Closed form expression for sum of exponentials $\sum_n e^{-\alpha n} \frac{1}{\sqrt{n}}$? Is there a closed form expression for the following sum
$$ \sum_{n=1}^{N} e^{-\alpha n } \frac{1}{\sqrt{n}}$$
with $\alpha >0$ and possibly $N \to \infty$?
 A: There is no closed form in terms of elementary functions. However, according to Wolfram Alpha, we have the following expression in the finite case:
$$\sum_{n=1}^N \frac{e^{-\alpha n}}{\sqrt{n}}=\operatorname{Li}_{1/2}(e^{-\alpha})-e^{-\alpha(N+1)}\Phi(e^{-\alpha},1/2,N+1),$$
where $\operatorname{Li}_s(x)$ is the polylogarithm
$$\operatorname{Li}_s(x):=\sum_{n=1}^{\infty} \frac{z^n}{n^s},$$
and $\Phi(z,s,a)$ is the Lerch transcendent
$$\Phi(z,s,a):=\sum_{n=0}^{\infty} \frac{z^n}{(n+\alpha)^s}.$$

To derive this result, note that
$$\begin{align} \sum_{n=1}^N \frac{e^{-\alpha n}}{\sqrt{n}}&=\sum_{n=1}^{\infty} \frac{e^{-\alpha n}}{\sqrt{n}}-\sum_{n=N+1}^{\infty} \frac{e^{-\alpha n}}{\sqrt{n}}\\&=\sum_{n=1}^{\infty} \frac{(e^{-\alpha})^n}{n^{1/2}}-\sum_{k=0}^{\infty} \frac{e^{-\alpha (k+N+1)}}{\sqrt{k+N+1}}\\&=\operatorname{Li}_{1/2}(e^{-\alpha})-e^{-\alpha (N+1)} \sum_{k=0}^{\infty} \frac{(e^{-\alpha})^k}{(k+(N+1))^{1/2}}\\&=\operatorname{Li}_{1/2}(e^{-\alpha})-e^{-\alpha(N+1)}\Phi(e^{-\alpha},1/2,N+1),\end{align}$$
where we have used the change of dummy index $k=n-(N+1)$ when going from the first to the second line. In the limit $N\to \infty$, we have that
$$\sum_{n=1}^{\infty} \frac{e^{-\alpha n}}{\sqrt{n}}=\operatorname{Li}_{1/2}(e^{-\alpha}).$$
