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How can I evaluate $\sum_{n=1}^\infty \frac{2n}{3^{n+1}}$
please help me with this $$\sum\limits_{k=0}^{\infty}\frac{k}{3^k}$$
I need just a hint, not a full answer. Thanks!
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1A hint? Sure! $$\frac{\mathrm d}{\mathrm dx}\frac1{1-x}=\frac1{(1-x)^2}$$ – J. M. ain't a mathematician Feb 09 '12 at 15:04
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3A hint is to do $\sum_{k=0}^\infty kx^k$ first, then substitute $x=1/3$. Is that enough? – GEdgar Feb 09 '12 at 15:05
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A similar question. Also, see this – David Mitra Feb 09 '12 at 15:08
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There are several ways to do this. One involves the fact that $$ \frac{k}{3^k} = kx^k = x\cdot kx^{k-1} = x\cdot \frac{d}{dx} x^k $$ where $x=1/3$, and $\displaystyle\sum_{k=0}^\infty x\cdot \frac{d}{dx} x^k$ can be found.
Another looks like this: $$ \sum_{k=0}^\infty \frac{k}{3^k} = \sum_{k=1}^\infty \frac{k}{3^k} = \left\{\begin{array}{cccccccccccccccc} & & 1/3 \\ & + & 1/9 & + & 1/9 \\ & + & 1/27 & + & 1/27 & + & 1/27 \\ & + & 1/81 & + & 1/81 & + & 1/81 & + & 1/81 \\ & + & \cdots \end{array}\right. $$
Then you can sum the first column, then the second column, etc., and finally find the sum of all of the column sums.
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$$\sum_{k=0}^{\infty}{kr^k}=\frac{r}{(r-1)^2}$$ In your case $r=\frac{1}{3}$
Riccardo.Alestra
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