac{v + p}{p + 1} - \frac{v - 1}{p + 1}, \]

if we multiply these equals by the product of all the factors \( v, \frac{v+1}{2}, \frac{v+2}{3}, \ldots \), &c., to \( \frac{v+p-1}{p} \), we get

\[ v(v+1)(v+2)\cdots(v+p-1) \]

Now, if in this identical equation we substitute the numbers 1, 2, 3, &c. to \( n \) successively, for \( v \), the results obtained from its first member

\[ \frac{v(v+1)(v+2)\cdots(v+p-1)}{1 \cdot 2 \cdot 3 \cdots (p+1)}, \]

will be a series having this function for its general term, and the terms of which will evidently be the difference between the terms of another series having the first part of the second member of the equation, viz.

\[ \frac{v(v+1)(v+2)\cdots(v+p)}{1 \cdot 2 \cdot 3 \cdots (p+1)}, \]

for its general term; Hence it will happen, as in the two foregoing problems, that the sum of all the terms of the former series will be equal to the last term of the latter; which conclusion may be expressed in the form of a theorem, as follows:

**Theorem.** The sum of \( n \) terms of a series having for its general term the function

\[ \frac{v(v+1)(v+2)\cdots(v+p-1)}{1 \cdot 2 \cdot 3 \cdots p}, \]

is equal to

\[ \frac{n(n+1)(n+2)\cdots(n+p)}{1 \cdot 2 \cdot 3 \cdots (p+1)}. \]

Or, setting aside the denominators of the terms, we may express the theorem thus: *The sum of \( n \) terms of a series, having for its general term the expression*

\[ v(v+1)(v+2)\cdots(v+p-1), \]

*is equal to*

\[ \frac{n(n+1)(n+2)\cdots(n+p)}{p+1}. \]

We shall give a few particular cases of this formula.

I. \( 1 + 2 + 3 + 4 + \cdots + n = \frac{n(n+1)}{2} \).

II. \( 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + 4 \cdot 5 + \cdots + n(n+1) = \frac{n(n+1)(n+2)}{3} \).

III. \( 1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + 3 \cdot 4 \cdot 5 + \cdots + n(n+1)(n+2) = \frac{n(n+1)(n+2)(n+3)}{4} \).

(10.) By means of the above general theorem, we may find the sum of any number of terms of a series composed of the powers of the terms of an arithmetical progression, the general term of which will, in the simplest case, be \( v^p \), \( p \) being a given number. The manner of doing this will appear from the following problems.

**Prin. III.** It is proposed to find the sum of \( n \) terms of the series of squares \( 1 + 4 + 9 + 16 + 25 + \cdots \), &c. or \( 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + \cdots \).

The general term of this series being \( v^2 \), we put it under this form, \( v(v+1) - v \); hence we get, by substituting 1, 2, 3, &c. for \( v \),

\[ 1^2 = 1 \cdot 2 - 1, \] \[ 2^2 = 2 \cdot 3 - 2, \] \[ 3^2 = 3 \cdot 4 - 3, \] \[ 4^2 = 4 \cdot 5 - 4, \] \[ \vdots \]

Therefore, adding, we find

\[ 1^2 + 2^2 + 3^2 + 4^2 + \cdots + n^2 = 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + n(n+1). \]

But by the general theorem (9.)

\[ 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + n(n+1) = \frac{n(n+1)(n+2)}{3}, \]

and \( 1 + 2 + 3 + 4 + \cdots + n = \frac{n(n+1)}{2} \);

therefore \( 1^2 + 2^2 + 3^2 + 4^2 + \cdots + n^2 = \frac{n(n+1)(n+2)}{3} - \frac{n(n+1)}{2} = \frac{n(n+1)(2n+1)}{6} \).

We might have arrived at the same conclusion by considering, that since \( v^2 \), the general term of the series, is equivalent to \( v(v+1) - v \), the series must be the difference between two others, one having \( v(v+1) \) and the other \( v \) for its general term; for the sake of perspicuity, however, we have put down the terms of all the three series.

**Prin. IV.** It is proposed to find the sum of \( n \) terms of the series

\[ 1^3 + 2^3 + 3^3 + 4^3 + 5^3 + \cdots, \]

The general term in this case is \( v^3 \); now, to transform this function, so as to deduce the sum of the series from the general theorem, we assume

\[ v^3 = v(v+1)(v+2) + Av(v+1) + Bv, \]

where \( A \) and \( B \) denote quantities which are to have such values as shall render the two sides of the equation identical, whatever be the value of \( v \). Taking now the product of the factors, we have

\[ v^3 = v^3 + (A + 3)v^2 + (A + B + 2)v; \]

therefore, by the theory of indeterminate co-efficients (Algebra, § 159),

\[ A + 3 = 0, \quad A + B + 2 = 0; \]

hence we find \( A = -3, \quad B = -A - 2 = 1 \); thus it appears that, \( v \) being any number whatever,

\[ v^3 = v(v+1)(v+2) - 3v(v+1) + v. \]

Now, let \( S \) denote the sum of \( n \) terms of the series under consideration, which has \( v^3 \) for its general term, and put \( P, Q, R \) for the like sums of the three series whose general terms are the functions \( v(v+1)(v+2), v(v+1), \) and \( v \) respectively; then it is evident that \( S = P - 3Q + R \).

But by the theorem (9.)

\[ P = \frac{n(n+1)(n+2)(n+3)}{4}, \] \[ Q = \frac{n(n+1)(n+2)}{3}, \] \[ R = \frac{n(n+1)}{2}; \] therefore \( S = \frac{n(n+1)(n+2)(n+3)}{4} \)

\[- n(n+1)(n+2) + \frac{n(n+1)}{2},\]

and, by proper reduction,

\[ S = 1^3 + 2^3 + 3^3 + 4^3 \cdots + n^3 = \frac{n^2(n+1)^2}{4}. \]

**Corollary.** We have found (Prop. I.) that

\[ 1 + 2 + 3 + 4 \cdots + n = \frac{n(n+1)}{2}, \]

therefore, comparing this with the result just now obtained, it is evident that

\[ (1 + 2 + 3 + 4 \cdots + n)^2 = 1^3 + 2^3 + 3^3 + 4^3 \cdots + n^3. \]

This is a very curious and elegant property of numbers.

**(II.)** It is manifest that, by the mode of proceeding employed in last problem, we may investigate the sum of \( n \) terms of the series

\[ 1^n + 2^n + 3^n + 4^n + \cdots, \]

\( n \) being any whole positive number whatever; and indeed in the very same way we may find the sum of any number of terms of a series whose general term is

\[ a + b v + c v^2 + d v^3 + \cdots, \]

where \( a, b, c, \ldots \) denote given numbers; namely, by transforming it into a function of the form

\[ A + B v + C v^2 + D v^3 + \cdots, \]

where \( A, B, C, \ldots \) denote constant quantities. Our limits, however, will not allow us to go into particulars.

**(III.)** The next class of series we shall consider, comprehends such as may be formed by the successive substitution of \( a, a+1, a+2, \ldots \) (a being put for any given quantity whatever) in the series of functions

\[ \frac{1}{v(v+1)}, \frac{1}{v(v+1)(v+2)}, \frac{1}{v(v+1)(v+2)(v+3)}, \ldots \]

We shall begin with the first of these.

