SERIES

SERIES, 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, &c.

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 class 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

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

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 two first terms, its general form will be

If we suppose $a=1$ and $b=\backslash frac\{1\}\{2\}$, we get

as a particular example of a harmonic series.

4. Recurring Series. Let its terms be denoted by

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

For example, let us suppose $A=1,\; B=2x,\; m=4x^2,\; n=3x$; then $C=10x^3,\; D=38x^4,\; E=154x^5,\; F=614x^6$, so that the first six terms of the series are

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 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

The two series of quantities and are both recurring, as is manifest from the law which connects the quantities one with another. (See ALGEBRA, § 358.)

(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

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

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

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. (See ALGEBRA, § 78, and § 260.)

2. By the Method of Indeterminate Coefficients. (ALGEBRA, § 261.)

3. By the Binomial Theorem. (ALGEBRA, § 263—§ 269.)

4. By Taylor's Theorem, (FLUXIONS, § 66—§ 72.) 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 axes 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 number that denotes the place or order of that term in the series. In conformity

conformity to this method, putting the symbols $T\_\{(1)\}$, $T\_\{(2)\}$, $T\_\{(3)\}$, &c. to denote the terms of any series whatever, we may express it generally thus.

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 (v) is put for any indefinite number.

The nature of the arithmetical series

will be defined by the equation

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

will be expressed by the equation

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

the general term is evidently $\backslash frac\{ab\}\{(v-1)a\; -\; (v-2)b\}$.

(8.) We shall now 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

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

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

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 $\backslash frac\{n(n+1)\}\{1\; \backslash cdot\; 2\}$, 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

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

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

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

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

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

(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

Series. 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 $\backslash frac\{v(v+1)\}\{1\; \backslash cdot\; 2\}$ for its general terms; and, again, the terms of this last are the differences of the terms of the series having $\backslash frac\{v(v+1)(v+2)\}\{1\; \backslash cdot\; 2\; \backslash cdot\; 3\}$ 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 $1+2+3+\backslash dots+n$, which has $v$ for its general term, is $\backslash frac\{n(n+1)\}\{1\; \backslash cdot\; 2\}$; and the sum of the series

The next series which has $\backslash frac\{v(v+1)(v+2)\}\{1\; \backslash cdot\; 2\; \backslash cdot\; 3\}$ 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

if we multiply these equals by the product of all the factors $v,\; \backslash frac\{v+1\}\{2\},\; \backslash frac\{v+2\}\{3\},\; \backslash dots$ to $\backslash frac\{v+p-1\}\{p\}$, we get

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

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.

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,

is equal to

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

is equal to

We shall here give a few particular cases of the last general formula.

(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.

PROB. III. It is proposed to find the sum of $n$ terms of the series of squares $1+4+9+16+25+\backslash dots$, &c. or $1^2+2^2+3^2+4^2+5^2+\backslash dots$, &c.

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 $v$,

Therefore adding, we find

Series. But by the general theorem (9.)

We might have arrived at the same conclusion by considering that since $v^3$, 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.

PROB. IV. It is proposed to find the sum of $n$ terms of the series

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

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

Therefore, by the theory of indeterminate coefficients, (ALGEBRA, § 261.)

Hence we find $A=-3,\; B=-A-2=1$; thus it appears that $v$ being any number whatever,

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 $S=P-3Q+R$. By the theorem (9.)

and by proper reduction, $S$, or

Corollary. We have found (PROB. I.) that

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

this is a very curious and elegant property of numbers.

(11.) It is manifest that by the mode of proceeding employed in last problem we may investigate the sum of $n$ terms of the series

$m$ 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

where $a$ and $b$, &c. denote given numbers; namely, by transforming it into the function of the form

where $A,\; B$, and $C$, &c. denote constant quantities. Our limits, however, will not allow us to go into particulars.

(12.) The next class of series we shall consider, comprehends such as may be formed by the successive substitution of ($a$ being put for any given quantity whatever) in the series of functions

We shall begin with the first of these.

PROB. V. It is proposed to find the sum of $n$ terms of the series

which is formed by substituting

successively for $v$ in the general term $\backslash frac\{1\}\{v(v+1)\}$.

