Calculate the value of $\sum\limits _{n=1}^{\infty }\:\dfrac{n}{2^n}$

Question

In a previous question it is asked to represent $f(x)=\dfrac{x}{1-x^2}$ as a power series. It gave me $\displaystyle\sum _{n=1}^{\infty \:}x\left(2x^2-x^4\right)^{n-1}$. Then they ask to use the last expression to calculate $\displaystyle\sum _{n=1}^{\infty }\:\frac{n}{2^n}$. Thanks!

Answer 1

Consider $$ \sum_{n=0}^\infty y^n=\frac{1}{1-y}\quad;\quad\text{for}\ |y|<1.\tag1 $$ Differentiating $(1)$ with respect to $y$ yields $$ \sum_{n=1}^\infty ny^{n-1}=\frac{1}{(1-y)^2}.\tag2 $$ Multiplying $(2)$ by $y$ yields $$ \sum_{n=1}^\infty ny^{n}=\frac{y}{(1-y)^2}.\tag3 $$ Now plug in $y=\dfrac12$ to $(3)$ yields $$ \large\color{blue}{\sum_{n=1}^\infty \frac{n}{2^n}=\frac{2}{(2-1)^2}=2}. $$

Answer 2

Here's a version without using power series:

Let $A$ = 1/2 + 2/4 + 3/8 + 4/16 + .... =

(1/2 + 1/4 + 1/8 + 1/16 + ...) + (0/2 + 1/4 + 2/8 + 3/16 + ... ) = 1 + $A/2$.

Solving gives $A = 2 $