**Prop. V.** It is proposed to find the sum of \( n \) terms of the series

\[ \frac{1}{a(a+1)} + \frac{1}{(a+1)(a+2)} + \frac{1}{(a+2)(a+3)} + \cdots, \]

which is formed by substituting \( a, a+1, a+2, \ldots \) successively for \( v \) in the general term

\[ \frac{1}{v(v+1)}. \]

Whatever be the value of \( v \), we have

\[ \frac{1}{v(v+1)} = \frac{1}{v} - \frac{1}{v+1}; \]

therefore, proceeding as in the foregoing problems, we get

\[ \frac{1}{a(a+1)} = \frac{1}{a} - \frac{1}{a+1}, \]

\[ \frac{1}{(a+1)(a+2)} = \frac{1}{a+1} - \frac{1}{a+2}, \]

\[ \frac{1}{(a+2)(a+3)} = \frac{1}{a+2} - \frac{1}{a+3}, \]

\[ \ldots \]

\[ \frac{1}{(a+n-2)(a+n-1)} = \frac{1}{a+n-2} - \frac{1}{a+n-1}, \]

\[ \frac{1}{(a+n-1)(a+n)} = \frac{1}{a+n-1} - \frac{1}{a+n}. \]

Here it is evident that the terms of the series to be summed are the differences betwixt every two adjoining terms of this other series,

\[ \frac{1}{a} + \frac{1}{a+1} + \frac{1}{a+2} + \frac{1}{a+3} + \cdots + \frac{1}{a+n}, \]

hence it immediately follows, that the sum of all the terms of the former is the difference between the two extreme terms of the latter; that is,

\[ \frac{1}{a(a+1)} + \frac{1}{(a+1)(a+2)} + \cdots + \frac{1}{(a+n-1)(a+n)} = \frac{1}{a} - \frac{1}{a+n}. \]

If we suppose the series to be continued ad infinitum, then, as \( n \) will be indefinitely great, and \( \frac{1}{a+n} \) indefinitely small, the sum will be simply \( \frac{1}{a} \); or, in other words, the fraction \( \frac{1}{a} \) is a limit to the sum of the series.

**Prop. VI.** Let it be required to find the sum of \( n \) terms of this series,

\[ \frac{1}{a(a+1)(a+2)} + \frac{1}{(a+1)(a+2)(a+3)} + \cdots + \frac{1}{(a+2)(a+3)(a+4)} + \cdots, \]

the general term in this case being

\[ \frac{1}{v(v+1)(v+2)}. \]

Because

\[ \frac{2}{v(v+2)} = \frac{1}{v} - \frac{1}{v+2}, \]

therefore, multiplying by \( \frac{1}{2(v+1)} \), we have

\[ \frac{1}{v(v+1)(v+2)} = \frac{1}{2} \left\{ \frac{1}{v(v+1)} - \frac{1}{(v+1)(v+2)} \right\}, \]

and hence, by substituting \( a, a+1, a+2, \ldots \) successively for \( v \),

\[ \frac{1}{a(a+1)(a+2)} = \frac{1}{2} \left\{ \frac{1}{a(a+1)} - \frac{1}{(a+1)(a+2)} \right\}, \]

\[ \frac{1}{(a+1)(a+2)(a+3)} = \frac{1}{2} \left\{ \frac{1}{(a+1)(a+2)} - \frac{1}{(a+2)(a+3)} \right\}, \]

\[ \frac{1}{(a+2)(a+3)(a+4)} = \frac{1}{2} \left\{ \frac{1}{(a+2)(a+3)} - \frac{1}{(a+3)(a+4)} \right\}, \]

\[ \ldots \]

\[ \frac{1}{(a+n-1)(a+n)(a+n+1)} = \frac{1}{2} \left\{ \frac{1}{(a+n-1)(a+n)} - \frac{1}{(a+n)(a+n+1)} \right\}. \]

Hence it appears that the terms of the series to be summed are the halves of the differences of the terms of the series

\[ \frac{1}{a(a+1)} + \frac{1}{(a+1)(a+2)} + \frac{1}{(a+2)(a+3)} + \cdots + \frac{1}{(a+n)(a+n+1)}; \]

consequently, the sum of all the terms of the former is half the difference between the extreme terms of the latter, or is

\[ \frac{1}{2} \left\{ \frac{1}{a(a+1)} - \frac{1}{(a+n)(a+n+1)} \right\}. \]

**(IV.)** From these two particular cases it is easy to see how we may sum the series when the general term is

\[ \frac{1}{v(v+1)(v+2) \cdots (v+p)}. \] Series. \( p \) being any whole number whatever: for since

\[ \frac{p}{v(v+p)} = \frac{1}{v} - \frac{1}{v+p}, \]

therefore, multiplying the denominators by all the factors which are intermediate between \( v \) and \( v + p \), we have

\[ \frac{p}{v(v+1)(v+2)\ldots(v+p)} = \frac{1}{v(v+1)(v+2)\ldots(v+p-1)} - \frac{1}{(v+1)(v+2)(v+3)\ldots(v+p)}. \]

Now the latter side of this equation is a general expression for the difference between any two adjacent terms of a series whose general term is

\[ \frac{1}{v(v+1)(v+2)\ldots(v+p-1)}, \]

therefore the difference between the first and last terms of this series must be the sum of the series whose general term is the function on the other side of the equation, viz.

\[ \frac{p}{v(v+1)(v+2)\ldots(v+p)}. \]

Hence we have the following very general theorem.

**Theorem.** Let \( a \) denote any number whatever, and let \( 1, 2, 3, \ldots, p \) be a series of numbers, each of which exceeds that before it by unity; the sum of \( n \) terms of a series formed by substituting the numbers \( a, a+1, a+2, \ldots, a+n-1 \) successively for \( v \) in the function

\[ \frac{1}{v(v+1)(v+2)\ldots(v+p)} \]

is equal to

\[ \frac{1}{p} \left[ \frac{a(a+1)(a+2)\ldots(a+p-1)}{(a+n)(a+n+1)(a+n+2)\ldots(a+n+p-1)} \right]. \]

**Corollary.** The same series continued ad infinitum is equal to

\[ \frac{1}{p} \cdot \frac{a(a+1)(a+2)\ldots(a+p-1)}{1}. \]

(14.) We shall now give examples of the application of this theorem.

**Example 1.** Required the sum of \( n \) terms of the series

\[ \frac{1}{2 \cdot 3 \cdot 4 \cdot 5} + \frac{1}{3 \cdot 4 \cdot 5 \cdot 6} + \frac{1}{4 \cdot 5 \cdot 6 \cdot 7} + \ldots, \]

The terms of this series are evidently produced by the successive substitution of the numbers \( 2, 3, 4, 5, \ldots \) for \( v \) in the function

\[ \frac{1}{v(v+1)(v+2)(v+3)}; \]

therefore, comparing this expression with the general formula, we have \( a = 2, p = 3 \), and the sum required

\[ = \frac{1}{2} \left\{ \frac{1}{2 \cdot 3 \cdot 4} - \frac{1}{(2+n)(3+n)(4+n)} \right\}. \]

**Ex. 2.** Required the sum of the series

\[ \frac{1}{1 \cdot 4 \cdot 7} + \frac{1}{4 \cdot 7 \cdot 10} + \frac{1}{7 \cdot 10 \cdot 13} + \frac{1}{10 \cdot 13 \cdot 16} + \ldots, \]

continued ad infinitum.

