# Evaluation of geometric-like series

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. – Ragib Zaman Jan 24 '12 at 11:04
@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. – Gerry Myerson Jan 24 '12 at 11:27
@GerryMyerson Sorry, you are correct, that is what I meant. – Ragib Zaman Jan 24 '12 at 11:49
But of course, we could just give it a name "the john_leo" function and the question would be simple. – Fabian Jan 24 '12 at 11:52
Actually, this seems to have been studied before. I plugged it into Wolfram|Alpha and it says this is basically the Lerch Transcendent. – Dejan Govc Jan 24 '12 at 12:51

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