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