Calculating the limit of the following series I want to prove that $$\lim_{x\to1^-}(1-x)\sum_{n=1}^{\infty}(-1)^{n-1}\frac{nx^n}{1-x^{2n}} = \frac14$$
So far I tried to manipulate the series for instance using $$\sum_{n=1}^{\infty}(-1)^{n-1}\frac{nx^n}{1-x^{2n}} =  -\sum_{n=1}^{\infty}(-1)^{n}nx^n\sum_{m=0}^{\infty}\left(x^{2n}\right)^m$$ since $x < 1$. Interchanging the two sums (not sure if allowed) I obtained, assuming I did not make mistakes, the sum $$\sum_{m=0}^{\infty}\frac{x^{2m+1}}{(1+x^{2m+1})^2}$$
I am unable to continue from this point. Perhaps my work isn't actually useful at all. Can you help me?
 A: Your computation is very close to the answer. If we denote the sum by $S(x)$, then
$$ S(x)
= (1 - x) \sum_{m=0}^{\infty} \frac{x^{2m+1}}{(1+x^{2m+1})^2}
= \frac{1}{1+x} \sum_{m=0}^{\infty} \frac{x^{2m+1} - x^{2m+3}}{(1 + x^{2m+1})^2}.
$$
Now the idea is that $S(x)$ can be regarded as a Riemann sum for $\frac{1}{(1+t)^2}$ over $[0, 1]$. To make use of this idea, note that $t \mapsto \frac{1}{(1+t)^2}$ is decreasing for $t \geq 0$, and so
$$ x^2 \int_{x^{2m+1}}^{x^{2m-1}} \frac{\mathrm{d}t}{(1+t)^2}
\leq \frac{x^{2m+1} - x^{2m+3}}{(1 + x^{2m+1})^2}
\leq \int_{x^{2m+3}}^{x^{2m+1}} \frac{\mathrm{d}t}{(1+t)^2}. $$
Summing this over $m = 0, 1, 2, \dots$, we obtain
$$ \frac{x^2}{1+x} \int_{0}^{1/x} \frac{\mathrm{d}t}{(1+t)^2} \leq S(x) \leq \frac{1}{1+x} \int_{0}^{x} \frac{\mathrm{d}t}{(1+t)^2}. $$
Therefore, letting $x \to 1^-$ yields
$$ \lim_{x \to 1^-} S(x) = \frac{1}{2} \int_{0}^{1} \frac{\mathrm{d}t}{(1+t)^2} = \frac{1}{4}. $$
A: The sum in question belongs more properly to the theory of theta functions and elliptic integrals and this is an approach which makes use of standard results from this theory.

Let's put $q=-x$ so that the expression under limit becomes $$-(1+q)\sum_{n=1}^{\infty} \frac{nq^n} {1-q^{2n}}$$ The sum above can be written as $$\sum_{n\geq 1}\left(\frac{nq^{n}}{1-q^{n}}-\frac{nq^{2n}}{1-q^{2n}}\right)$$ which in terms of Ramanujan function $$P(q)=1-24\sum_{n\geq 1}\frac{nq^n}{1-q^n} $$ becomes $$\frac{P(q^2)-P(q)}{24}$$ and it follows that the original expression under limit is $$(1-x)\cdot\frac{P(-x)-P(x^2)}{24}$$ Let's replace this variable $x$ again by $q$ (for convention)  and the limit we seek is $$\lim_{q\to 1^-}(1-q)\cdot\frac{P(-q)-P(q^2)} {24}$$ Treating $q$ as the nome we use the following standard results
\begin{align} 
q&=\exp\left(-\pi\frac{K'} {K} \right)\notag\\
P(-q) &=\left(\frac{2K}{\pi}\right)^2\left(\frac{6E}{K}+4k^2-5\right)\notag\\
P(q^2) &=\left(\frac{2K}{\pi}\right)^2\left(\frac{3E}{K}+k^2-2\right)\notag
\end{align}
(for proofs see this post) and the expression under limit equals $$(1-e^{-\pi K'/K}) \frac{1}{8}\left(\frac{2K}{\pi}\right)^2\left(\frac{E} {K} - k'^2\right)$$ where moduli $k, k'$ and elliptic integrals $K, K'$ correspond to nome $q$. As $q\to 1^-$ we have $k\to 1^-,k'\to 0^+$ and $K\to\infty, K'\to\pi/2,E\to 1$ and the desired limit equals the limit of $$\frac{\pi K'} {K} \cdot\frac{K} {2\pi^2}\cdot(E-k'^2K)$$ This works out to be $1/4$ if we can show that $k'^2K\to 0$. This is an easy consequence of the asymptotic $$K=\log(4/k')+o(1)$$ as $k\to 1^-$ (for proof see this post).