By a little attention it will appear that its terms are produced by the substitution of the numbers \( \frac{1}{3}, \frac{1}{13}, \frac{1}{23}, \ldots \) successively for \( v \) in the function

\[ \frac{1}{3v(3v+3)(3v+6)} = \frac{1}{27v(v+1)(v+2)}. \]

In this case, \( a = \frac{1}{3}, p = 2 \), therefore the sum is

\[ \frac{1}{2} \times \frac{1}{27} \times \frac{1}{1} \times \frac{1}{14} = \frac{1}{24}. \]

(15.) When the function from which the series is derived has not the very form required in the theorem, it may be brought to that form by employing suitable transformations, as in the two following examples.

**Ex. 3.** It is proposed to find the sum of the series

\[ \frac{1}{14} + \frac{1}{25} + \frac{1}{36} + \frac{1}{47} + \ldots, \]

continued ad infinitum.

This series is evidently formed by the substitution of the numbers \( 1, 2, 3, \ldots \) successively for \( v \) in the function

\[ \frac{1}{v(v+3)}. \]

This expression, however, does not in its present form agree with the general formula, because the factors \( v+1, v+2 \) are wanting; therefore, to transform it, we multiply its numerator and denominator by \( (v+1)(v+2) \), and it becomes

\[ \frac{(v+1)(v+2)}{v(v+1)(v+2)(v+3)}. \]

We next assume its numerator

\[ (v+1)(v+2) = A(v+2)(v+3) + B(v+3) + C, \]

and by multiplying get

\[ v^2 + 3v + 2 = Av^2 + (5A+B)v + (6A+3B+C); \]

therefore, that \( v \) may be indeterminate, we must make

\[ A = 1, 5A+B = 3, 6A+3B+C = 2, \]

from which equations we get \( A = 1, B = 3 - 5A = -2, C = 2 - 6A - 3B = 2 \), so that

\[ \frac{1}{v(v+3)} = \frac{(v+2)(v+3) - 2(v+3) + 2}{v(v+1)(v+2)(v+3)}. \]

Thus it appears that the proposed series is resolvable into three others, the general terms of which all agree with the theorem. Now the sum of the infinite series whose general term is

\[ \frac{1}{v(v+1)} \text{ appears by the theorem to be } \frac{1}{a} \text{ or } 1, \]

because \( a = 1 \), and the sum of the infinite series whose general term is

\[ \frac{-2}{v(v+1)(v+2)} \text{ is in like manner found to be } \frac{-2}{2} + \frac{1}{1 \cdot 2} = \frac{1}{2}; \]

and, lastly, the infinite series whose general term is

\[ \frac{2}{v(v+1)(v+2)(v+3)} \text{ is } \frac{2}{3 \cdot 1 \cdot 2 \cdot 3} = \frac{1}{9}; \]

therefore, collecting these into one, the sum of the proposed series is

\[ 1 - \frac{1}{2} + \frac{1}{3} = \frac{11}{18} \text{ the answer.} \]

**Ex. 4.** Required the sum of the infinite series

\[ \frac{1}{2 \cdot 3 \cdot 4} + \frac{2}{3 \cdot 4 \cdot 5} + \frac{3}{4 \cdot 5 \cdot 6} + \frac{4}{5 \cdot 6 \cdot 7} + \ldots, \]

The terms of this series are evidently formed by the substitution of the numbers \( 2, 3, 4, \ldots \) successively in the function

\[ \frac{v-1}{v(v+1)(v+2)}. \]

Now \( v-1 = v+2-3 \); therefore

\[ \frac{v-1}{v(v+1)(v+2)} = \frac{1}{v(v+1)} - \frac{3}{v(v+1)(v+2)}. \]

Thus it appears that the proposed series is reducible to two others, one having its terms produced by the substitution of 2, 3, &c. for \( v \) in the function \( \frac{1}{v(v+1)} \), and the other by a like substitution in the function \( \frac{-3}{v(v+1)(v+2)} \). Now, by our theorem, the sum of the first of these is \( \frac{1}{2} \), and that of the second is \( \frac{-3}{2} \cdot \frac{1}{2-3} = -\frac{1}{3} \); therefore the sum of the proposed series is \( \frac{1}{2} - \frac{1}{4} = \frac{1}{4} \).

From these examples it is sufficiently evident how the theorem is to be applied in other cases; and it appears also, that by means of it we can sum any series whatever whose general term is of the form

\[ \frac{A}{v(1+v)} + \frac{B}{v(1+v)(v+1)} + \frac{C}{v(1+v)(v+1)(v+2)(v+3)} + \ldots \]

or admits of being reduced to that form.

(16.) It deserves to be remarked, that the series

\[ \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \ldots \]

which is of a very simple form, and in appearance of the same nature as those we have summed, does not, however, admit of being treated in the same manner; and indeed if it be continued ad infinitum, its sum is infinite, that is, it exceeds any number which can be assigned. The truth of this assertion will be evident if we can show that a certain definite number of its terms, beginning with any proposed term, can always be found, the sum of which shall exceed an unit, or 1; for this being the case, as we can go on con- tinually in assigning such sets of terms, we can conceive as many to be taken as there are units in any proposed number, however great; and therefore their sum, and much more the sum of all the terms of the series, from its begin- ning to the end of the last sets of terms, will exceed that number. Now, that this can always be done may be proved as follows:

Let the term of the series from which we are to reckon be \( \frac{1}{a} \); then, if the thing be possible, and if \( n \) be the requi- site number of terms, we must have

\[ \frac{1}{a} + \frac{1}{a+1} + \frac{1}{a+2} + \frac{1}{a+3} + \ldots + \frac{1}{a+n-1} > 1. \]

Now, because

\[ a\left(1 + \frac{1}{a}\right)^2 = a + 2 + \frac{1}{a}, \]

\[ a\left(1 + \frac{1}{a}\right)^3 = a + 3 + \frac{3}{a} + \frac{1}{a^2}, \]

and in general,

\[ a\left(1 + \frac{1}{a}\right)^p = a + p + \frac{p(p-1)}{1 \cdot 2} \cdot \frac{1}{a} + \ldots, \]

therefore, \( p \) being any whole number,

\[ a\left(1 + \frac{1}{a}\right)^p > a + p, \text{ and consequently} \]

\[ \frac{1}{a+p} < \frac{1}{a\left(1 + \frac{1}{a}\right)^p}; \]

hence it follows, that the series

\[ \frac{1}{a} + \frac{1}{a+1} + \frac{1}{a+2} + \ldots + \frac{1}{a+n-1} \]

will be greater than the other series

\[ \frac{1}{a} + \frac{1}{a\left(1 + \frac{1}{a}\right)} + \frac{1}{a\left(1 + \frac{1}{a}\right)^2} + \ldots \]

Now, this last being evidently a geometrical series, of which the common ratio is \( \frac{1}{1 + \frac{1}{a}} \), its sum is

\[ 1 + \frac{1}{a} + \frac{1}{a\left(1 + \frac{1}{a}\right)^n-1}; \]

therefore the sum of the series

\[ \frac{1}{a} + \frac{1}{a+1} + \frac{1}{a+2} + \frac{1}{a+3} + \ldots + \frac{1}{a+n-1} \]

will always be greater than this expression; but if we sup- pose \( n \) so great that the quantity \( \left(1 + \frac{1}{a}\right)^{n-1} \) is equal to or exceeds \( a \), which is evidently always possible, then the above expression for the sum of the geometrical series will be equal to 1, or will exceed 1; therefore the same number of terms of the series \( \frac{1}{a} + \frac{1}{a+1} + \frac{1}{a+2} + \frac{1}{a+3} + \ldots \) will al- ways exceed 1; now this is the property of the series we proposed to demonstrate.

