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This question already has an answer here:

I have the series $\sum_{n=0}^\infty \frac{n}{2^n}$. I must show that it converges to 2.

I was given a hint to take the derivative of $\sum_{n=0}^\infty x^n$ and multiply by $x$ , which gives

$\sum_{n=1}^\infty nx^n$ , or $\sum_{n=0}^\infty nx^n$.

Clearly if I take $x=\frac{1}{2}$ , the series is $\sum_{n=0}^\infty \frac{n}{2^n}$. How do I proceed from here?

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marked as duplicate by Martin Sleziak, Claude Leibovici calculus Nov 18 '16 at 9:37

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.

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Notice that if $|x|<1$ then the original series converges with

$$ \sum_{n=0}^\infty x^n \;\; =\;\; \frac{1}{1-x}. $$

Computing the derivative and plugging in $x=\frac{1}{2}$ should hopefully seem easier now.

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