Is there a closed form for series of $\sum_k \frac{x^k}{1-x^{2k+1}}$

Question

is there a way to evaluate $\sum_{k=0}^{+\infty} \frac{x^k}{1-x^{2k+1}}$ in terms of popular functions or even in terms of the q-digamma function?

$0<x<1$

I tried to write the denominator as an geometric infinite series. Expression got cuter but didn't get easier. I can't see any integration or trick to work this one.

Wolfram gives me the partial sum in terms of the x-digamma function but it uses complex values in the digamma which makes hard for me to take the limit as I know nothing about this function.

So I gave it a better search in this forum and found that:

$\sum_{k=0}^{\infty} \frac{x^k}{1-x^{2k}} = L(x) - L(x^2)$

where $L(x) = \sum_{k=0}^{\infty} \frac{x^k}{1-x^{k}}$ is the Lambert series of $x$

Answer 1

Write $$H(x):=\sum_{k=0}^{+\infty} \frac{x^k}{1-x^{2k+1}},$$ and $$G(x):=xH(x^2)=\sum_{k=0}^{+\infty} \frac{x^{2k+1}}{1-x^{4k+2}}.$$

One can then calculate the Taylor expansion of $G$:

\begin{eqnarray*} G(x) &=& \sum_{k = 0}^{+\infty} \sum_{j = 1\\2\nmid j}^{+\infty}x^{j(2k+1)}\\ &=& \sum_{k = 1\\2\nmid k}^{+\infty}\sum_{j = 1\\2\nmid j}^{+\infty}x^{jk}\\ &=& \sum_{n = 1\\2\nmid n}^{+\infty} \sigma_0(n) x^n, \end{eqnarray*} where $\sigma_0(n)$ is the number of divisors of $n$.

Answer 2

This is not an answer.

The problem is that morever the functions involved in the expression are extremely expansive to compute.

Just an idea : consider $$\int_0^\infty \frac{x^k}{1-x^{2k+1}}\,dk=-\frac{\tanh ^{-1}\left(\sqrt{x}\right)}{\sqrt{x} \log (x)}\qquad \text{if} \qquad \Re(\log (x))<0$$

The summation and the integral are highly correlated. Computing the partial sum for $p=10^4$, some results $$\left( \begin{array}{ccc} x & \text{integral} &\text{summation} \\ 0.05 & 0.33955 & 1.10527 \\ 0.10 & 0.44971 & 1.22232 \\ 0.15 & 0.55613 & 1.35345 \\ 0.20 & 0.66857 & 1.50163 \\ 0.25 & 0.79248 & 1.67070 \\ 0.30 & 0.93279 & 1.86569 \\ 0.35 & 1.09508 & 2.09328 \\ 0.40 & 1.28642 & 2.36245 \\ 0.45 & 1.51634 & 2.68566 \\ 0.50 & 1.79825 & 3.08062 \\ 0.55 & 2.15187 & 3.57332 \\ 0.60 & 2.60743 & 4.20356 \\ 0.65 & 3.21372 & 5.03517 \\ 0.70 & 4.05453 & 6.17713 \\ 0.75 & 5.28602 & 7.83090 \\ 0.80 & 7.23316 & 10.4122 \\ 0.85 & 10.6911 & 14.9279 \\ 0.90 & 18.1929 & 24.5461 \\ 0.95 & 43.5702 & 56.2745 \end{array} \right)$$

Put them on a graph. Using a quick and dirty linear regression based on much more data points gives $(R^2=0.999683)$ $$\text{summation}=1.21847\,\, \text{integral}+1.14809$$ $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & 1.14809 & 0.07802 & \{0.99325,1.30293\} \\ b & 1.21847 & 0.00220 & \{1.21409,1.22285\} \end{array}$$

Answer 3

A closed form might not be available, but a Taylor series certainly is. Let's begin with the partial fraction decomposition$$\frac{x^k}{1-x^{2k+1}}=\sum_{j=0}^{2k}\frac{\exp\frac{2\pi j(k-1)i}{2k+1}}{\prod_{l\ne j}\left(\exp\frac{2\pi li}{2k+1}-\exp\frac{2\pi ji}{2k+1}\right)}\frac{1}{1-\exp\frac{-2\pi ji}{2k+1}\cdot x},$$where $l$ runs from $0$ to $2k$ except at $j$. Let $[x^n]f(x)$ denote the coefficient of $x^n$ in $f(x)$ so$$[x^n]\frac{x^k}{1-x^{2k+1}}=\sum_{j=0}^{2k}\frac{\exp\frac{2\pi j(k-n-1)i}{2k+1}}{\prod_{l\ne j}\left(\exp\frac{2\pi li}{2k+1}-\exp\frac{2\pi ji}{2k+1}\right)}.$$Thus$$\sum_k\frac{x^k}{1-x^{2k+1}}=\sum_{n\ge0}\left(\sum_k\sum_{j=0}^{2k}\frac{\exp\frac{2\pi j(k-n-1)i}{2k+1}}{\prod_{l\ne j}\left(\exp\frac{2\pi li}{2k+1}-\exp\frac{2\pi ji}{2k+1}\right)}\right)x^n.$$The denominator can be rewritten using $e^{ia}-e^{ib}=2ie^{i(a+b)/2}\sin\frac{a-b}{2}$, so$$\prod_{l\ne j}\left(\exp\frac{2\pi li}{2k+1}-\exp\frac{2\pi ji}{2k+1}\right)=(-1)^j4^k\exp\frac{-\pi j}{2k+1}\prod_{l\ne j}\sin\frac{\pi(l-j)}{2k+1}.$$Hence$$\sum_k\frac{x^k}{1-x^{2k+1}}=\sum_{n\ge0}\left(\sum_k\frac{1}{4^k}\sum_{j=0}^{2k}\frac{\exp\frac{-2\pi j(n+1)i}{2k+1}}{\prod_{l\ne j}\sin\frac{\pi(l-j)}{2k+1}}\right)x^n.$$But I'm not sure we can simplify it much more than that.