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The infinite series $$\sum_{n=0}^\infty n/2^{n}$$ converges to $2$. But how to show that the infinite series converges to $2$?

Please help. Thank you.

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Express $n/2^n$ as sum $$\sum_{n=0}^\infty n/2^{n} = \sum_{n=1}^\infty \sum_{k=1}^n 2^{-n}$$ change order of sums $$\sum_{k=1}^\infty \sum_{n=k}^\infty 2^{-n}$$ recall that $\sum_{n=k}^\infty 2^{-n} = 2^{1-k}$, and calculate $$\sum_{k=1}^\infty 2^{1-k} = 2\sum_{k=1}^\infty 2^{-k} = 2$$

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Since $\frac{1}{2}<1$, you can interchange summation and differentiation in an infinite sum (uniform convergence). Also, rewrite $ \frac{n}{2^n}$ as $p \frac{d}{dp}(p^n)$ where $p=\frac{1}{2}$. Can you handle from here?

EDIT You can hence rewrite the infinite sum as $$ p \frac{d}{dp}\sum_{k=0}^{\infty}p^{k} $$ You get a simple geometric sum that you need to differentiate and multiply by $p$

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  • $\begingroup$ No, Sir. I am not so intelligent. Please elaborate the ans! $\endgroup$ – Silent Sep 17 '13 at 11:47
  • $\begingroup$ Pls see the edit $\endgroup$ – Alex Sep 17 '13 at 11:51
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We know that for all $|x|<2$, $$\displaystyle \sum_{n=0}^{+\infty} \frac{x^n}{2^n} = \frac{1}{1-\frac{x}{2}}$$ Then, if you differentiate with respect to $x$ we get for all $|x|<2$ $$\displaystyle \sum_{n=0}^{+\infty} n\frac{x^{n-1}}{2^n} = \frac{1}{2(1-\frac{x}{2})^2}$$ so that $$\displaystyle \sum_{n=0}^{+\infty} n\frac{x^{n}}{2^n} = \frac{x}{2(1-\frac{x}{2})^2}$$ and then do $x=1$.

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