Elements of Algebra

CHAP. V. OF ALGEBRA. 93 $1+a$) $1$ ($1-a+a^2-a^3+a^4$ $1+a$ $-a$ $-a-a^2$ $a^2$ $a^2+a^3$ $-a^3$ $-a^3-a^4$ $a^4$ $a^4+a^5$ $-a^5, \&c.$ Whence it follows, that the fraction $\frac{1}{1+a}$ is equal to the series, $1-a+a^2-a^3+a^4-a^5+a^6-a^7, \&c.$ 299. If we make $a=1$, we have this remarkable comparison: $\frac{1}{1+a} = \frac{1}{2} = 1-1+1-1+1-1+1-1, \&c.$ to infinity; which appears rather contradictory; for if we stop at $-1$, the series gives $0$; and if we finish at $+1$, it gives $1$; but this is precisely what solves the difficulty; for since we must go on to infinity, without stopping either at $-1$ or at $+1$, it is evident, that the sum can neither be $0$ nor $1$, but that this result must lie between these two, and therefore be $\frac{1}{2}$*. 300. Let us now make $a=\frac{1}{2}$, and our fraction will be $\frac{1}{1+\frac{1}{2}} = \frac{2}{3}$, which must therefore express the value of the series $1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}+\frac{1}{64}, \&c.$ to infinity; here if we take only the two leading terms of this series, we have $\frac{1}{2}$, which is too small by $\frac{1}{6}$; if we take three terms, we have $\frac{3}{4}$, which is too much by $\frac{1}{12}$; if we take four terms, we have $\frac{5}{8}$, which is too small by $\frac{1}{24}, \&c.$ *It may be observed, that no infinite series is in reality equal to the fraction from which it is derived; unless the remainder be considered, which, in the present case, is alternately $+\frac{1}{2}$ and $-\frac{1}{2}$; that is, $+\frac{1}{2}$ when the series is $0$, and $-\frac{1}{2}$ when the series is $1$, which still gives the same value for the whole expression. Vid. Art. 293.