Question

Find the sum of the series $\frac{1}{2}+\frac{1}{3^2}+\frac{1}{2^3}+\frac{1}{3^4}+\frac{1}{2^5}+\frac{1}{3^6}+\dots \infty$

Answer 1

Given series is $\frac{1}{2}+\frac{1}{3^2}+\frac{1}{2^3}+\frac{1}{3^4}+\frac{1}{2^5}+\frac{1}{3^6}+\dots \infty$
The above series can be written as sum of two series as
$\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+.......+\frac{1}{3^2}+\frac{1}{3^4}+\frac{1}{3^6}+......\infty$
Since, we have the sum of the terms in G.P series $a+a^2+a^3+.......\infty$ is $S_{\infty }=\frac{a}{1-r}$, where $r<1$......(i)
Applying the above formula for given two series as follow:
Consider first series, $\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+.......$
Where, first term $a=\frac{1}{2}$ and common difference $r=\frac{\frac{1}{2^3}}{\frac{1}{2}}$
$=\frac{2}{2^3}$ [Since, $\frac{a^m}{a^n}=a^{m-n}$ ]
$=\frac{1}{2^{3-1}}$
$\therefore r=\frac{1}{2^2}$
Thus, for first series infinite sum $=\frac{\frac{1}{2}}{1-\frac{1}{2^2}}$ [From equation(i)]
$=\frac{\frac{1}{2}}{1-\frac{1}{4}}$
$=\frac{\frac{1}{2}}{\frac{4-1}{4}}$
$=\frac{1}{2}\times \frac{4}{3}$
$=\frac{2}{3}$
Similar for second series $\frac{1}{3^2}+\frac{1}{3^4}+\frac{1}{3^6}+......\infty$
Where, first term $a=\frac{1}{3^2}$ and common ratio $r=\frac{\frac{1}{3^4}}{\frac{1}{3^2}}$
$=\frac{1}{3^4}\times \frac{1}{3^2}$
$=\frac{1}{3^{4-2}}$
$=\frac{1}{3^2}$
$\therefore r=\frac{1}{9}$
Thus, Sum of infinite terms $=\frac{\frac{1}{3^2}}{1-\frac{1}{9}}$ [From equation(i)]
$=\frac{\frac{1}{9}}{\frac{9-1}{9}}$
$=\frac{1}{9}\times \frac{9}{8}$
$=\frac{1}{8}$
Therefore, the sum of infinite terms of given series is $S_{\infty }=\frac{2}{3}+\frac{1}{8}$
$=\frac{(2\times 8)+(1\times 3)}{24}$
$=\frac{16+3}{24}$
$\therefore S_{\infty }=\frac{19}{24}$