# How to solve infinite series $\sum_{n=0}^\infty\frac{n}{2^{(n+1)}}$? [duplicate]

This question already has an answer here:

$$\sum_{n=0}^{\infty} \frac{n}{2^{(n+1)}}$$

I put it in Wolfram Alpha and got the result that it converges to $1$

I know that the infinite series:

$$\sum_{n=1}^{\infty} \frac{1}{2^{n}}$$

Converges to $1$.

But my series is quite different as it has an additional $n$ term multiplied in, and I can't quite see how to solve it to arrive at result $1$.

## marked as duplicate by user61527, Sasha, Peter Woolfitt, Steven Stadnicki, user88595Jul 2 '14 at 18:11

Consider the infinite geometric progression $$\sum_{n=0}^\infty x^n=\frac1{1-x}.\tag1$$ Differentiating $(1)$ with respect to $x$ yields $$\sum_{n=0}^\infty nx^{n-1}=\frac1{(1-x)^2}.\tag2$$ Multiplying $(2)$ by $x^2$ yields $$\sum_{n=0}^\infty nx^{n+1}=\frac{x^2}{(1-x)^2}.\tag3$$ Setting $x=\dfrac12$ to $(3)$ yields $$\sum_{n=0}^\infty \frac{n}{2^{n+1}}=\frac{\left(\frac12\right)^2}{\left(1-\frac12\right)^2}=\large\color{blue}{1}.$$