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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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The Lerch transcendent is defined as

$$\Phi(z,s,a)=\sum_{k=0}^\infty \frac{z^k}{(k+a)^s}$$

Now,

$$\begin{align*} \sum_{p=p_0}^\infty \frac{x^p}{p^{3/2}}&=\sum_{p-p_0=0}^\infty \frac{x^p}{p^{3/2}}\\ &=\sum_{k=0}^\infty \frac{x^k x^{p_0}}{(k+p_0)^{3/2}}\\ &=x^{p_0}\sum_{k=0}^\infty \frac{x^k}{(k+p_0)^{3/2}}=x^{p_0}\Phi\left(x,\frac32,p_0\right) \end{align*}$$

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    Yes, I've understood that. I was trying to look for a closed form because I need it for a function asymptotic. Thanks though2012-01-27
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    ...the Lerch transcendent is a closed form. If what you wanted was an expression in terms of *elementary* functions, you're out of luck.2012-01-27
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    ... you're right, I was looking for elementary functions. Guess I'm out of luck, yes.2012-01-27