Is there a closed form of $\sum_{n=1}^{\infty} \frac{\sqrt{n} } {e^{2 \pi n}-1}$?

Question

Recently there were some discussions [1], [2], [3] of sums of the type

$$S(k) = \sum_{n=1}^{\infty} \frac{n^{k-1}}{e^{2 \pi n}-1}\tag{1}$$

Here $k$ was assumed to be a positive integer.

For a positive integer $k$ I have given here and here a closed expression with a slightly different definition of $S(m)$.

Now I would like to drop this restriction an allow $k$ to assume real values.

Questions

(1) as a typical example: is there a closed expression for $S(\frac{3}{2})$?

(2) what are the analytic properties of $S(z)$ in the complex $z$-plane?

My attempts regarding question (1)

It is more convenient to study the "generating" sum

$$S(z,t) = \sum_{n=1}^{\infty} \frac{n^{z-1} e^{-n t}}{e^{2 \pi n}-1}\tag{2}$$

so that $S(z) = S(z,t=0)$ and also

$$S(z+1,t) = - \frac{\partial S(z,t)}{\partial t}\tag{3}$$

I now replace the power of $n$ by an integral

$$n^{z-1} = \frac{1}{\Gamma(1-z)} \int_{0}^{\infty} s^{-z} e^{-n s} \tag{4}$$

which leads to an integral representation as follows

$$S(z,t)=\sum _{n=1}^{\infty } \frac{e^{-n t} n^{z-1}}{e^{2 \pi n}-1}=\frac{1}{\Gamma (1-z)} \int_0^{\infty } s^{-z} f(s,t) \, ds, \Re(z)<1 \tag{5}$$

where

$$\begin{align} f(s,t) & = \sum _{n=1}^{\infty } \frac{e^{ - n (s+t)}}{e^{2 \pi n}-1}\\ & =\sum _{n=1}^{\infty } \sum _{m=1}^{\infty } e ^{ -n (s+t)-2 \pi n m}\\ & = \sum _{m=1}^{\infty } \sum _{n=1}^{\infty } e^{ -n(2 \pi m + s+t)}\\ & =\sum _{m=1}^{\infty } \frac{1}{e^{2 \pi m+s+t}-1}\\ & =-\frac{\psi _{e^{-2 \pi }}^{(0)}\left(\frac{s+t}{2 \pi }+1\right)+\log \left(1-e^{-2 \pi }\right)}{2 \pi } \end{align}\tag{6}$$

Inserting this into $(5)$ we obtain

$$\begin{align} S(z,t) & =\sum _{n=1}^{\infty } \frac{e^{-n t} n^{z-1}}{e^{2 \pi n}-1}\\ & =\frac{-1}{2 \pi\Gamma (1-z)} \int_0^{\infty } s^{-z} \left(\psi _{e^{-2 \pi }}^{(0)}\left(\frac{s+t}{2 \pi }+1\right)+\log \left(1-e^{-2 \pi }\right)\right) \, ds, \Re(z)<1 \end{align}\tag{7}$$

where $\psi _{q }^{(0)}\left(x\right)$ is the q-polygamma function.

Now we can calculate the sum in question as

$$\sum _{n=1}^{\infty } \frac{\sqrt{n}}{e^{2 \pi n}-1} =\frac{1}{\sqrt{\pi}}\left(\frac{1}{2 \pi }\right)^2 \int_0^{\infty } \frac{1}{\sqrt{s}} \psi _{e^{-2 \pi }}^{(1)}\left(\frac{s}{2 \pi }+1\right) \, ds\tag{8}$$

here we have used

$$S\left(\frac{3}{2}\right)=-\frac{\partial }{\partial t}S\left(\frac{1}{2},t\right)|_{t\to 0}\tag{9}$$

and

$$\frac{\partial }{\partial t}\psi _{e^{-2 \pi }}^{(0)}\left(\frac{s+t}{2 \pi }+1\right)=\frac{1}{2 \pi } \psi _{e^{-2 \pi }}^{(1)}\left(\frac{s+t}{2 \pi }+1\right)\tag{10}$$

Here I am stuck, as I'm not sure if $(8)$ can be further simplified. The analytic properties should be deducible from $(7)$.

Answer 1

The following formula is due to Ramanujan: for $\text{Re}(s)>2$ and $\alpha,\beta>0$ with $\alpha \beta =4\pi^2$ we have $$ \alpha^{s/2} \left(\frac{\Gamma(s)\zeta(s)}{(2\pi)^s} + \cos \left(\frac{\pi s}{2} \right) \sum_{n=1}^{\infty} \frac{n^{s-1}}{e^{\alpha n}-1} \right) = \beta^{s/2} \left(\cos \left(\frac{\pi s}{2} \right) \frac{\Gamma(s)\zeta(s)}{(2\pi)^s} + \sum_{n=1}^{\infty} \frac{n^{s-1}}{e^{\beta n}-1} -\sin \left(\frac{\pi s}{2} \right) \, PV \int_0^{\infty} \frac{t^{s-1}}{e^{\beta t}-1}\cot \left(\frac{\beta t}{2} \right) \, dt \right) $$

