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I have a series of the form $$\sum\limits_{k=0}^\infty \left( \frac{5}{6} \right)^k k$$

I come across it using generating functions to find an expectation for the geometric distribution. I was wondering what's the easiest way to evaluate series of this form. More generally $$\sum\limits_{k=0}^\infty p^k k \quad p \in (0,1)$$

Thanks

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    $\begingroup$ Here's a very similar problem. $\endgroup$ – David Mitra Feb 2 '15 at 11:48
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I am renaming $p$ to $x$. Call your series $f(x)$. Divide your series by $x$, which gives $$ f(x)/x = \sum_{k=0}^\infty kx^{k-1}. $$ An antiderivative of this is $$ G(x) = \sum_{k=0}^\infty x^k = \frac{1}{1-x}. $$ Now what you want is $$ f(x) = xG'(x) = \frac{x}{(1-x)^2}. $$

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  • $\begingroup$ grr, made some errors. editing now.... should be correct; check me! wolfram alpha agrees for x = 5/6 that you should get 30. $\endgroup$ – hunter Feb 2 '15 at 11:45
  • $\begingroup$ Yeah that makes a lot of sense, didn't consider differentiating the closed form. Thanks $\endgroup$ – HBeel Feb 2 '15 at 11:50
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$$\sum_{k=0}^{\infty}p^k=\frac1{1-p}\tag{$|p|<1$}$$ So: $$\frac{{\rm d}}{{\rm d}p}\frac1{1-p}=\frac{{\rm d}}{{\rm d}p}\sum_{k=0}^{\infty}p^k=\sum_{k=0}^{\infty}kp^{k-1}=\frac1p\sum_{k=0}^{\infty}kp^k\implies\sum_{k=0}^{\infty}kp^k=\frac p{(1-p)^2}$$

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