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I am trying to evaluate the following sum:

$$\sum_{p = p_0}^{\infty} \frac{x^p}{p^{3/2}} $$

where $p_0$ is some integer larger than one and $x$ is smaller than one.

Sums like $\sum_{p = p_0}^{\infty} px^p$ or $\sum_{p = p_0}^{\infty} \frac{x^p}{p} $ can be evaluated by switching sums and integrals, but I don't know how to deal with the $p^{3/2}$. Can anyone help me?

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    Asking for the general evaluation is probably too optimistic - even with $x=1$ and $p_0=1$ the sum is $\zeta (1/2) $ which has no nicer form.2012-01-24
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    @Ragib, don't you mean $\zeta(3/2)$? But your point is absolutely right, there's no reason to expect an evaluation in terms of the usual functions.2012-01-24
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    @GerryMyerson Sorry, you are correct, that is what I meant.2012-01-24
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    But of course, we could just give it a name "the john_leo" function and the question would be simple.2012-01-24
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    Actually, this seems to have been studied before. I plugged it into [Wolfram|Alpha](http://www.wolframalpha.com/input/?i=Sum%5Bx%5Ep/p%5E%283/2%29,%7Bp,k,Infinity%7D%5D) and it says this is basically the [Lerch Transcendent](http://mathworld.wolfram.com/LerchTranscendent.html).2012-01-24
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    @Dejan Govc thanks, that helps a lot. Oh, and actually I think $\zeta(3/2)$ is a quite nice form.2012-01-24

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