See Theorem 9.1 of this paper (the condition $\text{Re}(s)>0$ is stated but this appears to be a mistake). Letting $\alpha=\beta=2\pi$ and rearranging gives an expression for the series you ask about in terms of a principal-value integral: $$ \sum_{n=1}^{\infty} \frac{n^{s-1}}{e^{2\pi n}-1} = \frac{\Gamma(s)\zeta(s)}{(2\pi)^s} + \cot\left(\frac{\pi s}{4} \right) \, PV \int_0^{\infty} \frac{t^{s-1}}{e^{2\pi t}-1} \cot(\pi t) \, dt, \hspace{1cm} \text{Re}(s)>2 \hspace{0.5cm} (*) $$ This is a natural extension of the well-known formula $$ \sum_{n=1}^{\infty} \frac{n^{4k+1}}{e^{2\pi n}-1} = \frac{B_{4k+2}}{2(4k+2)}, \hspace{1cm} k \geq 1. $$

EDIT: To answer your question in the title we need to extend the range of validity of $(*)$. Observe that for $\text{Re}(s)>2$ we have $$ PV \int_0^{\infty} \frac{t^{s-1}}{e^{2\pi t}-1} \cot(\pi t) \, dt = PV \int_0^{\infty} \frac{t^{s-1}}{e^{2\pi t}-1} \left(\cot(\pi t)-\frac{1}{\pi t} \right) dt + \frac{1}{\pi} \int_0^{\infty} \frac{t^{s-2}}{e^{2\pi t}-1} \, dt, $$ and it is well-known that $$ \int_0^{\infty} \frac{t^{s-2}}{e^{2\pi t}-1} \, dt = \frac{\Gamma(s-1)\zeta(s-1)}{(2\pi)^{s-1}} = \frac{\zeta(2-s)}{2\sin(\frac{\pi s}{2})}, \hspace{1cm} \text{Re}(s)>2, $$ the second expression following from the functional equation of the zeta function. Substituting into $(*)$ gives $$ \sum_{n=1}^{\infty} \frac{n^{s-1}}{e^{2\pi n}-1} = \frac{\Gamma(s)\zeta(s)}{(2\pi)^s} + \cot\left(\frac{\pi s}{4} \right) \, PV \int_0^{\infty} \frac{t^{s-1}}{e^{2\pi t}-1} \left(\cot(\pi t) -\frac{1}{\pi t} \right) dt + \frac{\zeta(2-s)}{4\pi \sin^2(\frac{\pi s}{4})}, $$ and the new principal value integral is now a convergent (and analytic) function for all $s$ with positive real part, so by analytic continuation the above holds for $\text{Re}(s)>0$. Special cases are $$ \sum_{n=1}^{\infty} \frac{n}{e^{2\pi n}-1} = \frac{1}{24}-\frac{1}{8\pi}, $$ $$ \sum_{n=1}^{\infty} \frac{\sqrt{n}}{e^{2\pi n}-1} = \frac{\zeta(\frac{3}{2})}{4\pi \sqrt{2}} + \frac{\zeta(\frac{1}{2})}{\pi(2+\sqrt{2})} + (\sqrt{2}-1)\; PV \int_0^{\infty} \frac{\sqrt{t}}{e^{2\pi t}-1} \left(\cot(\pi t) -\frac{1}{\pi t} \right) dt, $$ $$ \sum_{n=1}^{\infty} \frac{1}{e^{2\pi n}-1} = \frac{\gamma}{\pi} + PV \int_0^{\infty} \frac{1}{e^{2\pi t}-1} \left(\cot(\pi t) -\frac{1}{\pi t} \right) dt , $$ and $$ \sum_{n=1}^{\infty} \frac{n\log n}{e^{2\pi n}-1} = \frac{1}{24} - \frac{1}{2}\log A + \frac{\log(2\pi)}{8\pi} - \frac{\pi}{4} \; PV \int_0^{\infty} \frac{t}{e^{2\pi t}-1} \left(\cot(\pi t) -\frac{1}{\pi t} \right) dt $$ where $\gamma$ and $A$ are the Euler-Mascheroni and Glaisher-Kinkelin constants, respectively.

Answer 2

Not exactely a closed form.

Using $$\frac{1}{e^{2 \pi n}-1}=\sum_{k=0}^\infty e^{-2 \pi (k+1) n}$$ $$S=\sum_{n=1}^\infty\frac{n^a}{e^{2 \pi n}-1}= \sum_{k=0}^\infty \text{Li}_{-a}\left(e^{-2 \pi (k+1)}\right)$$

$$\frac{\text{Li}_{-a}\left(e^{-2 (k+2) \pi }\right)}{\text{Li}_{-a}\left(e^{-2 (k+1) \pi }\right)}\sim e^{-2 \pi }$$ make the summation a bit faster convergent.

Using, as in title $a=\frac 1 2$, computing the partial sums $$\left( \begin{array}{cc} p & \sum_{k=0}^p \text{Li}_{-\frac{1}{2}}\left(e^{-2 \pi (k+1)}\right) \\ 0 & 0.001872385883 \\ 1 & 0.001875873242 \\ 2 & 0.001875879755 \\ 3 & 0.001875879767 \\ \end{array} \right)$$

Using $$\text{Li}_{-\frac{1}{2}}\left(e^{-2 \pi (k+1)}\right)\sim e^{-2 \pi (k+1)}+\sqrt{2} e^{-4 \pi (k+1)}$$ a good approximation is

$$S\sim\sum _{k=0}^p \text{Li}_{-\frac{1}{2}}\left(e^{-2 \pi (k+1)}\right)+(1+A)\,\text{Li}_{-\frac{1}{2}}\left(e^{-2 \pi (p+2)}\right) $$ where $$A=\frac{e^{-2 \pi (p+1)}}{e^{2 \pi }-1}+\sqrt 2\,\frac{e^{-4 \pi (p+1)}}{e^{4 \pi }-1}$$

In practice, using $p=1$ or $p=2$ is more than sufficient.