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?
 A: $$\sum_{n=1}^{\infty} \frac{\sqrt{n}}{a^{\frac{n}{2}}}=\text{Li}_{-\frac{1}{2}}\left(\frac{1}{\sqrt{a}}\right)$$ where appears the polylogarithm function.
For $a=2$, the result is
$$4.1450444026566332049672175338821574521236562661674935536$$
A: 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 $\sqrt{3/8} + \sqrt{2} + \frac{1}{2-2^{3/4}} = 5.169{\dots}$.
A: The polylogarithm is defined $$\mathrm{Li}_s(z)=\sum_{n\ge1}\frac{z^n}{n^s}$$
for $s\in\Bbb C$ and $|z|<1$.This being the case, we can write
$$S=\sum_{n\ge1}\frac{\sqrt n}{2^{n/2}}=\mathrm{Li}_{-1/2}(\tfrac{1}{\sqrt2}).$$
From here, we have
$$\mathrm{Li}_s(z)=\frac{z}{2}+\frac{\Gamma(1-s,-\ln z)}{(-\ln z)^{1-s}}+2z\int_0^\infty \frac{\sin(s\arctan t-t\ln z)}{(1+t^2)^{s/2}(e^{2\pi t}-1)}dt$$
for complex $z,s$, where $\Gamma(a,z)$ is the upper incomplete gamma function. So,
$$S=\frac{1}{2\sqrt2}+\left(\frac{2}{\ln2}\right)^{3/2}\Gamma(\tfrac32,\tfrac12\ln2)+\sqrt2\int_0^\infty\frac{\sin(\tfrac{t}2\ln2-\tfrac12\arctan t)}{e^{2\pi t}-1}(1+t^2)^{1/4}dt.$$
The $\Gamma$ integral is
$$\Gamma(\tfrac32,\tfrac12\ln2)=\int_{\tfrac12\ln2}^\infty t^{1/2}e^{-t}dt=\left(\int_0^\infty-\int_0^{\tfrac12\ln2}\right)t^{1/2}e^{-t}dt=\frac{\sqrt\pi}{2}-\int_0^{\tfrac12\ln2}t^{1/2}e^{-t}dt=\frac{\sqrt\pi}2-I_1.$$
The other integral is, with $t\mapsto \tan x$,
$$I_2=\int_0^\infty\frac{\sin(\tfrac{t}2\ln2-\tfrac12\arctan t)}{e^{2\pi t}-1}(1+t^2)^{1/4}dt=\int_0^{\pi/2}\frac{\sin(\tfrac{\ln2}{2}\tan x-x/2)}{e^{2\pi\tan x}-1}\sec(x)^{5/2}dx.$$
So,
$$S=\frac1{2\sqrt2}+\sqrt{\frac{2\pi}{\ln^32}}-\left(\frac{2}{\ln2}\right)^{3/2}I_1+\sqrt2 I_2.$$
Then from here, we have
$$-\left(\frac{2}{\ln2}\right)^{3/2}I_1+\sqrt2 I_2\approx −0.552128172429,$$
so
$$S<\frac1{2\sqrt2}+\sqrt{\frac{2\pi}{\ln^32}}-\frac12.$$
I would be very surprised if there is a closed form for $S$ which does not rely on $\mathrm{Li}$ special values.
