or **Serjeant at Law**, or of the **Coif**, is the highest degree taken at the common law, as that of Doctor is of the civil law; and as these are supposed to be the most learned and experienced in the practice of the courts, there is one court appointed for them to plead in by themselves, which is the common pleas, where the common law of England is most strictly observed: but they are not restricted from pleading in any other court, where the judges, who cannot have that honour till they have taken the degree of serjeant at law, call them **brothers**.

**Sergeant at Arms**, or **Mace**, an officer appointed to attend the person of the king; to arrest traitors, and such persons of quality as offend; and to attend the lord high steward, when sitting in judgment on a traitor.

Of these, by statute 13 Rich. II. cap. 6, there are not to be above 30 in the realm. There are now nine at court at L.100 per annum salary each; they are called the king's sergeants at arms, to distinguish them from others: they are created with great ceremony, the person kneeling before the king, his majesty lays the mace on his right shoulder, and says, *Rise up, sergeant at arms, and esquire for ever*. They have, besides, a patent for the office, which they hold for life.

They have their attendance in the presence-chamber, where the band of gentlemen-pensioners wait; and, receiving the king at the door, they carry the maces before him to the chapel door, whilst the band of pensioners stand foremost, and make a lane for the king, as they also do when the king goes to the house of lords.

There are four other sergeants at arms, created in the same manner; one, who attends the lord chancellor; a second, the lord treasurer; a third, the speaker of the house of commons; and a fourth, the lord mayor of London on solemn occasions.

They have a considerable share of the fees of honour, and travelling charges allowed them when in waiting, viz. five shillings per day when the court is within ten miles of London, and ten shillings when twenty miles from London. The places are in the lord chamberlain's gift.

There are also sergeants of the mace of an inferior kind, who attend the mayor or other head officer of a corporation.

**Common Sergeant**, an officer in the city of London, who attends the lord mayor and court of aldermen on court days, and is in council with them on all occasions, within and without the precincts or liberties of the city. He is to take care of orphan estates, either by taking account of them, or to sign their indentures, before their passing the lord mayor and court of aldermen: and he was likewise to let and manage the orphan estates, according to his judgment to their best advantage. See Recorder.

**Sergeant**, in war, is an uncommissioned officer in a company of foot or troop of dragoons, armed with an halbert, and appointed to see discipline observed, to teach the soldiers the exercise of their arms, to order, straighten, and form their ranks, files, &c. He receives the orders from the adjutant, which he communicates to his officers. Each company generally has two sergeants.

**Sergeantry** (*Serjeantia*), signifies, in law, a service that cannot be due by a tenant to any lord but the king; and this is either **grand sergeantry**, or **petit**. The first is a tenure by which the one holds his lands of the king by such services as he ought to do in person to the king at his coronation; and may also concern matters military, or services of honour in peace; as to be the king's butler, carver, &c. **Petit sergeantry** is where a man holds lands of the king, to furnish him yearly with some small thing towards his wars; and in effect payable as rent. Though all tenures are turned into socage by the 12 Car. II. cap. 24, yet the honorary services of grand sergeantry still remain, being therein excepted. See Knight-Service.

**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.

**Series**, in arithmetic and algebra, a rank or number of terms in succession, increasing or diminishing in some certain ratio or proportion. There are several various kinds of series; as arithmetical, geometrical, infinite, &c. The two first of these are, however, more generally known or distinguished by the names of arithmetical and geometrical progression. These series have already been explained and illustrated in the article Algebra, particularly the two first; it therefore only remains, in this place, to add a little to what has already been done to the last of these; namely,

Infinite Series,

Is formed by dividing the numerator of a fraction by its denominator, that denominator being a compound quantity; or by extracting the root of a surd.

An infinite series is either converging or diverging. A converging series is that in which the magnitude of the several terms gradually diminish; and a diverging series is that in which the successive terms increase in magnitude.

The law of an infinite series is the order in which the terms are observed to proceed. This law is often easily discovered from a few of the first terms of the series; and then the series may be continued as far as may be thought necessary, without any farther division or evolution.

An infinite series, as has already been observed, is obtained by division or evolution; but as that method is very tedious, various other methods have been proposed for performing the same in a more easy manner; as, by assuming a series with unknown coefficients, by the binomial theorem, &c.

I. Of the Method of Series by Division and Evolution.

Rule.

Let the division or evolution of the given fraction, containing which is to be converted into an infinite series, be performed as in Chapters I. and IV. of our article Algebra, and the required series will be obtained.

