Find the sum $\sum_{n=1}^\infty\frac{\sqrt{n}}{2^{n/2}}$ Find the sum $\sum_{n=1}^\infty\frac{\sqrt{n}}{2^{n/2}}$. Is it possible to find a closed form for this sum? I was trying to upper bound this sum by another series, but could not find one.
Can anyone help?

The closed form using polylogarithms is given elsewhere.  An alterantive for (upper and lower) bounds...
I note that for $n > 4$, $\displaystyle \frac{2}{2^{n/2}} < \frac{\sqrt{n}}{2^{n/2}} < \frac{1}{2^{n/4}}$.
The left inequality is easy: For $4 < n$, $2 < \sqrt{n}$, then divide through by $2^{n/2}$.  Using the first three terms of the original series and then the lower bounding term subsequently, we get the lower bound $(1/4)(4 + 6\sqrt{2} + \sqrt{6} = 3.733{\dots}$.
For the right inequality, start with $n = 4$, observing $\displaystyle \sqrt{4} = 2 = 2^{4/4}$.  Then $\frac{\mathrm{d}}{\mathrm{d}n}\sqrt{n} < 1/4$ for $n > 4$ and $\frac{\mathrm{d}}{\mathrm{d}n} 2^{n/4} > 1/4$ when $2^{n/4} \ln 2 > 1$, which is certainly the case for $n > 4$.  Therfore,
$$  \frac{\sqrt{n}}{2^{n/4}} < 1 \qquad (n > 4)  \text{.}  $$
Now divide through by $2^{n/4}$.  Splicing the three terms from the original series into the series of these upper bounding terms gives the upper bound $\sqrt{3/8} + \sqrt{2} + \frac{1}{2-2^{3/4}} = 5.169{\dots}$.
