How can i prove that $ \sum\limits_{n=0}^\infty \frac{n}{3^n}=\frac{3}{4}$?
I tried to find some pattern but didn't succeed
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11Differentiate $\sum x^n$ – L. F. Oct 27 '13 at 20:44
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2$\sum_{n=0}^\infty \frac{n}{3^n} = \sum_{n=0} \sum_{m=1}^n \frac{1}{3^n}.$ Now rearrange this sum and use facts about geometric series. A picture might help on how to rearrange the sum. – Zach L. Oct 27 '13 at 20:50
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@L.F. how differentiate helps here? – mori Oct 28 '13 at 06:42
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Another approach: As the series converges, let L denote its limit. Now you have $$L=\sum_{n=0}^\infty \frac{n}{3^n}=\sum_{n=1}^\infty \frac{n}{3^n}= \frac{1}{3}\sum_{n=1}^\infty \frac{n}{3^{n-1}}= \frac{1}{3}\sum_{n=0}^\infty \frac{n+1}{3^{n}}= \frac{1}{3}\left[\sum_{n=0}^\infty \frac{n}{3^{n}}+\sum_{n=0}^\infty \frac{1}{3^{n}}\right] =\frac{1}{3}\left[L+ \sum_{n=0}^\infty \frac{1}{3^{n}}\right].$$ Now just use the usual formula for the limit of a geometric series, and solve algebraically for $L$.
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