Examples.

1. Convert the fraction \(\frac{1}{1-x}\) into an infinite series?

\[ \begin{align*} \frac{1}{1-x} &= (1 + x + x^2 + x^3 + x^4 + \ldots) \\ &= 1 + x + x^2 + x^3 + x^4 + \ldots \end{align*} \]

Hence the fraction \(\frac{1}{1-x} = 1 + x + x^2 + x^3 + x^4 + \ldots\).

From inspection of the terms of this series, it appears that each term is formed by multiplying the preceding term by \(x\); and hence it may be continued as far as may be thought necessary without continuing the division.

2. Let the fraction \(\frac{ay}{1+x}\) be converted into an infinite series?

\[ \begin{align*} \frac{ay}{1+x} &= ay - ax + ayx^2 - ax^3 + ayx^4 - ax^5 + \ldots \\ &= ay - ax + ayx^2 - ax^3 + ayx^4 - ax^5 + \ldots \end{align*} \]

Hence \(\frac{ay}{1+x} = ay \times 1 - x + x^2 - x^3 + x^4 + \ldots\), &c., and the law of the series is obvious.

3. Reduce the fraction \(\frac{m^2 + x^2}{m + x}\) into an infinite series?

\[ \begin{align*} \frac{m^2 + x^2}{m + x} &= m - x + \frac{2x^2}{m} - \frac{2x^3}{m^2} + \frac{2x^4}{m^3} + \ldots \\ &= m - x + \frac{2x^2}{m} - \frac{2x^3}{m^2} + \frac{2x^4}{m^3} + \ldots \end{align*} \]

Hence \(\frac{m^2 + x^2}{m + x} = m - x + \frac{2x^2}{m} - \frac{x^3}{m^2} + \frac{x^4}{m^3} + \ldots\), &c., and the law of the series is evident.

4. Convert the quantity \(\frac{a^2}{a^2 + 2ay + y^2}\) into an infinite series?

\[ \begin{align*} \frac{a^2}{a^2 + 2ay + y^2} &= \left(1 - \frac{2y}{a} + \frac{3y^2}{a^2} - \frac{4y^3}{a^3} + \ldots\right) \\ &= 1 - \frac{2y}{a} + \frac{3y^2}{a^2} - \frac{4y^3}{a^3} + \ldots \end{align*} \]

Whence Series.

Whence \(\frac{a^2}{a^2 + 2ay + y^2} = 1 - \frac{2y}{a} + \frac{3y^2}{a^2} - \frac{4y^3}{a^3}, \ldots\) and each term is found by multiplying the preceding by \(\frac{y}{a}\) and increasing the coefficient by unity.

5. Let \(a^2 + x^2\) be converted into an infinite series?

\[ \begin{align*} &\frac{a^2 + x^2}{a^2} + \frac{x^2}{2a} - \frac{x^4}{8a^2} + \frac{x^6}{16a^3} - \frac{x^8}{128a^4} \\ &+ \frac{2a + x^2}{2a} \left( \frac{x^2}{a^2} + \frac{x^4}{8a^2} \right) \\ &- \frac{x^4}{4a^2} + \frac{x^6}{8a^3} + \frac{x^8}{64a^4} \\ &+ \frac{2a + x^2}{a^2} \left( \frac{x^2}{4a^2} - \frac{x^4}{16a^3} \right) \\ &+ \frac{x^6}{8a^4} + \frac{x^8}{64a^5} - \frac{x^{10}}{256a^6} \\ &+ \frac{5x^8}{64a^6} + \frac{x^{10}}{64a^7} + \frac{x^{12}}{256a^8} \\ &+ \frac{x^8}{128a^9} + \ldots \end{align*} \]

Hence the square root of \(a^2 + x^2 = a + \frac{x^2}{2a} - \frac{x^4}{8a^2} + \frac{x^6}{16a^3} - \frac{x^8}{128a^4} + \ldots\)

In continuing the operation, those terms may be neglected whose dimensions exceed those of the last term to which the root is to be continued.

II. Of the Method of Series by assuming a Series with unknown Coefficients.

Rule. Assume a series with unknown coefficients to represent that required. Let this series be multiplied or involved, according to the nature of the question; and the quantities of the same dimension being put equal to each other, the coefficients will be determined; and hence the required series will be known.

