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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?

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2 Answers 2

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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

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

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  • $\begingroup$ Wonderful explanation! As always! $\endgroup$
    – Peanut
    Nov 23, 2020 at 16:31
  • $\begingroup$ Simplicity at its best (+1)! The sum in question belongs to the theory of theta as discussed in my answer. $\endgroup$
    – Paramanand Singh
    Nov 24, 2020 at 7:47
  • $\begingroup$ @ParamanandSingh, Thank you! I am aware that this kind of series opens up a door to theta functions, Lambert series, etc. But I am by no means an expert to those topics, so I always resort to some elementary means. $\endgroup$ Nov 24, 2020 at 7:50
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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).

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  • $\begingroup$ Can you please suggest some reference books for this topic? $\endgroup$
    – PNDas
    Nov 24, 2020 at 7:57
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    $\begingroup$ @PNDas: see the last part of one of my answers which lists many references. It also has a reference to my blog where you can search for theta, elliptic, Ramanujan, Jacobi, Gauss, AGM. $\endgroup$
    – Paramanand Singh
    Nov 24, 2020 at 8:21
  • $\begingroup$ @PNDas: btw this is a difficult topic and requires a hell lot of patience to understand coherently. My blog was a result of my efforts to understand this topic. $\endgroup$
    – Paramanand Singh
    Nov 24, 2020 at 8:27
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    $\begingroup$ @PNDas: thanks for your kind words. I am an engineering graduate (btech in computer science) and work in IT. So I don't have dedicated math education (like MSc, PhD). But I have great interest in mathematics and have learnt some stuff from various online sources (this website being my favorite). $\endgroup$
    – Paramanand Singh
    Nov 24, 2020 at 9:34
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    $\begingroup$ @PNDas: wish you best of luck for your higher studies! $\endgroup$
    – Paramanand Singh
    Nov 24, 2020 at 9:35

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