How can we show that $\ln{2}=\sum_{n=1}^{\infty}{(-1)^{n-1}\over n}\left({12\over e^{n\pi}-1}+{4\over e^{n\pi}+1}\right)$

Question

$$\ln{2}=\sum_{n=1}^{\infty}{(-1)^{n-1}\over n}\left({12\over e^{n\pi}-1}+{4\over e^{n\pi}+1}\right)\tag1$$

Any hints?

Answer 1

We have that $${1\over e^{n\pi}-1}={e^{-n\pi}\over 1-e^{-n\pi}}=\sum_{k\geq 1}e^{-nk\pi}$$ and $${1\over e^{n\pi}+1}={e^{-n\pi}\over 1+e^{-n\pi}}=-\sum_{k\geq 1}(-1)^ke^{-nk\pi}.$$ Hence \begin{align*} \sum_{n=1}^{\infty}{(-1)^{n-1}\over n}&\left(12\sum_{k\geq 1}e^{-nk\pi}-4\sum_{k\geq 1}(-1)^ke^{-nk\pi}\right)\\ &=4\sum_{k\geq 1}\left(3-(-1)^k \right)\sum_{n=1}^{\infty}{(-1)^{n-1}\over n}e^{-nk\pi}\\ &=4\sum_{k\geq 1}\left(3-(-1)^k \right)\ln(1+e^{-k\pi})\\ &=8\sum_{k\geq 1}\ln(1+e^{-k\pi})+8\sum_{k\geq 1}\ln(1+e^{-(2k-1)\pi})\\ &=8\ln\left(\prod_{k\geq 1}(1+e^{-k\pi})(1+e^{-(2k-1)\pi})\right)\\ &=8\ln\left(\prod_{k\geq 1}\frac{(1+e^{-(2k-1)\pi})}{(1-e^{-(2k-1)\pi})}\right) +8\ln\left(\prod_{k\geq 1}(1+e^{-k\pi})(1-e^{-(2k-1)\pi})\right)\\ &=\ln(2) \end{align*} where in the last step we use Prove $\left(\frac{e^{\pi}+1}{e^{\pi}-1}\cdot\frac{e^{3\pi}+1}{e^{3\pi}-1}\cdot\frac{e^{5\pi}+1}{e^{5\pi}-1}\cdots\right)^8=2$ and $$\prod_{k\geq 1}(1+e^{-k\pi})(1-e^{-(2k-1)\pi})= \prod_{k\geq 1}\frac{(1-e^{-2k\pi})(1-e^{-(2k-1)\pi})}{(1-e^{-k\pi})}=1.$$

Answer 2

This is a result of theory of theta functions. Let $0 < q < 1$ and $$a(q) = \sum_{n = 1}^{\infty}\frac{q^{n}}{n(1 - q^{n})}$$ Next we can see that \begin{align} b(q) &= \sum_{n = 1}^{\infty}(-1)^{n - 1}\cdot\frac{q^{n}}{n(1 - q^{n})}\notag\\ &= \sum_{n \text{ odd}}\frac{q^{n}}{n(1 - q^{n})} - \sum_{n \text{ even}}\frac{q^{n}}{n(1 - q^{n})}\notag\\ &= \sum_{n = 1}^{\infty}\frac{q^{n}}{n(1 - q^{n})} - 2\sum_{n \text{ even}}\frac{q^{n}}{n(1 - q^{n})}\notag\\ &= a(q) - a(q^{2})\notag \end{align} Similarly we can prove that $$c(q) = \sum_{n = 1}^{\infty}\frac{q^{n}}{n(1 + q^{n})} = a(q) - 2a(q^{2})$$ and $$d(q) = \sum_{n = 1}^{\infty}(-1)^{n - 1}\frac{q^{n}}{n(1 + q^{n})} = c(q) - c(q^{2})$$ i.e $$d(q) = \sum_{n = 1}^{\infty}(-1)^{n - 1}\frac{q^{n}}{n(1 + q^{n})} = a(q) - 3a(q^{2}) + 2a(q^{4})$$ The sum in question is $$S = 12b(q) + 4d(q) = 16a(q) - 24a(q^{2}) + 8a(q^{4})\tag{1}$$ where $q = e^{-\pi}$. We can see from this post that \begin{align} a(q) &= -\frac{\log kk'^{4}}{12} - \frac{\log 2}{3} - \frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi K'}{24K}\tag{2}\\ a(q^{2}) &= -\frac{\log kk'}{6} - \frac{\log 2}{6} - \frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi K'}{12K}\tag{3}\\ a(q^{4}) &= -\frac{\log k^{4}k'}{12} + \frac{\log 2}{6} - \frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi K'}{6K}\tag{4} \end{align} Since $q = e^{-\pi}$ it follows from theory of theta functions and elliptic integrals that $K' = K$ and $k = k' = 1/\sqrt{2}$. Thus $$a(q) = -\frac{\log 2}{8} - \frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi}{24}$$ and $$a(q^{2}) = -\frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi}{12}$$ and $$a(q^{4}) = \frac{3\log 2}{8} - \frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi}{6}$$ and then from $(1)$ we get $S = \log 2$.

Answer 3

Suppose we seek to verify that

$$\log 2 = \sum_{n\ge 1} \frac{(-1)^{n+1}}{n} \left(\frac{12}{\exp(n\pi)-1} + \frac{4}{\exp(n\pi)+1}\right).$$

I present some functional equations which however do not suffice for a proof but may perhaps serve as a starting point. The reader is invited to supply the missing equations as conjectures for eventual proof. The computational details are available on request.

We make the following definitions:

$$S_1(x) = \sum_{n\ge 1} \frac{(-1)^{n+1}}{n} \frac{\exp(nx)}{\exp(2nx)-1}$$

and

$$S_2(x) = \sum_{n\ge 1} \frac{(-1)^{n+1}}{n} \frac{1}{\exp(2nx)-1}$$

with $x\ge 0.$ We are interested in

$$16S_1(\pi) + 8S_2(\pi) \quad\text{i.e.}\quad T(x) = 2S_1(x)+S_2(x).$$

We also use

$$S_3(x) = \sum_{n\ge 1} \frac{1}{n} \frac{\exp(nx)}{\exp(2nx)-1}$$

and

$$S_4(x) = \sum_{n\ge 1} \frac{1}{n} \frac{1}{\exp(nx)-1}$$ so that

$$S_3(x) = S_4(x)-S_4(2x).$$

We then obtain

$$\bbox[5px,border:2px solid #00A000] {S_1(x) = \frac{\pi^2}{24x} -\frac{x}{24} + S_1(\pi^2/x)}$$

and

$$\bbox[5px,border:2px solid #00A000] {S_3(x) = \frac{\pi^2}{12x} -\frac{1}{2}\log 2 + \frac{x}{24} - S_3(2\pi^2/x)}$$

and

$$\bbox[5px,border:2px solid #00A000] {S_4(x) = \frac{\pi^2}{6x} -\frac{1}{2}\log (2\pi) + \frac{1}{2}\log x - \frac{x}{24} + S_4(4\pi^2/x).}$$

Most importantly we get

$$\bbox[5px,border:2px solid #00A000] {T(x) = \frac{\pi^2}{8x} -\frac{1}{2}\log 2 + 2T(\pi^2/x) - 3S_3(\pi^2/x)}$$

which has the alternate form

$$\bbox[5px,border:2px solid #00A000] {T(x) = \frac{\pi^2}{8x} -\frac{1}{2}\log 2 + S_1(\pi^2/x) - S_3(2\pi^2/x).}$$