When \( a = \left(1 + \frac{1}{a}\right)^{n-1} \), then \( a^2 = a\left(1 + \frac{1}{a}\right)^{n-1} \); but this quantity is greater than \( a + n - 1 \), the denominator of the last term of the series

\[ \frac{1}{a} + \frac{1}{a+1} + \frac{1}{a+2} + \frac{1}{a+3} + \ldots + \frac{1}{a+n-1}, \]

the sum of which, we have proved, will upon that hypoth- esis exceed unity; much more then will the sum exceed unity if we suppose the series continued until the deno- minator of its last term be equal to or greater than \( a^2 \). Hence, beginning with the term \( \frac{1}{a} \), it appears that

\[ \frac{1}{a} + \frac{1}{a+1} + \frac{1}{a+2} + \frac{1}{a+3} + \ldots + \frac{1}{a+n-1} > 1, \]

\[ \frac{1}{a} + \frac{1}{a+1} + \frac{1}{a+2} + \frac{1}{a+3} + \ldots + \frac{1}{a+25} > 1, \]

\[ \frac{1}{a} + \frac{1}{a+1} + \frac{1}{a+2} + \frac{1}{a+3} + \ldots + \frac{1}{a+676} > 1, \]

\[ \frac{1}{a} + \frac{1}{a+1} + \frac{1}{a+2} + \frac{1}{a+3} + \ldots + \frac{1}{a+458929} > 1, \]

&c.

Although the sum of the series we have been considering is infinite, yet it evidently increases very slowly; indeed it is a limit to all such as have a finite sum; for every in- finite series the terms of which decrease faster than the re- ciprocals of an arithmetical progression is always finite.

(17.) We have already explained what is meant by a recurring series (2.). We shall now treat briefly, first, of their origin; next, of the way in which they may be sum- med; and, lastly, of the manner of determining the general term of any particular series.

The series which is produced by the development of a rational algebraic fraction has always the property which constitutes the characteristic of the class called recurring (2.); and, on the other hand, any series having that property being proposed, an algebraic fraction may be found, by the expansion of which the series shall be produced.

The fraction \( \frac{1 + 2x}{1-x-x^2} \), for example, by dividing the numerator by the denominator, is converted into the infi- nite series

\[ 1 + 3x + 4x^2 + 7x^3 + 11x^4 + 18x^5 + \ldots \] which is of such a nature that if \( T, T', T'' \) denote any three of its succeeding terms, their relation to one another is expressed by the equation

\[ T' = T + T'. \]

If we employ algebraic division to convert the fraction into a series, the law of its terms will not appear so readily as if we were to use the method of indeterminate co-efficients. By this method we assume the fraction equal to

\[ A + Bx + Cx^2 + Dx^3 + Ex^4 + \ldots, \]

and hence, multiplying by the denominator, and bringing all the terms to one side, as explained in Algebra, § 159, we have

\[ \begin{align*} (A + B) & + (C + D) \\ -1 & -A \\ -2 & -B \\ & x^2 + x^3 + \ldots = 0; \end{align*} \]

and hence

\[ \begin{align*} A - 1 & = 0, \\ C - B & = A = 0, \\ B - A & = 2 = 0, \\ D - C & = B = 0, \\ & \ldots \end{align*} \]

From these equations it appears that the law of the series is such as we have assigned.

The equation expressing the relation which subsists among a certain number of succeeding terms of a recurring series is called its scale of relation. The same name is also sometimes given to the equation expressing the connection of the co-efficients of the terms. Thus the scale of relation of the foregoing series is either

\[ T' = T + T', \]

where \( T, T', \) and \( T'' \) denote any three succeeding terms of the series; or it is

\[ R = P + Q, \]

where \( P, Q, \) and \( R \) denote their numeral co-efficients.

(18.) We come next to show how the sum of any proposed number of terms of a recurring series may be found. Let the series continued to \( n \) terms be

\[ T_1 + T_2 + T_3 + \ldots + T_{n-1} + T_n, \]

where the characters \( T_1, T_2, \ldots \) denote the successive terms, and the numbers \( (1), (2), \ldots \) their order or place; and as whatever number of terms is contained in the scale, the manner of summing the series is the same, we shall in what follows, for the sake of brevity, suppose that it consists of three, in which case it may be expressed thus,

\[ pT_{n-2} + qT_{n-1} + rT_n = 0, \]

where \( p, q, r \) denote certain given quantities.

The scale of relation affords the following series of equations,

\[ \begin{align*} pT_1 + qT_2 + rT_3 & = 0, \\ pT_2 + qT_3 + rT_4 & = 0, \\ pT_3 + qT_4 + rT_5 & = 0, \\ & \ldots \\ pT_{n-2} + qT_{n-1} + rT_n & = 0. \end{align*} \]

Taking now the sum of these equations, we get

\[ \begin{align*} p\{T_1 + T_2 + T_3 + \ldots + T_{n-1}\} \\ + q\{T_2 + T_3 + T_4 + \ldots + T_n\} \\ + r\{T_3 + T_4 + T_5 + \ldots + T_n\} \\ & = 0. \end{align*} \]

But, putting \( s \) for the sum of \( n \) terms of the series, this equation may manifestly be expressed thus,

\[ \begin{align*} p\{s - T_{n-1} - T_n\} \\ + q\{s - T_1 - T_2\} \\ + r\{s - T_1 - T_2\} \\ & = 0. \end{align*} \]

Hence, after reduction, we find \( s = \)

\[ \frac{p\{T_{n-1} + T_n\} + q\{T_1 + T_2\} + r\{T_1 + T_2\}}{p + q + r}; \]

from which it appears that in this case the sum depends only on the first two and the last two terms of the series.

Example. It is proposed to find from this formula the sum of \( n \) terms of the series

\[ 1 + 2x + 3x^2 + 4x^3 + 5x^4 + \ldots \]

its scale of relation being

\[ x^2 T_{n-2} - 2x T_{n-1} + T_n = 0. \]

Here \( p = x^2, q = -2x, r = 1, \) therefore, observing that the last two terms of the series must be \((n-1)x^{n-2}\) and \(nx^{n-1}, \) we have, after substituting and reducing,

\[ s = \frac{1 - (n+1)x^n + x^{n+1}}{1 - 2x + x^2}. \]

This formula will not apply in the case of \( x = 1, \) because then the numerator and denominator are each \( = 0; \) but in such cases as this we may find the value of the function which expresses the sum by what is delivered at § 55, Fluxions.

