This question already has an answer here:

Can anyone please help me solve an infinite series:

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

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, user88595 Jul 2 '14 at 18:11

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


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 $$ \sum_{n=0}^\infty \frac{n}{2^{n+1}}=\frac{\left(\frac12\right)^2}{\left(1-\frac12\right)^2}=\large\color{blue}{1}. $$

  • $\begingroup$ Your DE MAN!!!!! $\endgroup$ – bodacydo Aug 20 '15 at 19:23

Not the answer you're looking for? Browse other questions tagged or ask your own question.