Question

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

Answer 1

Given,

$\frac{1}{2}+\frac{1}{3^2}+\frac{1}{2^3}+\frac{1}{3^4}+\frac{1}{2^5}+\frac{1}{3^6}+...\dots \infty$

$=(\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+.....\infty )+(\frac{1}{3^2}+\frac{1}{3^4}+\frac{1}{3^6}+.....\infty )$

$=\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+.....\infty$ $\to$ series with first term$=\frac{1}{2}$, common ratio$=\frac{1}{2^2}$

$=\frac{\frac{1}{2}}{1-\frac{1}{2^2}}=\frac{\frac{1}{2}}{\frac{3}{4}}=\frac{2}{3}$

$=S_{\infty }=\frac{a}{1-r}$

$\frac{1}{3^2}+\frac{1}{3^4}+\frac{1}{3^6}+....\infty$ $\to$ series with first term$=\frac{1}{3^2}$, common ratio$=\frac{1}{3^2}$

$=\frac{\frac{1}{3^2}}{1-\frac{1}{3^2}}=\frac{\frac{1}{9}}{\frac{8}{9}}=\frac{1}{8}$

$\therefore$ Sum of given series$=\frac{2}{3}+\frac{1}{8}=\frac{19}{24}$.