Elements of Algebra

94 ELEMENTS SECT. II. 301. Suppose again $a = \frac{1}{2}$, our fraction will then be $\frac{1}{1+\frac{1}{2}} = \frac{2}{3}$, which must be equal to this series $1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \frac{1}{16} - \frac{1}{32} + \ldots$ continued to infinity. Now, by considering only two terms, we have $\frac{1}{2}$, which is too small by $\frac{1}{6}$; three terms make $\frac{3}{4}$, which is too much by $\frac{1}{12}$; four terms give $\frac{5}{8}$, which is too small by $\frac{1}{24}$, and so on. 302. The fraction $\frac{1}{1+a}$ may also be resolved into an infinite series another way; namely, by dividing $1$ by $a+1$, as follows: $a+1) \ 1 \quad * \ (\frac{1}{a} - \frac{1}{a^2} + \frac{1}{a^3}, \ldots$ $\quad \underline{1 + \frac{1}{a}}$ $\quad \quad - \frac{1}{a}$ $\quad \quad \underline{- \frac{1}{a} - \frac{1}{a^2}}$ $\quad \quad \quad \frac{1}{a^2}$ $\quad \quad \quad \underline{\frac{1}{a^2} + \frac{1}{a^3}}$ $\quad \quad \quad \quad - \frac{1}{a^3}, \ldots$ Consequently, our fraction $\frac{1}{a+1}$ is equal to the infinite series $\frac{1}{a} - \frac{1}{a^2} + \frac{1}{a^3} - \frac{1}{a^4} + \frac{1}{a^5} - \frac{1}{a^6}, \ldots$ Let us make $a=1$, and we shall have the series $1 - 1 + 1 - 1 + 1 - 1, \ldots = \frac{1}{2}$, as before: and if we suppose $a=2$, we shall have the series $\frac{1}{2} - \frac{1}{4} + \frac{1}{8} - \frac{1}{16} + \frac{1}{32} - \frac{1}{64} \ldots = \frac{1}{3}$. * It is unnecessary to carry the actual division any farther, as the series may be continued to any length, from the law observable in the terms already obtained; for the signs are alternately *plus* and *minus*, and any subsequent term may be obtained by multiplying that immediately preceding it by $\frac{1}{a}$.