(19.) The process by which we have determined the value of \( n \) terms of the series \( T_1 + T_2 + T_3 + \ldots \) will also apply to the finding the rational fraction from which the series may be deduced, which is also the sum of the series continued ad infinitum. For in this case the equation from which we have deduced the sum being

\[ p\{T_1 + T_2 + T_3 + \ldots\} + q\{T_2 + T_3 + T_4 + \ldots\} + r\{T_3 + T_4 + T_5 + \ldots\} = 0, \]

that is,

\[ ps + q\{s - T_1\} + r\{s - T_1 - T_2\} = 0, \]

we have

\[ s = \frac{q + r}{p + q + r}. \]

For example, let it be required to find the fraction which, being developed, produces the series

\[ 1 + 2x + 3x^2 + 4x^3 + \ldots \]

the scale of relation of which is

\[ x^2 T_{n-2} - 2x T_{n-1} + T_n = 0. \]

Here \( p = x^2, q = -2x, r = 1, T_1 = 1, T_2 = 2x; \) therefore, substituting in the formula, we get

\[ \frac{1}{1 - 2x + x^2} = \frac{1}{(1-x)^2} \]

for the fraction required, or for the sum of the series continued ad infinitum.

(20.) We come now to the last branch of the theory of recurring series which we proposed to consider, namely, how to find in any case the general term.

We shall begin with the most simple, and suppose the fraction to be \( \frac{a}{1-px}; \) which being expounded into a series by division, is

\[ a + apx + ap^2x^2 + ap^3x^3 + \ldots \]

Here it is immediately manifest that the general term is \( ap^{n-1}x^{n-1}. \)

Next let us suppose the fraction to be \( \frac{a + bx}{1 - ax - bx}. \)

Let the two roots of the quadratic equation \( 1 - ax - bx = 0 \) be \( x = \frac{1}{p}, x = \frac{1}{q}, \) so that \( 1 - px = 0, \) and \( 1 - qx = 0; \) therefore \( 1 - ax - bx = (1 - px)(1 - qx); \) thus we have

\[ \frac{a + bx}{1 - ax - bx} = \frac{a + bx}{(1 - px)(1 - qx)}. \]

Let us assume this expression equal to

\[ \frac{P}{1 - px} + \frac{Q}{1 - qx}, \]

where \( P \) and \( Q \) denote quantities which are to be independent of \( x; \) then, reducing to a common denominator, we have

\[ \frac{a + bx}{(1 - px)(1 - qx)} = \frac{P + Q - (qP + pQ)x}{(1 - px)(1 - qx)}. \]

Hence, that \( x \) may remain indeterminate, we must make P + Q = a, qP + pQ = -b,

and from these equations we get

\[ P = \frac{ap + b}{p - q}, \quad Q = \frac{-aq + b}{p - q}. \]

Now, by the operation of division, we find

\[ \frac{P}{1 - px} = P + Ppx + Pp^2x^2 + \ldots, \] \[ \frac{Q}{1 - qx} = Q + Qqx + Qq^2x^2 + \ldots; \]

therefore, since \( \frac{a + bx}{1 - ax - \beta x^2} = \frac{P}{1 - px} + \frac{Q}{1 - qx}, \) it follows that the development of the fraction \( \frac{a + bx}{1 - ax - \beta x^2} \)

which proceeds according to the powers of \( x, \) is

\[ (P + Q) + (Pp + Qq)x + (Pp^2 + Qq^2)x^2 + \ldots, \]

And here it is evident that the general term is \( (Pp^{n-1} + Qq^{n-1})x^n. \)

Let us take as a particular example the fraction

\[ \frac{1 - x}{1 - x - 2x^2}, \]

which when expanded into a series, becomes

\[ 1 + 0x + 2x^2 + 2x^3 + 6x^4 + 10x^5 + \ldots \]

Here, from the equation \( 1 - x - 2x^2 = 0, \) we get \( x = \frac{1}{2} \) and \( x = -1, \) so that \( 1 - 2x \) and \( 1 + x \) are divisors of the function \( 1 - x - 2x^2, \) that is, \( 1 - x - 2x^2 = (1 + x)(1 - 2x); \) hence \( p = -1, q = 2; \) and since \( a = 1, b = -1, \) therefore \( P = \frac{2}{3}, Q = \frac{1}{3}, \) and the general term \( (Pp^{n-1} + Qq^{n-1})x^n \) becomes, by substituting,

\[ \left\{ \frac{2}{3}(-1)^{n-1} + \frac{1}{3}2^{n-1} \right\} x^n = \frac{2^n - 1}{3} x^n, \]

where the sign \( + \) is to be taken when \( n \) is an odd number; but the sign \( - \) when \( n \) is even.

Sometimes the values of \( p \) and \( q \) will come out imaginary quantities; these, however, will be found always to destroy one another when substituted in the general term.

Let us next suppose the fraction which produces a recurring series to be

\[ \frac{a + bx + cx^2}{1 - ax - \beta x^2 - \gamma x^3}. \]

Let \( x = \frac{1}{p}, x = \frac{1}{q}, x = \frac{1}{r} \) be the three roots of the cubic equation \( 1 - ax - \beta x^2 - \gamma x^3 = 0, \) then the denominator of the fraction will be the product of the three factors

\[ 1 - px, 1 - qx, 1 - rx. \]

We must now assume the fraction equal to the expression

\[ \frac{P}{1 - px} + \frac{Q}{1 - qx} + \frac{R}{1 - rx}; \]

in which \( P, Q, R \) denote quantities which are independent of \( x. \)

The three terms of this expression are next to be reduced to a common denominator and collected into one, and the coefficients of the powers of \( x \) in the numerator of the result are to be put equal to the like powers of \( x \) in the proposed fraction; we shall then have

\[ P + Q + R = a, \] \[ (q + r)P + (p + r)Q + (p + q)R = -b, \] \[ qP + pQ + qR = c, \]

and by these equations the values of \( P, Q, R \) may be found.

Let \( \frac{P}{1 - px} + \frac{Q}{1 - qx} + \frac{R}{1 - rx} \) be now resolved into series by division; then, adding the like powers of \( x \) in each, we have

\[ (P + Q + R) + (Pp + Qq + Rr)x + (Pp^2 + Qq^2 + Rr^2)x^2 + \ldots \]

for the series which is the development of the fraction

\[ \frac{d + bx + cx^2}{1 - ax - \beta x^2 - \gamma x^3} \]

and here the general term is evidently

\[ (Pp^{n-1} + Qq^{n-1} + Rr^{n-1})x^n; \]

and in the very same manner may the general term be found in every case in which the denominator of the fraction admits of being resolved into unequal factors.

