Question

The value of $\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+\cdots \infty$ is

Answer 1

$L=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+\cdots \infty$
$\frac{L}{3}=\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+\frac{4}{3^5}+\cdots \infty \Rightarrow \frac{2L}{3}=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\cdots \infty \Rightarrow \frac{2L}{3}=\frac{1/3}{1-\frac{1}{3}}$
$\Rightarrow L=\frac{3}{4}=0.75$

Alternate Solution:

We know that
$\frac{1}{1-x}=\sum _{n=0}^{\infty }{x}^n,\text{if}|x|<1$
Differentiate both the sides w.r.t. $x$, we get
$\frac{1}{(1-x{)}^2}=\sum _{n=0}^{\infty }n{x}^{n-1}$
$\frac{1}{(1-x{)}^2}=\sum _{n=1}^{\infty }n{x}^{n-1}$
(as first term is zero at $n=0)$
Now, put $x=\frac{1}{3}$
$\frac{1}{{(1-\frac{1}{3})}^2}=\sum _{n=1}^{\infty }\frac{n}{3^{n-1}}$
$\Rightarrow \frac{9}{4}=3\sum _{n=1}^{\infty }\frac{n}{3^n}$
$\Rightarrow \sum _{n=1}^{\infty }\frac{n}{3^n}=\frac{3}{4}=0.75$