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Possible Duplicate:
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!

  • 1
    A hint? Sure! $$\frac{\mathrm d}{\mathrm dx}\frac1{1-x}=\frac1{(1-x)^2}$$2012-02-09
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    A hint is to do $\sum_{k=0}^\infty kx^k$ first, then substitute $x=1/3$. Is that enough?2012-02-09
  • 0
    A [similar](http://math.stackexchange.com/questions/90637/what-is-the-limit-of-sum-limits-n-1-inftyn2-3n/90723) question. Also, see [this](http://math.stackexchange.com/questions/30732/how-can-i-evaluate-sum-n-1-infty-frac2n3n1)2012-02-09

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