Sum of series 2.

$\displaystyle \;\;\sum_{n=1}^{\infty} n\left(\frac{1}{2}\right)^{n}$

SOLUTION : We have series represantation of $\;(1-x)^{-n}\;$ as

$\displaystyle(1-x)^{-n}=1+n x+\frac{n(n-1)}{2 !} x^{2}+\frac{n(n-1)(n-2)}{3 !} x^{3}+\cdots$

$\Rightarrow\displaystyle(1-x)^{-2}=1+2 x+3 x^{2}+4 x^{3}+5 x^{4}+\ldots$

Replace $x$ with $\;\displaystyle \frac{1}{2}\;$ we will get

$\Rightarrow \displaystyle\left(1-\frac{1}{2}\right)^{-2}=1+2 . \frac{1}{2}+3 .\frac{1}{2^2}+4 . \frac{1}{2^{3}}+5 . \frac{1}{24}+\cdots$

$\Rightarrow \displaystyle\left(1-\frac{1}{2}\right)^{-2}=\sum_{n=1}^{\infty} n\left(\frac{1}{2}\right)^{n-1}\;\;\;\;\cdots(1)$

        or we can write above as

$\displaystyle\left(1-\frac{1}{2}\right)^{-2}=2 \sum_{n=1}^{\infty} n\left(\frac{1}{2}\right)^{n}$

Use this in $1$

$\Rightarrow \displaystyle\sum_{n=1}^{\infty} n\left(\frac{1}{2}\right)^{n}=\frac{1}{2}\left(1-\frac{1}{2}\right)^{-2}$

$\Rightarrow \displaystyle\sum_{n=1}^{\infty} n\left(\frac{1}{2}\right)^{n}=\frac{1}{2}\left(\frac{1}{2}\right)^{-2}$

$\Rightarrow \displaystyle\sum_{n=1}^{\infty} n\left(\frac{1}{2}\right)^{n}= 2$

ANSWER : 2