(21.) Let us now suppose the fraction to have the form

\[ \frac{a + bx}{(1 - px)^2}, \]

the denominator being the product of two equal factors; this fraction cannot be decomposed into other fractions the denominators of which are the simple factors of its denominator. We may, however, transform it into two, which shall have their numerators constant quantities by proceeding as follows: Assume the numerator \( a + bx = P + Q(1 - px), \) then, that \( x \) may remain indeterminate, we must have \( P + Q = a, -pQ = b, \) therefore

\[ Q = -\frac{b}{p}, P = a + \frac{b}{p}. \]

The assumption of \( a + bx = P + Q(1 - px) \) gives us therefore

\[ \frac{a + bx}{(1 - px)^2} = \frac{P}{(1 - px)^2} + \frac{Q}{1 - px}. \]

Now, putting the first term of the latter side of this equation under the form \( P(1 - px)^{-2}, \) it is resolved by the binomial theorem into the series

\[ P(1 + 2px + 3p^2x^2 + 4p^3x^3 + \ldots); \]

the other fraction, \( \frac{Q}{1 - px}, \) being expanded into a series, is

\[ Q + Qpx + Qp^2x^2 + \ldots, \]

Therefore the complete development of \( \frac{a + bx}{(1 - px)^2} \) is

\[ P + Q + (2P + Q)px + (3P + Q)p^2x^2 + \ldots, \]

and here the general term is manifestly

\[ (nP + Q)p^{n-1}x^n, \] or, substituting for \( P \) and \( Q \) their values,

\[ \{npa + (n - 1)b\}p^{n-2}x^n. \]

(22.) In general, whatever be the form of the fraction from which a recurring series is derived, to determine the general term we must decompose the fraction into others which may be as simple as possible; and provided it be rational, and the highest power of \( x \) in the numerator at least one degree less than the highest power in the denominator, it may be always decomposed into others having one or other of these two forms

\[ \frac{P}{1 - px} + \frac{Q}{1 - qx}, \]

in which expressions \( P, Q, p, \) and \( q, \) denote quantities independent of \( x. \) Each partial fraction gives a recurring series, the general term of which will be sufficiently obvious; and as the series belonging to the original fraction is the sum of these series, so also its general term will be the sum of all their general terms.

We have now treated of some of the more general methods of summing series which admit of being explained by the common principles of algebra; but the subject is of great extent, and to treat of it so as to give a tolerable notion of its various branches, would require more room than could with propriety be spared in such a work as ours.

(23.) The method of fluxions or differential calculus affords a method, almost the only general one we possess, of summing series. The general principles upon which it is applied may be stated briefly as follows. Since the integral of any differential containing one variable quantity may always be expressed by a series, on the contrary, every series may be regarded as the expression of an integral: when any series then is proposed, we must endeavour to find the differential expression of which that series is the integral; and as we can always find the integral of a differential, at least by approximation, within given limits, we may thence determine, if not the exact, at least the approximate value of any infinite series. We shall now show how this principle may be applied in some particular cases.

**Problem I.** It is proposed to find the sum of \( n \) terms of the series

\[ x + 2x^2 + 3x^3 + 4x^4 + \cdots + nx^n. \]

Let the sum be denoted by \( s \). Then, multiplying all the terms by \( \frac{dx}{x} \), we have

\[ \frac{sdx}{x} = dx + 2xdx + 3x^2dx + \cdots + nx^ndx. \]

Let the integral of both sides be now taken, and the result is

\[ \int \frac{sdx}{x} = x + x^2 + x^3 + x^4 + \cdots + x^n. \]

Now the series on the right-hand side of this equation is a geometrical progression, the sum of which is known to be \( \frac{x - x^{n+1}}{1-x} \) (Algebra, § 56). Therefore

\[ \int \frac{sdx}{x} = \frac{x - x^{n+1}}{1-x}, \]

and, taking the differentials,

\[ \frac{dx}{x} = \frac{dx - (n+1)x^n dx + nx^{n+1} dx}{(1-x)^2}. \]

Hence we find

\[ s = \frac{x - (n+1)x^{n+1} + nx^{n+2}}{(1-x)^2}. \]

This result agrees with that formerly found (17) of this article.

**Problem II.** It is proposed to sum the infinite series

\[ 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \frac{1}{11} + \cdots \]

We may consider this series as a particular case of the more general series

\[ x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots, \]

namely, that in which \( x = 1 \). Putting therefore the sum \( s = \arctan(x) + c \), radius being unity. (Fluxions, § 139.) Now, when \( x = 0 \), all the terms of the series vanish, so that in this case \( s = 0 \); and when \( x = 0 \), \( \arctan(x) = 0 \); therefore \( c \), the constant quantity added to complete the fluent, is 0, and we have simply \( s = \arctan(x) \), and when \( x = 1 \), then \( s = \frac{\pi}{4} \approx 0.7853982 \).

**Problem III.** Required the sum of the infinite series

\[ \frac{x}{1 \cdot 2} + \frac{x^2}{2 \cdot 3} + \frac{x^3}{3 \cdot 4} + \frac{x^4}{4 \cdot 5} + \cdots \]

Putting \( s \) for the sum, and taking the differentials, we get

\[ ds = \frac{dx}{x} \left( \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \cdots \right). \]

Now the series in the parenthesis is evidently equal to

\[ -x - \log(1-x), \text{(see Algebra, sect. xix.)}; \text{ therefore } \]

\[ ds = \frac{dx}{x} \times \log(1-x). \]

To find the integral, let us put \( v \) for the function \( \frac{1}{x} \log(1-x) \), then, taking its fluxion, we have

\[ dv = \frac{dx}{x} \times \log(1-x) - \frac{dx}{x(1-x)}; \]

therefore, substituting, we get

\[ ds = dv + \frac{dx}{x(1-x)} - \frac{dx}{x}; \]

and taking the integrals,

\[ s = v - \log(1-x) + c = \log(1-x) - \log(1-x) + c. \]

To determine the constant quantity \( c \), let us take \( x = 0 \); then in this case all the terms of the series vanish, so that \( s = 0 \), also \( \log(1-x) = \log 1 = 0 \); and since in general

\[ \log(1-x) = \frac{1}{x} \left( -x - \frac{x^2}{2} - \frac{x^3}{3} - \cdots \right), \]

when \( x = 0 \), then \( \log(1-x) = -1 \); therefore \( 0 = -1 + c \), and \( c = 1 \); hence it appears that

\[ s = \frac{\log(1-x)}{x} - \log(1-x) + 1 = \frac{(1-x)\log(1-x)}{x} + 1. \]

**Example.** Let \( x = \frac{1}{2} \), then our formula gives

\[ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \frac{1}{4 \cdot 5} + \cdots = 1 - \log_2 3 = 0.688528. \]

**Problem IV.** Let the series to be summed be

\[ 1 + \frac{m}{n} x + \frac{m+1}{n+1} x^2 + \frac{m+2}{n+2} x^3 + \cdots. \]

Putting \( s \) for this series, let all its terms be multiplied by \( x^{n-1} \), so that the exponent of \( x \) in each may be identical with its denominator; the result is

\[ x^{n-1} = x^{n-1} + \frac{m}{n} x^n + \frac{m+1}{n+1} x^{n+1} + \frac{m+2}{n+2} x^{n+2} + \cdots, \]

and hence, taking the differentials

\[ x^{n-1} dx + (n-1)x^{n-2} dx = (n-1)x^{n-2} dx + mx^{n-1} dx + (m+1)x^n dx + (m+2)x^{n+1} dx + \cdots. \]