Whatever be the value of $v$, we have

therefore, proceeding as in the foregoing problems, we get

Here it is evident that the terms of the series to be summed

Series. summed are the differences betwixt every two adjoining terms of this other series.

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

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

PROB. VI. Let it be required to find the sum of $n$ terms of this series.

the general term in this case being $\backslash frac\{1\}\{v(v+1)(v+2)\}$.

Because $\backslash frac\{2\}\{v(v+2)\}\; =\; \backslash frac\{1\}\{v\}\; -\; \backslash frac\{1\}\{v+2\}$, therefore, multiplying by $\backslash frac\{1\}\{2(v+1)\}$, we have

and hence, by substituting successively for $v$,

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

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

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

$p$ being any whole number whatever: for since

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

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

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.

Hence we have the following very general theorem.

THEOREM. Let $a$ denote any number whatever, and let $1,\; 2,\; 3,\; \backslash dots,\; p$ 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 to $a+n-1$ successively for $v$ in the function

is equal to

COROLLARY. The same series continued ad infinitum is equal to

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

Example 1. Required the sum of $n$ terms of the series

The terms of this series are evidently produced by the successive substitution of the numbers 2, 3, 4, 5, &c. for $v$ in the function

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

Ex. 2. Required the sum of the series

continued ad infinitum.

By a little attention it will appear that its terms are produced by the substitution of the numbers $\backslash frac\{1\}\{7\}$, $\backslash frac\{1\}\{10\}$, $\backslash frac\{1\}\{13\}$, &c. successively for $v$ in the function

In this case then $a=\backslash frac\{1\}\{7\}$, $p=2$, therefore the sum is

(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

continued ad infinitum.

This series is evidently formed by the substitution of the numbers 1, 2, 3, &c. successively for $v$ in the function $\backslash frac\{1\}\{v(v+3)\}$. 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

we next assume its numerator

VOL. XIX. Part I.

and by multiplying get

$v^3\; +\; 3v^2\; +\; 2v\; =\; A(v^2\; +\; 5v\; +\; 6)\; +\; B(v\; +\; 3)\; +\; C$;
therefore, that $v$ may be indeterminate, we must make

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

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 $\backslash frac\{1\}\{v(v+1)\}$ appears by the theorem

to be $\backslash frac\{1\}\{a\}$, or 1, because $a=1$, and the sum of the infinite

series whose general term is $\backslash frac\{-2\}\{v(v+1)(v+2)\}$, is in like

manner found to be $\backslash frac\{-2\}\{2\}\; +\; \backslash frac\{1\}\{1\; \backslash cdot\; 2\}\; =\; \backslash frac\{-1\}\{2\}$; and lastly, the in-

finite series whose general term is $\backslash frac\{2\}\{v(v+1)(v+2)(v+3)\}$

is $\backslash frac\{2\}\{3\}\; \backslash cdot\; \backslash frac\{1\}\{1\; \backslash cdot\; 2\; \backslash cdot\; 3\}\; =\; \backslash frac\{1\}\{9\}$; therefore, collecting these into one,

the sum of the proposed series is $1\; -\; \backslash frac\{1\}\{2\}\; +\; \backslash frac\{1\}\{9\}\; =\; \backslash frac\{11\}\{18\}$, the answer.

Ex. 4. Required the sum of the infinite series

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

Now $v-1=v+2-3$; therefore,

thus it appears that the proposed series is reducible to two others, one having its terms produced by the sub-

stitution of 2, 3, &c. for $v$ in the function $\backslash frac\{1\}\{v(v+1)\}$, and the other by a like substitution in the function

$\backslash frac\{-3\}\{v(v+1)(v+2)\}$. Now, by our theorem, the sum of

the first of these is $\backslash frac\{1\}\{2\}$, and that of the second is $\backslash frac\{-3\}\{2\}$

Series. $\backslash frac\{1\}\{2\; \backslash cdot\; 3\}\; =\; -\backslash frac\{1\}\{4\}$, therefore the sum of the proposed series is

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

or admits of being reduced to that form.

