Calculating the limit of the following series

Question

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?

Answer 1

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

$$ \lim_{x \to 1^-} S(x) = \frac{1}{2} \int_{0}^{1} \frac{\mathrm{d}t}{(1+t)^2} = \frac{1}{4}. $$

Answer 2

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.

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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).