<s>[INST] 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?
 [/INST] 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 $\sqrt{3/8} + \sqrt{2} + \frac{1}{2-2^{3/4}} = 5.169{\dots}$.
</s>