Elements of Algebra

96 ELEMENTS SECT. II. manner. Thus, if the fraction $\frac{1}{1-a+a^2}$ were proposed, the infinite series, to which it is equal, will be found as follows: $1-a+a^2) \ 1 \quad * (1+a-a^3-a^4+a^6, \ldots$ $\underline{1-a+a^2}$ $\qquad a-a^2$ $\qquad \underline{a-a^2+a^3}$ $\qquad \qquad -a^3$ $\qquad \qquad \underline{-a^3+a^4-a^5}$ $\qquad \qquad \qquad -a^4+a^5$ $\qquad \qquad \qquad \underline{-a^4+a^5-a^6}$ $\qquad \qquad \qquad \qquad a^6$ $\qquad \qquad \qquad \qquad \underline{a^6-a^7+a^8}$ $\qquad \qquad \qquad \qquad \qquad a^7-a^8$ $\qquad \qquad \qquad \qquad \qquad \underline{a^7-a^8+a^9}$ $\qquad \qquad \qquad \qquad \qquad \qquad -a^9$ We have therefore the equation $\frac{1}{1-a+a^2}=1+a-a^3-a^4+a^6+a^7, \ldots$; where, if we make $a=1$, we have $1=1+1-1-1+1+1-1-1, \ldots$ which series contains twice the series found above $1-1+1-1+1, \ldots$ Now, as we have found this to be $\frac{1}{2}$, it is not extraordinary that we should find $\frac{2}{2}$, or $1$, for the value of that which we have just determined. By making $a=\frac{1}{2}$, we shall have the equation $\frac{1}{\frac{3}{4}}=\frac{4}{3}= 1+\frac{1}{2}-\frac{1}{8}-\frac{1}{16}+\frac{1}{64}+\frac{1}{128}-\frac{1}{512}, \ldots$ If $a=\frac{1}{3}$, we shall have the equation $\frac{1}{\frac{7}{9}}=\frac{9}{7}=1+\frac{1}{3}- \frac{1}{27}-\frac{1}{81}+\frac{1}{729}, \ldots$ and if we take the four leading terms of this series, we have $1\frac{23}{81}$, which is only $\frac{1}{567}$ less than $\frac{9}{7}$. Suppose again $a=\frac{2}{3}$, we shall have $\frac{1}{\frac{7}{9}}=\frac{9}{7}=1+\frac{2}{3}- \frac{8}{27}-\frac{16}{81}+\frac{64}{729}, \ldots$ This series is therefore equal to the preceding one; and, by subtracting the one from the other, we obtain $\frac{1}{3}-\frac{7}{27}-\frac{15}{81}+\frac{63}{729}, \ldots$ which is necessarily $=0$. 305. The method, which we have here explained, serves to resolve, generally, all fractions into infinite series; which is often found to be of the greatest utility. It is also re-