Elements of Algebra

CHAP. V. OF ALGEBRA. 95 303. In the same manner, by resolving the general fraction $\frac{c}{a+b}$ into an infinite series, we shall have, $a+b) c (*(\frac{c}{a} - \frac{bc}{a^2} + \frac{b^2c}{a^3} - \frac{b^3c}{a^4}*)$ $c + \frac{bc}{a}$ \_\_\_\_\_\_\_\_\_ $-\frac{bc}{a}$ $-\frac{bc}{a} - \frac{b^2c}{a^2}$ \_\_\_\_\_\_\_\_\_ $\frac{b^2c}{a^2}$ $\frac{b^2c}{a^2} + \frac{b^3c}{a^3}$ \_\_\_\_\_\_\_\_\_ $-\frac{b^3c}{a^3}$ Whence it appears, that we may compare $\frac{c}{a+b}$ with the series $\frac{c}{a} - \frac{bc}{a^2} + \frac{b^2c}{a^3} - \frac{b^3c}{a^4}, \&c.$ to infinity. Let $a=2, b=4, c=3$, and we shall have $\frac{c}{a+b} = \frac{3}{2+4} = \frac{3}{6} = \frac{1}{2} = \frac{3}{2} - 3 + 6 - 12, \&c.$ If $a=10, b=1$, and $c=11$, we shall have $\frac{c}{a+b} = \frac{11}{10+1} = 1 = \frac{11}{10} - \frac{11}{100} + \frac{11}{1000} - \frac{11}{10000}, \&c.$ Here if we consider only one term of the series, we have $\frac{11}{10}$, which is too much by $\frac{1}{10}$; if we take two terms, we have $\frac{99}{100}$, which is too small by $\frac{1}{100}$; if we take three terms, we have $\frac{1001}{1000}$, which is too much by $\frac{1}{1000}, \&c.$ 304. When there are more than two terms in the divisor, we may also continue the division to infinity in the same \* Here again the law of continuation is manifest; the signs being alternately $+$ and $-$, and each succeeding term is formed by multiplying the foregoing one by $-\frac{b}{a}$.