Examples. 1. Let \(\frac{1}{a-x}\) be converted into an infinite series? Assume \(\frac{1}{a-x} = A + Bx + Cx^2 + Dx^3 + Ex^4, \ldots\)

Then this assumed series, multiplied by \(\frac{a-x}{a-x}\), gives

\[ \begin{align*} &1 = aA + aBx + aCx^2 + aDx^3 + aEx^4, \ldots \\ &- Aa - Ba^2 - Ca^3 - Da^4, \ldots \end{align*} \]

Now, by equating the coefficients of the same powers of \(x\), we have \(aA = 1, aB - A = 0, aC - B = 0, aD - C = 0, aE - D = 0, \ldots\). Hence \(A = \frac{1}{a}, B = \frac{a}{a^2}, C = \frac{a^2}{a^3}, D = \frac{a^3}{a^4}, E = \frac{a^4}{a^5}, \ldots\); whence, by substitution, we have \(\frac{1}{a-x} = \frac{1}{a} + \frac{x}{a^2} + \frac{x^2}{a^3} + \frac{x^3}{a^4} + \frac{x^4}{a^5}, \ldots\).

2. Convert the quantity \(\frac{c^2}{c^2 + 2cy - y^2}\) into an infinite series?

Let the assumed series be \(A + By + Cy^2 + Dy^3, \ldots\), which multiplied by \(c^2 + 2cy - y^2\), gives

\[ \begin{align*} &c^2 = c^2A + c^2By + c^2Cy^2 + c^2Cy^3, \ldots \\ &+ 2cAy + 2cBy^2 + 2cCy^3, \ldots \\ &- Ay^2 - By^3, \ldots \end{align*} \]

Now, by equating the coefficients of the homologous terms, we have \(c^2 = c^2A, c^2B + 2cA = 0, c^2C + 2cB - A = 0, c^2D + 2cC - B = 0, \ldots\); whence \(A = 1, B = -\frac{2}{c}, C = -\frac{2}{c^2}, D = -\frac{2}{c^3}, \ldots\); hence \(\frac{c^2}{c^2 + 2cy - y^2} = 1 - \frac{2y}{c} + \frac{5y^2}{c^2} - \frac{12y^3}{c^3}, \ldots\).

3. Required the square root of \(a^2 - x^2\)?

Let \(a^2 - x^2 = A + Bx + Cx^2 + Dx^3, \ldots\), which being squared gives

\[ \begin{align*} &a^2 - x^2 = A^2 + 2ABx + B^2x^2 + 2ADx^3 + \ldots \\ &+ 2ACx^2 + 2BCx^3, \ldots \end{align*} \]

Hence \(A = a, 2AB + 1 = 0, B^2 + 2AC = 0, 2AD + 2BC = 0, \ldots\). Then \(A = a, B = -\frac{1}{2a}, C = -\frac{B^2}{8a^3}, D = -\frac{BC}{16a^5}, \ldots\); hence \(a^2 - x^2 = a - \frac{x^2}{2a} - \frac{x^4}{8a^3} - \frac{x^6}{16a^5}, \ldots\).

III. Of the Method of reducing a fractional Quantity into an Infinite Series by the Binomial Theorem.

As this method has already been illustrated in the article Algebra, we shall therefore briefly state the Binomial theorem, and add a few examples.

Binomial Theorem.

\[ (a + b)^n = a^n + \frac{m}{n}a^{m-n}b + \frac{m}{n}a^{m-n}b^2 + \frac{m}{n}a^{m-n}b^3 + \ldots \]

Or \(a^n \times 1 + \frac{b}{a} = a^n \times 1 + \frac{m}{n}a^{m-n}b + \frac{m}{n}a^{m-n}b^2 + \frac{m}{n}a^{m-n}b^3 + \ldots\)

Examples.

1. Let \(\frac{a}{ax - x^2}\) be converted into an infinite series? Now \(\frac{a}{ax - x^2} = a \times \frac{1}{ax - x^2} = \frac{a}{ax - x^2} \times \frac{1}{1 - \frac{x}{a}} = \frac{a}{ax - x^2} \times \frac{1}{1 - \frac{x}{a}} \times \frac{1}{1 - \frac{x}{a}}\). And this last expression, being compared with the general theorem, gives \(\frac{b}{a} = \frac{x}{a}, m = -1, n = 2\). Hence, by substitution, we have \(\frac{a}{ax - x^2} = \frac{a}{ax - x^2} \times \frac{1}{1 - \frac{x}{a}} \times \frac{1}{1 - \frac{x}{a}} \times \frac{1}{1 - \frac{x}{a}}\). 2. Required the square root of \(a^2 + x^3\).