Let both sides of this equation be now multiplied by \( x^{m-n} \), and it becomes

\[ x^{m-n} dx + (n-1)x^{m-n-1} dx = (n-1)x^{m-n-1} dx + mx^{m-n} dx + (m+1)x^{m-n} dx + (m+2)x^{m-n} dx + \cdots. \]

Putting now the single character \( dp \) for the fluxional expression which forms the first member of this equation, we get, by taking the fluents of both sides,

\[ p = \frac{n-1}{m-1} x^{m-1} + x^n + x^{n+1} + x^{n+2} + \cdots. \]

but the series in the parenthesis is the development of

\[ \frac{1}{1-x}, \text{ therefore } \]

\[ p = \frac{n-1}{m-1} x^{m-1} + \frac{x^n}{1-x}. \] Taking now the fluxions, and substituting instead of \( dp \) the expression it was put to represent, we get

\[ \frac{2m-1}{x} + (n-1)x^{m-2}dx = \frac{mx^{m-1}dx}{1-x} + \frac{x^mdx}{(1-x)^2}; \]

and this, after reduction, becomes

\[ ds + \frac{n-1}{x} \cdot \frac{mx^{m-1}dx}{1-x} + \frac{mx^mdx}{(1-x)^2}. \]

This fluxional equation being of the first degree and first order, its primitive equation may be found (from the general formula given in Fluxions, § 171) to be

\[ s = \frac{1}{x^{m-1}} \times f \left\{ (n-1)x^{m-2}dx + \frac{mx^{m-1}dx}{1-x} + \frac{x^mdx}{(1-x)^2} \right\}; \]

and this again, by remarking that \( f(n-1)x^{m-2}dx = x^{m-1} \), and that

\[ \int \frac{mx^{m-1}dx}{1-x} = \frac{mx^m}{n(1-x)} - \int \frac{mx^mdx}{(1-x)^2}, \]

may be reduced to

\[ s = 1 + \frac{mx}{n(1-x)} + \frac{n-m}{nx^n} \int \frac{x^mdx}{(1-x)^2}. \]

The remaining fluent \( \int \frac{x^mdx}{(1-x)^2} \) may be found by § 101.

Fluxions, and it must be so taken, that after being multiplied by \( \frac{n-m}{nx^n} \), it shall vanish when \( x = 0 \); for then this hypothesis will make the whole function which expresses the value of \( s \) vanish, except its first term 1, as it ought to do.

Example. Let us suppose \( n = 2 \), then

\[ \int \frac{x^mdx}{(1-x)^2} = x + \frac{x}{1-x} + 2 \log.(1-x), \]

and

\[ \frac{2-m}{2x} \int \frac{x^mdx}{(1-x)^2} = \frac{(2-m)x}{2(1-x)} + \frac{2-m}{x} \log.(1-x), \]

the fluent being here taken as directed. In this case, after collecting the terms, we get \( s \), or

\[ 1 + \frac{m}{2}x + \frac{m+1}{3}x^2 + \frac{m+2}{4}x^3 + \ldots, \]

\[ = \frac{1}{1-x} + \frac{(2-m)}{x} \log.(1-x). \]

(24.) There is a branch of the doctrine of series which is of considerable importance in pure mathematics, as well as in many physical inquiries, and in the science of astronomy; it is called the Interpolation of series.

To interpolate a series is to interpose among its terms others which shall be subject to the same law, or which shall be formed in the same manner as the original terms of the series; or, in other words, it is to find the value of one or more terms by means of others which are given, and which may be either at equal or unequal intervals from one another, the places of the given terms as well as of those sought being supposed known.

It is easy to see that this problem may be applied to the construction of logarithmic tables; for we may regard the logarithms of the natural numbers 1, 2, 3, 4, &c. ad infinitum, as the terms of a particular series of which the numbers themselves are the indices; thus having given the logarithms of some numbers, we may by interpolating deduce from them the logarithms of others.

Again, in astronomy we may consider the numbers which express the successive observed positions of a celestial body as the terms of a series, their indices being the intervals of time between the observations and some assumed epoch, and the problem we are considering will enable us to determine the position at any instant different from the times of actual observation, provided the intervals between the observations be small, and the instant for which the position is sought not very remote from those at which the observations were made.

(25.) To illustrate the nature of the problem to be resolved, let us consider some particular case, as, for example, the arithmetical series

\[ a, a+d, a+2d, a+3d, a+4d, \ldots \]

Let \( t \) and \( t' \) be two given terms of the series which are at any distance from one another, and let \( n \) and \( n' \) be their indices, or numbers which denote their places in the series. Also let \( y \) be any term whatever, and \( x \) its index. Then, by the nature of an arithmetical series,

\[ t = a + (n-1)d, \quad t' = a + (n'-1)d, \]

\[ y = a + (x-1)d. \]

Now, as there are here three equations, each involving the quantities \( a \) and \( d \), we may eliminate both these quantities by the common rules (Algebra, sect. vii.); and this being done, we get

\[ (x-n')(t-t') = (n-n')(y-t); \]

and hence we find this expression,

\[ y = \frac{x-n'}{n-n'} + \frac{x-n}{n'-n} t', \]

which is a general formula for interpolating any arithmetical series; and it is observable, that it is entirely independent both of the first term and common difference.

Example. The 7th term of an arithmetical series is 15, and the 12th term is 25: it is required to find the 10th term.

Here \( n = 7, n' = 12, x = 10; \)

\( t = 15, t' = 25, y \) is sought.

Therefore, by the formula,

\[ y = \frac{2}{3} \times 15 + \frac{3}{5} \times 25 = 21, \text{ the answer.} \]

(26.) The mode of investigation by which we have found a formula for the interpolation of an arithmetical series will apply also to others, if the law according to which the terms are formed be known; in general, however, the law of a series to be interpolated is either not known, or it is not taken into account, and we only consider the absolute magnitudes of certain terms, and the numbers expressing their places in the series. To resolve the problem generally with these data, it is usual to proceed as follows: Let a straight line \( AB \), and a point \( A \) in it, be assumed as given in position, and let there be taken the segments \( AD, AD', AD'', AD''', \ldots \), proportional to the numbers denoting the places of the terms of a series reckoned from any term assumed as a fixed origin; and at the points \( D, D', D'' \), let there be erected perpendiculars proportional to the terms themselves. Let us now suppose a curve to pass through \( C, C', C'', C''', \ldots \); then, if it be so chosen that its curvature may vary gradually in its progress from point to point, without any abrupt changes of inflection, and, moreover, if the terms (which we may suppose to be either at equal or unequal distances) are pretty near to one another, it is easy to conceive, that if \( AP \) be taken equal to the number expressing the place of a term between \( C'D^n, C''D^m \), any two others, the term itself will, if not exactly, at least be nearly expressed by \( PQ \), the ordinate to the curve.

As an infinite variety of curves may be found that shall pass through the same given points, in this respect the problem is unlimited; it is, however, convenient to assume such as are simple and tractable. The parabolic class possess these properties, and accordingly they are commonly employed.

