Question

The value of the sum $\sum _{k=1}^{\infty }\sum _{n=1}^{\infty }\frac{k}{2^{n+k}}$ is

Answer 1

$\sum _{k=1}^{\infty }\sum _{n=1}^{\infty }\frac{k}{2^{n+k}}=\sum _{k=1}^{\infty }\sum _{n=1}^{\infty }\frac{k}{2^k}\cdot \frac{1}{2^n}=\sum _{k=1}^{\infty }\frac{k}{2^k}\left(\sum _{n=1}^{\infty }\frac{1}{2^n}\right)\cdots \left(1\right)$
Now $\sum _{n=1}^{\infty }\frac{1}{2^n}$ is sum of infinite G.P. with $a=\frac{1}{2},r=\frac{1}{2}$
$\therefore \sum _{n=1}^{\infty }\frac{1}{2^n}=\frac{1/2}{1-1/2}=1$
So $\left(1\right)$ will be
$=\sum _{k=1}^{\infty }\frac{k}{2^k}$
Now let $S_{\infty }=\sum _{k=1}^{\infty }\frac{k}{2^k}$
$\therefore S_{\infty }=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\cdots \infty \dots \left(2\right)\frac{1}{2}{S}_{\infty }=\frac{1}{2^2}+\frac{2}{2^3}+\frac{3}{2^4}+\cdots \infty \dots \left(3\right)$
From $\left(2\right)-\left(3\right)$
$\frac{1}{2}{S}_{\infty }=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots \infty =\frac{1/2}{1-1/2}=1\Rightarrow S_{\infty }=2$
$\therefore \sum _{k=1}^{\infty }\sum _{n=1}^{\infty }\frac{k}{2^{n+k}}=2$