By comparing this with the general theorem, we have \(a = a^2, b = x^3, m = 1, n = 2\). Hence, by substitution, the series becomes

\[ \frac{a}{x} \times 1 + \frac{1}{2} \times \frac{x^3}{a^2} + \frac{1}{2} \times \frac{x^6}{a^4} + \ldots \]

In order to apply this to numbers, let the square root of 85 be required? Now, the square root of 85 = \(\sqrt{81 + 4}\); hence \(a = 9\), and \(x^3 = 4\).

Then

\[ \begin{align*} \frac{1}{a^2} &= \frac{1}{81} = 0.0123456789 \\ \frac{x^3}{a^2} &= \frac{4}{81} = 0.0487890123 \\ \frac{x^6}{a^4} &= \frac{16}{81} = 0.1977979589 \\ \end{align*} \]

Square root of 85 = 9.219546, true except the last decimal.

3. Required the cube root of \(a^3 + b^3\).

This being compared with the general theorem gives \(a = x^3, b = y^3, m = 1, n = 3\). Hence \(a^3 + b^3\) is

\[ \frac{a}{x} \times 1 + \frac{1}{3} \times \frac{y^3}{x^3} + \frac{1}{3} \times \frac{y^6}{x^6} + \ldots \]

Let the cube root of 600 be required? Now 600 = \(8 \times 1 + \frac{8}{1}\). Then \(y^3 = 88, x^3 = 512, m = 1\), and \(n = 3\).

Then

\[ \begin{align*} \frac{1}{x^3} &= \frac{1}{512} = 0.001953125 \\ \frac{y^3}{x^3} &= \frac{88}{512} = 0.171875 \\ \frac{y^6}{x^6} &= \frac{729}{512} = 1.4306640625 \\ \end{align*} \]

Now Now each term of the given series is to be compared with the correspondent terms in the first part of the above theorem; and by substitution in the second, the several terms of the required series will be obtained.

**Examples.**

1. What is the square of the series \( y - y^3 + y^5 - y^7 + \ldots \)?

By comparing this with the general theorem, we find \( z = y \), \( a = 1 \), \( b = 0 \), \( c = -1 \), \( d = 0 \), \( g = -1 \), &c., and \( m = 2 \); whence \( y - y^3 + y^5 - y^7 + \ldots = y^2 \times (1 - 2ax^2 + c^2x^4 - 2ce^2x^6) + \ldots \).

\[ \begin{align*} &= y^2 - 2y^4 + 3y^6 - 4y^8 + \ldots \\ &= y^2 - 2y^4 + 3y^6 - 4y^8 + \ldots \end{align*} \]

2. Required the fourth power of the series \( 1 + x + x^2 + x^3 + \ldots \).

Here \( z = 1 \), \( a = 1 \), \( b = 1 \), \( c = 1 \), \( d = 1 \), and \( m = 4 \).

Then \( 1 + x + x^2 + x^3 + \ldots = 1 + 4b + 6b^2 + 4b^3 + b^4 + \ldots \).

\[ \begin{align*} &= 1 + 4x + 10x^2 + 20x^3 + 35x^4 + \ldots \\ &= 1 + 4x + 10x^2 + 20x^3 + 35x^4 + \ldots \end{align*} \]

3. What is the square of \( \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} + \ldots \)?

In this case \( z = \frac{1}{x} \), \( a = 1 \), \( b = 1 \), \( c = 1 \), \( d = 1 \), and \( m = 2 \).

Then \( \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} + \ldots = \frac{1}{x^2} \times \left(1 + 2b + b^2 + \frac{1}{x} + 2bc + \frac{1}{x^2} + 2bd + \frac{1}{x^3} + \ldots \right) \).

\[ \begin{align*} &= \frac{1}{x^2} \times \left(1 + \frac{2}{x} + \frac{3}{x^2} + \frac{4}{x^3} + \frac{5}{x^4} + \ldots \right) \\ &= \frac{1}{x^2} + \frac{2}{x^3} + \frac{3}{x^4} + \frac{4}{x^5} + \frac{5}{x^6} + \ldots \end{align*} \]

4. What is the square root of \( \frac{1}{r^2} - \frac{z^2}{2} + \frac{z^4}{4r^2} - \frac{z^6}{6r^4} + \frac{z^8}{8r^6} + \ldots \)?

The quantity reduced is \( \frac{1}{r^2} \times \left(1 - \frac{z^2}{2r^2} + \frac{z^4}{4r^4} - \frac{z^6}{6r^6} + \frac{z^8}{8r^8} + \ldots \right) \).