Let us then express the ordinates \( CD, CD', CD'', CD''' \), &c. which are the given terms of the series, by

\[ t, t', t'', t''', \ldots \]

and the abscissae \( AD, AD', AD'', AD''' \), or the numbers denoting the order of the terms, by

\[ n, n', n'', n''', \ldots \]

Put \( y \) for \( PQ \), a term to be interpolated, and \( x \) for \( AP \) its place. Then, considering \( x \) and \( y \) as indefinite co-ordinates, a parabolic curve that shall pass through the points \( C, C', C'', C''' \), &c., will have for its equation

\[ y = A + Bx + Cx^2 + Dx^3 + \ldots \]

the number of terms on the right-hand side being supposed equal to that of the given points, and \( A, B, C, D, \ldots \) put to denote constant quantities. To determine these, we must consider that when \( x = n \), then \( y = t \); and that when \( x = n' \), then \( y = t' \), and so on; therefore, substituting the successive corresponding values of \( x \) and \( y \), we get

\[ \begin{align*} t &= A + Bn + Cn^2 + Dn^3 + \ldots \\ t' &= A + Bn' + Cn'^2 + Dn'^3 + \ldots \\ t'' &= A + Bn'' + Cn''^2 + Dn''^3 + \ldots \\ &\vdots \\ t''' &= A + Bn''' + Cn'''^2 + Dn'''^3 + \ldots \\ &\vdots \end{align*} \]

This series of equations must be continued until their number be the same as that of the co-efficients \( A, B, C, D, \ldots \). If we now consider \( t, t', t'', \ldots \) and \( n, n', n'', \ldots \) as known, and \( A, B, C, \ldots \) as unknown quantities, we may determine these last by eliminating them one after another from the above equations, as is taught in Algebra. And the values of \( A, B, C, \ldots \) being thus determined and substituted in the general equation, we shall have a general expression for \( y \) in terms of \( x \), the number denoting its place and known quantities; and this is in substance the solution originally given of the problem by Sir Isaac Newton, who proposed it in the third book of his *Principia*, with a view to its application in astronomy.

A celebrated foreign mathematician (Lagrange) has, in the *Cahiers de l'Ecole Normale*, given a different form to the expression for \( y \). He has observed that since, when \( x \) becomes \( n, n', n'', n''' \), &c. successively, then \( y \) becomes \( t, t', t'', t''' \), &c. It follows that the expression for \( y \) must have this form,

\[ y = \alpha t + \beta t' + \gamma t'' + \delta t''' + \ldots \]

where the quantities \( \alpha, \beta, \gamma, \delta, \ldots \) must be such functions of \( x \), that if we put \( x = n \), then \( \alpha = 1 \) and \( \beta = 0, \gamma = 0, \delta = 0, \ldots \); and if we put \( x = n' \), then \( \alpha = 0, \beta = 1, \gamma = 0, \delta = 0, \ldots \); and again, if we make \( x = n'' \), then \( \alpha = 0, \beta = 0, \gamma = 1, \delta = 0, \ldots \), and so on. Hence it is easy to conclude that the values of \( \alpha, \beta, \gamma, \delta, \ldots \) must have the form

\[ \begin{align*} \alpha &= \frac{(x - n)(x - n')(x - n'')}{(n - n')(n - n'')(n - n''')} \\ \beta &= \frac{(x - n)(x - n')(x - n'')}{(n' - n)(n' - n'')(n' - n''')} \\ \gamma &= \frac{(x - n)(x - n')(x - n'')}{(n'' - n)(n'' - n')(n'' - n'')} \\ \delta &= \frac{(x - n)(x - n')(x - n'')}{(n''' - n)(n''' - n')(n''' - n'')} \\ &\vdots \end{align*} \]

and here the number of factors in the numerator and denominator must be each equal to the number of given points in the curve. This formula would be found to be identical with that which may be obtained by the method indicated in last article, if we were to take the actual product of the factors and arrange the whole expression according to powers of \( x \).

It possesses however one advantage over the other, viz. that of admitting of the application of logarithms.

We shall now show the application of this formula.

**Ex. 1.** Having given the logarithms of 101, 102, 104, and 105, it is required to find the logarithm of 103.

In this case we may reckon the terms of the series forward from the first given term, viz. log. 101, so that we have

\[ \begin{align*} t &= \log. 101 = 2.0043214, & n &= 0, \\ t' &= \log. 102 = 2.0086002, & n' &= 1, \\ y &= \log. 103 = \text{term sought}, & x &= 2, \\ t'' &= \log. 104 = 2.0170333, & n'' &= 3, \\ t''' &= \log. 105 = 2.0211893, & n''' &= 4. \end{align*} \]

Substituting now in the general formula, we get

\[ \begin{align*} \alpha &= \frac{1 \times 1 \times 2}{-1 \times -3 \times -4} = \frac{1}{6}, \\ \beta &= \frac{2 \times 1 \times -2}{1 \times -2 \times -3} = \frac{2}{3}, \\ \gamma &= \frac{2 \times 1 \times -1}{4 \times 3 \times 1} = \frac{1}{6}. \end{align*} \]

Therefore \( y = \frac{t}{6} + \frac{2t'}{3} + \frac{2t''}{3} - \frac{t'''}{6} \)

\[ = \frac{3}{2}(t' + t'') - \frac{1}{6}(t + t''') \]

\[ = 2.0128372, \text{ the answer.} \]

**Ex. 2.** Given a comet's distance from the sun on the following days in December, at twelve at night, to find its distance December 20.

December 12. distance 301, Dec. 24. distance 715, 21. 620, 26. 772.

Here we shall estimate the places of the terms from the time of the first position, viz. December 12. Therefore

\[ \begin{align*} t &= 301, & n &= 0, \\ y &= \text{sought}, & x &= 8, \\ t' &= 620, & n' &= 9, \\ t'' &= 715, & n'' &= 12, \\ t''' &= 772, & n''' &= 14. \end{align*} \]

In this case the general formula gives us

\[ \begin{align*} \alpha &= \frac{1}{63}, \beta = \frac{6}{45}, \gamma = -\frac{2}{3}, \delta = \frac{8}{33}, \end{align*} \]

therefore

\[ y = \frac{t}{63} + \frac{64t'}{45} - \frac{2t''}{3} + \frac{8t'''}{33} \]

\[ = 586.3, \text{ the answer.} \]

We shall conclude this article with a brief enumeration of the best works on the subject which we have been treating of.

Bernoulli, *Ars Conjectandi*; Newton, *Methodus Differentialis*; Taylor, *Methodus Incrementorum*; Stirling, *Methodus Differentialis, sive Tractatus de Summatione et Interpolatione Serierum*; Euler, *Institutiones Calculi Differentialis*; Emerson's Method of Increments; Emerson's Differential Method; De Moivre, *Miscellanea Analytica*; Lagrange, *Théorie des Fonctions Analytiques*; *Traité des Différences et des Séries*, a sequel to Lacroix's work on the Calcul Différentiel; Dr Hutton's Mathematical and Philosophical Tracts; Spence's Essay on the Theory of the various orders of Logarithmic Transcendents, with an Inquiry into their Applications to the Integral Calculus, and the Summation of Series.