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 kinds of series; as arithmetic, geometrical, infinite, &c. kinds of.

Series. The two first of these are, however, more generally known or distinguished by the names of arithmetic 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,

3. 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.

4. Converging and diverging series. 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.

5. Law of an infinite series. 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.

6. Method of converting Let the division or evolution of the given fraction, 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.

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

\begin{array}{r} 1-x \\ 1-x \overline{) 1} \\ \hline x \\ x-x^2 \\ \hline x^2 \\ x^2-x^3 \\ \hline x^3 \\ x^3-x^4 \\ \hline x^4 \\ x^4-x^5 \\ \hline x^5 \end{array}

Hence the fraction \frac{1}{1-x} = 1 + x + x^2 + x^3 + x^4, \&c.

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{array}{r} 1+x \overline{) ay} \\ \hline ay + ayx \\ \hline -ayx \\ \hline -ayx - ayx^2 \\ \hline ayx^2 \\ \hline ayx^2 + ayx^3 \\ \hline -ayx^3 \\ \hline -ayx^3 - ayx^4 \\ \hline ayx^4 \\ \hline ayx^4 + ayx^5 \\ \hline -ayx^5 \end{array}

Hence \frac{ay}{1+x} = ay \times 1 - x + x^2 - x^3 + x^4, \&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{array}{r} m+x \overline{) m^2+x^2} \\ \hline m^2 + mx \\ \hline -mx + x^2 \\ \hline -mx + x^2 \\ \hline 2x^2 \\ \hline 2x^2 + \frac{2x^3}{m} \\ \hline -\frac{2x^3}{m} \\ \hline -\frac{2x^3}{m} - \frac{2x^4}{m^2} \\ \hline \frac{2x^4}{m^2}, \&c. \end{array}

Hence \frac{m^2+x^2}{m+x} = m - x + \frac{2x}{m} \times \frac{x}{1} - \frac{x^2}{m} + \frac{x^3}{m^2}, \&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{array}{r} a^2+2ay+y^2 \overline{) a^2} \\ \hline a^2+2ay+y^2 \\ \hline -2ay-y^2 \\ \hline -2ay-4y^2 - \frac{2y^3}{a} \\ \hline 3y^3 + \frac{2y^4}{a} \\ \hline 3y^3 + \frac{6y^4}{a} + \frac{3y^5}{a^2} \\ \hline -\frac{4y^5}{a} - \frac{3y^6}{a^2} \end{array}

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

And evolution. 5. Let \sqrt{a^2+x^2} be converted into an infinite series?

\begin{aligned} & \frac{a^2+x^2}{a^2} + \frac{x^2}{2a} - \frac{x^4}{8a^3} + \frac{x^6}{16a^5} - \frac{x^8}{128a^7} \\ & \frac{2a + \frac{x^2}{2a}}{2a + \frac{x^2}{2a} + \frac{x^4}{4a^2}} + \frac{x^2}{4a^2} \\ & \frac{x^4}{4a^2} - \frac{x^6}{8a^4} + \frac{x^8}{64a^6} \\ & \frac{2a + \frac{x^2}{2a} - \frac{x^4}{4a^2} + \frac{x^6}{16a^4}}{2a + \frac{x^2}{2a} - \frac{x^4}{4a^2} + \frac{x^6}{16a^4} + \frac{x^8}{64a^6}} + \frac{x^8}{64a^6} \\ & \frac{x^{10}}{8a^4} + \frac{x^{12}}{16a^6} - \frac{x^{14}}{64a^8} + \frac{x^{16}}{256a^{10}} \\ & \frac{5x^8}{64a^6} + \frac{x^{10}}{64a^8} - \frac{x^{12}}{256a^{10}} \end{aligned}

Hence the square root of a^2+x^2 = a + \frac{x^2}{2a} - \frac{x^4}{8a^3} + \frac{x^6}{16a^5} - \frac{x^8}{128a^7} &c.

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.

By means of an assumed series. 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.