In this example \( z = \frac{1}{r^2} \), \( x = z^2 \), \( a = 1 \), \( b = -\frac{1}{2r^2} \), \( c = \frac{1}{4r^4} \), \( d = -\frac{1}{6r^6} \), &c., and \( m = -\frac{1}{2} \), \( m - 1 = -\frac{1}{2} \).

\[ \begin{align*} &= \frac{1}{r^2} \times \left(1 + \frac{z^2}{4r^2} + \frac{3z^4}{32r^4} + \frac{5z^6}{128r^6} + \ldots \right) \\ &= \frac{1}{r^2} + \frac{z^2}{4r^2} + \frac{3z^4}{32r^4} + \frac{5z^6}{128r^6} + \ldots \end{align*} \]

Harmonic Series, a series of terms formed in harmonical proportion. It has been already observed in the article Proportion, that if three numbers be in harmonical proportion, the first is to the third as the difference between the first and second is to the difference between the second and third.

Let \( a \), \( b \), and \( x \) be three terms in harmonical proportion; then \( a : x : : a - b : b - x \).

whence \( ax - bx = ab - ax \).

and \( 2ax - bx = ab \)

then \( x = \frac{ab}{a - b} \). Hence the three

terms of this series is \( a, b, \frac{ab}{a - b} \).

Again, let \( x \) be the fourth term, to find which in terms of \( a \) and \( b \), we have

\[ \begin{align*} b : x : : b - \frac{ab}{a - b} : \frac{ab}{a - b} - x \\ \text{Then } b - \frac{ab}{a - b} : x = \frac{ab}{a - b} - x \\ \frac{3ab}{2a - b} : x = \frac{ab}{a - b} \\ \frac{ab}{2a - b} : x = \frac{ab}{a - b} \\ \frac{ab}{2a - b} : \frac{ab}{3a - 2b} = \frac{ab}{3a - 2b} \\ \text{therefore the four first terms are } a, b, \frac{ab}{a - b}, \frac{ab}{3a - 2b}. \end{align*} \]

Whence the law of the series is obvious, and it may be continued. Reversion of Series is the method of finding the value of the quantity whose several powers are involved in a series, in terms of the quantity which is equal to the given series.

In order to this, a series must be assumed, which being involved and substituted for the quantity equal to the series, and its powers, neglecting those terms whose powers exceed the highest power to which it is proposed to extend the series.

Let it be required to revert the series \(a + bx + cx^2 + dx^3 + ex^4 + \ldots\), &c., or to find \(x\) in an infinite series expressed in the powers of \(y\).

Substitute \(y^n\) for \(x\), and the indices of the powers of \(y\) in the equation will be \(n, 2n, 3n, \ldots\), &c., and therefore \(n = 1\); and the differences are \(0, 1, 2, 3, 4, 5, \ldots\), &c.

Hence, in this case, the series to be assumed is \(Ay + By^2 + Cy^3 + Dy^4, \ldots\), &c., which being involved and substituted for the respective powers of \(x\), then we have

\[ \begin{align*} ax &= aA + aBy^2 + aCy^3 + aDy^4, \ldots \\ bx &= bA + bBy^2 + bCy^3 + bDy^4, \ldots \\ cx &= cA + cBy^2 + cCy^3 + cDy^4, \ldots \\ dx &= dA + dBy^2 + dCy^3 + dDy^4, \ldots \end{align*} \]

Whence, by comparing the homologous terms, we have \(aA = y\); therefore \(A = \frac{1}{a}, B = \frac{b}{a^2}, C = \frac{2b - ac}{a^3}, D = \frac{2bAC + bB^2 + 3cA^2B + dA^4}{a}\)

\[ \begin{align*} &= \frac{5abc - 5b^3 - ad^2}{a^7} \\ &= \frac{2b^2 - ac}{a^5} \times y^3 - \frac{5b^3 - 5abc + ad^2}{a^7} \times y^4, \ldots \end{align*} \]

Examples.

1st, Let \(x = \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4}, \ldots = y\). There being in this case equal to \(1, b = -\frac{1}{2}, c = \frac{1}{3}, d = -\frac{1}{4}, \ldots\), &c., we shall, by substituting these values, have \(x = y + \frac{y^2}{2} + \frac{y^3}{6} + \frac{y^4}{24}, \ldots\).

2nd, Let \(x = x^2 + x^3 + x^4 + x^5, \ldots = y\); to find \(x\).