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

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 shew that a certain definite number of its terms, beginning with any proposed term, can always be found, the sum of which shall exceed a unit or 1; for this being the case, as we can go on continually 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 beginning 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 $\backslash frac\{1\}\{a\}$, then if the thing be possible, and if $n$ be the requisite number of terms, we must have

Now because

and in general,

therefore, being any whole number,

hence it follows that the series

will be greater than the other series

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

therefore, the sum of the series

will always be greater than this expression; but if we suppose $n$ so great that the quantity $\backslash left(1\; +\; \backslash frac\{1\}\{a\}\backslash right)^\{n-1\}$ 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 $\backslash frac\{1\}\{a\}\; +\; \backslash frac\{1\}\{a+1\}\; +\; \backslash frac\{1\}\{a+2\}\; +\; \backslash frac\{1\}\{a+3\}\; +\; \backslash dots$ will always exceed 1; now this is the property of the series we proposed to demonstrate.

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

the sum of which, we have proved, will upon that hypothesis exceed unity; much more then will the sum exceed unity if we suppose the series continued until the denominator of its last term be equal to, or greater than $a^2$.

Hence, beginning with the term $\backslash frac\{1\}\{2\}$, it appears that

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

every infinite series, the terms of which decrease faster than the reciprocals 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 summed, 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 $\backslash frac\{1+2x\}\{1-x-x^2\}$, for example, by dividing the numerator by the denominator is converted into the infinite series

which is of such a nature that if $T,\; T\text{'},\; T\text{'}\text{'}$, denote any three of its succeeding terms, their relation to one another is expressed by the equation

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 use the method of indeterminate coefficients. By this method we assume the fraction

and hence, multiplying by the denominator, and bringing all the terms to one side, as explained in ALGEBRA, § 261, we have

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 coefficients of the terms. Thus the scale of relation of the foregoing series is either

where $T,\; T\text{'},$ and $T\text{'}\text{'}$ denote any three succeeding terms of the series, or it is

where $P,\; Q$ and $R$ denote their numeral coefficients.

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

where the characters $T\_\{(1)\},\; T\_\{(2)\},\; \backslash dots$ denote the successive terms, and the numbers $(1),\; (2),\; \backslash dots$ 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,

where $p,\; q,\; r$ denote certain given quantities.

The scale of relation affords the following series of equations,

Taking now the sum of these equations, we get

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

Hence, after reduction, we find $s\; =$

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

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

its scale of relation being

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,

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 § 92, FLUXIONS.

(19.) The process by which we have determined the value of $n$ terms of the series $T\_\{(1)\}\; +\; T\_\{(2)\}\; +\; T\_\{(3)\}\; +\; \backslash dots$ 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

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

the scale of relation of which is

Here $p\; =\; x^2$, $q\; =\; -2x$, $r\; =\; 1$, $T\_\{(1)\}\; =\; 1$, $T\_\{(2)\}\; =\; 2x$; therefore, substituting in the formula, we get

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 $\backslash frac\{a\}\{1-px\}$, which being expounded into a series by division, is

here it is immediately manifest that the general term is $ap^\{n-1\}x^\{n-1\}$.

Next let us suppose the fraction to be $\backslash frac\{a+bx\}\{1-ax-\backslash beta\; x^2\}$. Let the two roots of the quadratic equation $1-ax-\backslash beta\; x^2\; =\; 0$ be $x\; =\; \backslash frac\{1\}\{p\}$, $x\; =\; \backslash frac\{1\}\{q\}$, so that $1-px\; =\; 0$, and $1-qx\; =\; 0$; therefore, $1-ax-\backslash beta\; x^2\; =\; (1-px)(1-qx)$, thus, we have

Let us assume this expression equal to

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

Hence, that $x$ may remain indeterminate, we must make

and from these equations we get

Now, by the operation of division, we find

therefore, since $\backslash frac\{a+bx\}\{1-ax-\backslash beta\; x^2\}\; =\; \backslash frac\{P\}\{1-px\}\; +\; \backslash frac\{Q\}\{1-qx\}$, it follows that the development of the fraction $\backslash frac\{a+bx\}\{1-ax-\backslash beta\; x^2\}$ which proceeds according to the powers of $x$, is

