Elements of Algebra

92 ELEMENTS SECT. II. finity. Now, if we take only two terms of this series, we shall have $1+\frac{1}{2}$, and it wants $\frac{1}{2}$ of being equal to $\frac{1}{1-a}=2$. If we take three terms, it wants $\frac{1}{4}$; for the sum is $1\frac{3}{4}$. If we take four terms, we have $1\frac{7}{8}$, and the deficiency is only $\frac{1}{8}$. Therefore, the more terms we take, the less the difference becomes; and, consequently, if we continue the series to infinity, there will be no difference at all between its sum and the value of the fraction $\frac{1}{1-a}$, or $2$. 295. Let $a = \frac{1}{3}$; and our fraction $\frac{1}{1-a}$ will then be = $\frac{1}{1-\frac{1}{3}} = \frac{1}{\frac{2}{3}} = 1\frac{1}{2}$, which, reduced to an infinite series, becomes $1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}, \ldots$ which is consequently equal to $\frac{1}{1-a}$. Here, if we take two terms, we have $1\frac{1}{3}$, and there wants $\frac{1}{6}$. If we take three terms, we have $1\frac{4}{9}$, and there will still be wanting $\frac{1}{18}$. If we take four terms, we shall have $1\frac{13}{27}$, and the difference will be $\frac{1}{54}$; since, therefore, the error always becomes three times less, it must evidently vanish at last. 296. Suppose $a = \frac{2}{3}$; we shall have $\frac{1}{1-a} = \frac{1}{1-\frac{2}{3}} = 3$, $= 1+\frac{2}{3}+\frac{4}{9}+\frac{8}{27}+\frac{16}{81}+\frac{32}{243} \ldots$ to infinity; and here, by taking first $1\frac{2}{3}$, the error is $1\frac{1}{3}$; taking three terms, which make $2\frac{1}{9}$, the error is $\frac{8}{9}$; taking four terms, we have $2\frac{13}{27}$, and the error is $\frac{16}{27}$. 297. If $a = \frac{1}{4}$, the fraction is $\frac{1}{1-a} = \frac{1}{1-\frac{1}{4}} = \frac{1}{\frac{3}{4}} = 1\frac{1}{3}$; and the series becomes $1+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\frac{1}{256}, \ldots$ The first two terms are equal to $1\frac{1}{4}$, which gives $\frac{1}{12}$ for the error; and taking one term more, we have $1\frac{5}{16}$, that is to say, only an error of $\frac{1}{48}$. 298. In the same manner we may resolve the fraction $\frac{1}{1+a}$ into an infinite series by actually dividing the numerator $1$ by the denominator $1+a$, as follows *. * After a certain number of terms have been obtained, the law by which the following terms are formed will be evident; so that the series may be carried to any length without the trouble of continual division, as is shewn in this example.