In this example we have \(x = x, a = 1, b = -1, c = 1, d = -1, \ldots\), &c.; whence \(x = \frac{y}{1} + \frac{1}{1} y^2 + \frac{2 - 1}{1} y^3 + \frac{-5 + 5 - 1}{1} y^4, \ldots\), &c.

3rd, Let \(a = r - \frac{x^2}{2r} + \frac{x^4}{24r^3} - \frac{x^6}{720r^5} + \frac{x^8}{4032r^7}, \ldots\), &c., to find \(x\).

Put \(r = a = v\); then \(v = \frac{x^2}{2r} - \frac{x^4}{24r^3} + \frac{x^6}{720r^5} - \frac{x^8}{4032r^7}, \ldots\), &c. By comparison we find \(x = x, y = v, a = \frac{1}{2r}, b = \frac{-1}{24r^3}, c = \frac{1}{720r^5}, d = \frac{-1}{4032r^7}, \ldots\), &c.

Hence Hence \( x^2 = 2rv - \frac{1}{8r^3}v^3 + \frac{1}{32r^5}v^5 + \cdots \)

\( v^3, \text{etc.} = 2rv + \frac{1}{3}v^2 + \frac{4}{45}v^3 + \frac{1}{35r^2}v^4 + \cdots \)

whence \( x = \sqrt{2rv} \times \left( 1 + \frac{v}{12r} + \frac{3v}{160r^2} + \frac{5v}{896r^3} + \cdots \right) \)

Summation of Series is the method of finding the sum of the terms of an infinite series produced to infinity, or the sum of any number of terms of such a series.

The value of any arithmetical series, as \( 1^2 + 2^2 + 3^2 + \cdots + n^2 \), varies according as (a) the number of its terms varies; and therefore, if it can be expressed in a general manner, it must be explicable by \( n \) and its powers with determinate coefficients; and those powers, in this case, must be rational, or such whose indices are whole positive numbers; because the progression, being a whole number, cannot admit of surd quantities. Lastly, it will appear that the greatest of the said indices cannot exceed the common index of the series by more than unity; for, otherwise, when \( n \) is taken indefinitely great, the highest power of \( n \) would be indefinitely greater than the sum of all the rest of the terms.

Thus the highest power of \( n \), in an expression exhibiting the value of \( 1^2 + 2^2 + 3^2 + \cdots + n^2 \), cannot be greater than \( n^3 \); for \( 1^2 + 2^2 + 3^2 + \cdots + n^2 \) is manifestly less than \( n^3 \), or \( n^2 + n^2 + n^2 + \cdots \), continued to \( n \) terms; but \( n^3 \), when \( n \) is indefinitely great, is indefinitely greater than \( n^2 \), or any other inferior power of \( n \), and therefore cannot enter into the equation. This being premised, the method of investigation may be as follows:

Examples.

1. Required the sum of \( n \) terms of the series \( 1 + 2 + 3 + 4 + \cdots + n \)?

Let \( An^2 + Bn \) be assumed, according to the foregoing observations, as an universal expression for the value of \( 1 + 2 + 3 + 4 + \cdots + n \), where \( A \) and \( B \) represent unknown but determinate quantities. Therefore, since the equation is supposed to hold universally, whatsoever is the number of terms, it is evident, that if the number of terms be increased by unity, or, which is the same thing, if \( n + 1 \) be written therein instead of \( n \), the equation will still subsist; and we shall have

\[ A \times n + 1 + B \times n + 1 = 1 + 2 + 3 + 4 + \cdots + n + 1. \]

From which the first equation being subtracted, there remains \( A \times n + 1 - An^2 + B \times n - 1 - Bn = n + 1; \)

this contracted will be \( 2An + A + B = n + 1; \)

whence we have \( 2A - 1 \times n + A + B - 1 = 0; \)

Wherefore, by taking \( 2A - 1 = 0 \), and \( A + B - 1 = 0 \), we have \( A = \frac{1}{2} \), and \( B = \frac{1}{2} \); and consequently

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

What is the sum of the ten first terms of the series \( 1 + 2 + 3, \text{etc.} \)?

In this case \( n = 10 \), then \( \frac{n(n+1)}{2} = \frac{10 \times 11}{2} = 55. \)

2. Required the sum of the series \( 1^2 + 2^2 + 3^2 + \cdots + n^2 \),

or \( 1 + 4 + 9 + 16 + \cdots + n^2 \)?