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

Let us take as a particular example the fraction $\backslash frac\{1-x\}\{1-x-2x^2\}$, which when expanded into a series, becomes

Here, from the equation $1-x-2x^2=0$, we get $x\; =\; \backslash frac\{1\}\{2\}$ 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 $P\; =\; \backslash frac\{2\}\{3\}$, $Q\; =\; \backslash frac\{1\}\{3\}$, and the general term $(Pp^\{n-1\}\; +\; Qq^\{n-1\})x^\{n-1\}$ becomes by substituting

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

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

We must now assume the fraction equal to the expression

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

and by these equations the values of $P$, $Q$, $R$, may be found.

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

for the series which is the development of the fraction

and here the general term is evidently

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 $\backslash frac\{a+bx\}\{(1-px)^2\}$, 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

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

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

the other fraction $\backslash frac\{Q\}\{1-px\}$ being expanded into a series is

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

and here the general term is manifestly $(nP+Q)p^\{n-1\}x^\{n-1\}$, or, substituting for $P$ and $Q$ their values,

(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

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 fluxionary 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 fluent of any fluxion containing one variable quantity may always be expressed by a series, on the contrary every series may be regarded as the expression of a fluent: when any series then is proposed, we must endeavour to find the fluxional expression of which that series is the fluent; and as we can always find the fluent of a fluxion, 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 shew 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

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

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

Now the series on the right-hand side of this equation is a geometrical progression, the sum of which is known to be $\backslash frac\{x-x^\{n+1\}\}\{1-x\}$, (ALGEBRA, § 106.) Therefore

Series. and, taking the fluxions,

Hence we find

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

PROBLEM II. It is proposed to sum the infinite series

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

namely, that in which $x=1$. Putting, therefore, the sum $=s$, and taking the fluxions, we have

Now the series in the parenthesis is obviously the development of the rational fraction $\backslash frac\{1\}\{1+x^2\}$; therefore,

$+c$, radius being unity. (FLUXIONS, § 60.) Now when $x=0$, all the terms of the series vanish, so that in this case $s=0$; and as when $x=0$, $\backslash text\{arc.\}(\backslash tan.\; =\; x)\; =\; 0$; therefore $c$, the constant quantity added to complete the fluent is 0, and we have simply $s\; =\; \backslash text\{arc.\}(\backslash tan.\; =\; x)$, and when $x=1$, then $s\; =\; \backslash frac\{1\}\{2\}$ a quadrant $=\; 7853982$.

PROBLEM III. Required the sum of the infinite series

Putting $s$ for the sum, and taking the fluxions, we get

Now the series in the parenthesis is evidently equal to $-\backslash text\{Nap.\; log.\}(1-x)$, (see LOGARITHMS, page 76. column 1.); therefore

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

therefore, substituting, we get

and taking the fluents,

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 $\backslash log.\; (1-x)\; =\; \backslash log.\; 1\; =\; 0$; and since in general when $x=0$, then $\backslash frac\{\backslash log.\; (1-x)\}\{x\}\; =\; -1$: therefore $0\; =\; -1\; +\; c$, and $c=1$; hence it appears that

Example. Let $x\; =\; \backslash frac\{1\}\{2\}$, then our formula gives

PROBLEM IV. Let the series to be summed be

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

and hence taking the fluxions

Let both sides of this equation be now multiplied by $x^\{n-1\}$, and it becomes

Putting now the single character $p$ for the fluxional expression which forms the first member of this equation, we get by taking the fluents of both sides,

but the series in the parenthesis is the development of $\backslash frac\{1\}\{1-x\}$, therefore

taking now the fluxions, and substituting instead of $p$ the expression it was put to represent, we get

and this, after reduction, becomes

This fluxional equation being of the first degree, and first order, its primitive equation may be found (from the general formula given in FLUXIONS, § 182.) to be

and this again, by remarking that $\backslash int\; (n-1)\backslash dot\{x\}x^\{n-1\}\; =\; x^\{n-1\}$, and that

may be reduced to

The remaining fluent $\backslash int\; \backslash frac\{\backslash dot\{x\}x^n\}\{(1-x)^2\}$ may be found by § 109.