Let \( An^3 + Bn^2 + Cn \), according to the aforesaid observations, be assumed \( = 1^2 + 2^2 + 3^2 + \cdots + n^2 \); then, as in the preceding case, we shall have \( A \times n + 1^3 + B \times n + 1^2 + C \times n + 1 = 1^2 + 2^2 + 3^2 + \cdots + n^2 \);

that is, by involving \( n + 1 \) to its several powers, \( An^3 + 3An^2 + 3An + A + Bn + B + Cn + C = 1^2 + 2^2 + 3^2 + \cdots + n^2 + n + 1; \)

from which subtracting the former equation, we obtain \( 3An^2 + 3An + A + 2Bn + B + C = n^2 + 2n + 1; \)

and consequently \( 3A - 1 \times n^2 + 3A + 2B - 2 \times n + A + B + C - 1 = 0; \)

whence \( 3A - 1 = 0, 3A + 2B - 2 = 0, \) and \( A + B + C - 1 = 0; \)

therefore \( A = \frac{1}{3}, B = \frac{2 - 3A}{3} = \frac{1}{3}, C = 1 - A - B = \frac{1}{3}; \)

and consequently \( 1 + 4 + 9 + 16 + \cdots + n^2 = \frac{n^2}{3} + \frac{n^2}{2} + \frac{n}{6} \),

or \( \frac{n(n+1)(2n+1)}{6}. \)

What is the sum of the ten first terms of the series \( 1^2 + 2^2 + 3^2, \text{etc.} \)?

Here \( n = 10 \), then \( \frac{n(n+1)(2n+1)}{6} = \frac{10 \times 11 \times 21}{6} = 385. \)

3. Required the sum of the series \( 1^3 + 2^3 + 3^3 + 4^3 + \cdots + n^3 \),

or \( 1 + 8 + 27 + 64 + \cdots + n^3 \)?

By putting \( An^4 + Bn^3 + Cn^2 + Dn = 1 + 8 + 27 + 64 + \cdots + n^3 \); and proceeding as above, we shall have

\[ 4An^3 + 6An^2 + 4An + A + 3Bn^2 + 3Bn + B + 2Cn + C + D = n^3 + 3n^2 + 1, \]

and therefore \( A - 1 \times n^3 + 6A + 3B - 3 \times n^2 + 4A + 3B + 2C - 3 \times n + A + B + C + D - 1 = 0. \)

Hence \( A = \frac{1}{4}, B = \frac{3 - 6A}{3} = \frac{1}{4}, C = \frac{3 - 4A - 3B}{2} = \frac{1}{4}, \)

\( D = 1 - A - B - C = 0; \) and therefore \( 1^3 + 2^3 + 3^3 + 4^3 + \cdots + n^3 = \frac{n^4}{4} + \frac{n^3}{2} + \frac{n^2}{4}, \) or \( \frac{n^2(n+1)^2}{4}. \)

In the very same manner it will be found, that

\[ 1^4 + 2^4 + 3^4 + \cdots + n^4 = \frac{n^5}{5} + \frac{n^4}{2} + \frac{n^3}{3} - \frac{n^2}{30}, \]

\[ 1^5 + 2^5 + 3^5 + \cdots + n^5 = \frac{n^6}{6} + \frac{n^5}{2} + \frac{5n^4}{12} - \frac{n^3}{12}, \]

\[ 1^6 + 2^6 + 3^6 + \cdots + n^6 = \frac{n^7}{7} + \frac{n^6}{2} + \frac{n^5}{2} - \frac{n^3}{6} + \frac{n^2}{42}. \]

What is the sum of the ten first terms of the series \( 1^2 + 2^2 + 3^2, \text{etc.} \)?

Here \( n = 10 \), then \( \frac{n^2(n+1)^2}{4} = \frac{100 \times 121}{4} = 25 \times 121 = 3025. \)

4. Required the sum of \( n \) terms of the series of triangular numbers \( 0, 1, 3, 6, 10, \ldots, n \)?

Let \( An^2 + Bn^3 + Cn = 0, 1, 2, 3, \ldots, n. \)

Now the \( n + 1 \)th term of this series, by Example 2., is \( \frac{n^2}{2} + \frac{n}{2}. \)

Then \( A \times n + 1^3 + B \times n + 1^2 + C \times n + 1 = 0, 1, 2, 3, \ldots, n. \)

\( n + 1 = s + \frac{n^2}{2} + \frac{n}{2}. \)

Now, the first equation being subtracted from this, we have \( 3An^2 + 3A + 2Bn + B + C = \frac{n^2}{2} + \frac{n}{2}. \)

Or, \( 3An^2 + 3An + A = \frac{n^2}{2} + \frac{n}{2}. \) A + C = \frac{n^2}{2} + \frac{1}{4} - 2B \times n - B.

Whence, by equating the homologous terms, we have \(3A = \frac{1}{2}\), and \(A = \frac{1}{2} - 2B = 3A\); hence \(2B = \frac{1}{2} - \frac{1}{2} = 0\), \(A + C = -B\). Hence \(C = -\frac{1}{2}\).

Now, these values being substituted in the above equation, gives the sum \(= \frac{n^3}{6} - \frac{n}{6} = \frac{n(n-1)(n+1)}{6}\); and if \(n + 1\) be put for \(n\), the sum of \(n\) terms of this series will be \(\frac{n(n+1)(n+2)}{6}\).

By proceeding in the same manner, the sum of \(n\) terms of pyramidal numbers, \(1, 4, 10, 20, 35, \ldots\), \(n\) will be found \(= \frac{n(n+1)(n+2)}{6}\). And the sum of any series of figurate numbers is determined by a like formula, the law of continuation being obvious.

What is the sum of the ten first terms of triangular numbers \(1, 3, 6, 10, 15, \ldots\)? Here \(n = 10\); then \(\frac{n(n+1)(n+2)}{6} = \frac{10 \times 11 \times 12}{6} = 220\).

5. Let the sum of the series \(\frac{1}{R} + \frac{2}{R^2} + \frac{3}{R^3}\) continued to \(n\) terms be required?

If we multiply this series indefinitely continued by \(R - 1\), or \(R^2 - 2R + 1\), the product is \(R\); therefore the amount of the indefinite series is \(\frac{R}{R-1}\), and the sum of \(n\) terms may be found by subtracting the terms after the \(n\)th from that amount. Now, the terms after the \(n\)th are \(\frac{n+1}{R^n+1} + \frac{n+2}{R^n+2}, \ldots\), which may be divided into the two following series:

First, \(\frac{1}{R^n} \times \frac{1}{R} + \frac{1}{R^2} + \frac{1}{R^3}, \ldots = \frac{n}{R^n} \times \frac{1}{R-1}\).

Second, \(\frac{1}{R^n} \times \frac{1}{R} + \frac{2}{R^2} + \frac{3}{R^3}, \ldots = \frac{1}{R^n} \times \frac{R}{R-1}\).

Now, if we write \(a\) for \(\frac{1}{R^n}\) and \(r\) for \(R - 1\), and subtract the sum of these two series from the amount of the proposed series indefinitely continued, the remainder will be found \(= \frac{1-a}{r} \times R - \frac{na}{r}\).

6. Let the sum of the series \(\frac{n-1}{nR} + \frac{n-2}{nR^2} + \frac{n-3}{nR^3} \ldots\) be required?

This series is equal to the difference of the two following.

First, \(\frac{n}{nR} + \frac{n}{nR^2} + \frac{n}{nR^3}, \ldots = \frac{1}{R} + \frac{1}{R^2} + \frac{1}{R^3}, \ldots = \frac{1-a}{r}\).

Second, \(\frac{1}{nR} + \frac{2}{nR^2} + \frac{3}{nR^3}, \ldots = \frac{1}{n} \times \frac{1}{R} + \frac{1}{R^2} + \frac{1}{R^3}, \ldots = \frac{1}{n} \times \frac{1-a}{r} \times R - \frac{a}{r}\).

The difference of these series is \(\frac{1-a}{r} - \frac{R}{n} \times \frac{1-a}{r} + \frac{a}{r}\), which reduces becomes \(\frac{n+a-1 \times r + a - 1}{nr^2}\).

To proceed farther would lead us far beyond the limits assigned for this article; we must therefore refer those who require more information on this subject to the following authors.—Bertrand's Développement, &c., vol. i; Dodson's Mathematical Repository, vol. i; Emerson's Algebra; Appendix to Gravesend's Algebra; Hutton's Paper on Cubic Equations and Infinite Series, in the Philosophical Transactions for 1780; MacLaurin's Fluxions; Malcolm's Arithmetic; Mafer's Antiquities; and Scriptores Logarithmici, &c.; De Moivre's Doctrine of Chances, and a Paper by the same author in the Philosophical Transactions, no 240; Simpson's Algebra, Essays, Fluxions, and Miscellanies; Sterling's Summatio et Interpolatio Serierum; Syntagma Mathematicum, &c.