# Prove $\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^32^k {2k\choose k}}=\frac1{4}\zeta(3)-\frac1{6}\ln^32$

How to prove that $$\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^32^k {2k\choose k}}=\frac1{4}\zeta(3)-\frac1{6}\ln^32?$$

A friend posted this nice problem on my FB group and I managed to evaluate it using the $$\arcsin^2 x$$ identity. I would like to see different approaches. Thanks.

My solution: Using the following identity: (see here) $$\arcsin^2z=\frac12\sum_{k=1}^\infty\frac{(2z)^{2k}}{k^2{2k \choose k}}$$

Set $$\ z=\sqrt{\frac{x}{8}}$$ then divide both sides by $$x$$ and integrate from $$x=0$$ to $$-1$$, to get \begin{align} S&=\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^32^k {2k\choose k}}=-2\underbrace{\int_0^{-1}\frac{\arcsin^2\left(\sqrt{\frac x8}\right)}{x}\ dx}_{\large\arcsin\left(\sqrt{\frac x8}\right)=y}\\ &=-4\int_0^{\frac{\ln2}{2}i} y^2\cot y\ dy\overset{y=ix}{=}4\int_0^{\frac{\ln2}{2}} x^2\coth x\ dx \end{align} Lets find the antiderivative of the integral: \begin{align} I&=\int x^2\coth x\ dx\overset{IBP}{=}x^2\ln(\text{arcsinh}(x))-2\int x\ln(\text{arcsinh}(x))\ dx\\ &=x^2\ln(\text{arcsinh}(x))-2\int x\left\{x-\ln2-\ln(1-e^{-2x})\right\}\ dx\\ &=x^2\ln(\text{arcsinh}(x))-\frac23x^3+\ln2\ x^2-2\int x\ln(1-e^{-2x})\ dx\\ &=x^2\ln(\text{arcsinh}(x))-\frac23x^3+\ln2\ x^2+2\sum_{n=1}^\infty\frac1n\int xe^{-2nx}\ dx\\ &=x^2\ln(\text{arcsinh}(x))-\frac23x^3+\ln2\ x^2+2\sum_{n=1}^\infty\frac1n\left(-\frac{e^{-2nx}}{4n^2}-\frac{xe^{-2nx}}{2n}\right)\\ &=x^2\ln(\text{arcsinh}(x))-\frac23x^3+\ln2\ x^2-\frac12\sum_{n=1}^\infty\frac{(e^{-2x})^n}{n^3}-x\sum_{n=1}^\infty\frac{(e^{-2x})^n}{n^2}\\ &=x^2\left\{\ln x-\ln2-\ln(1-e^{-2x})\right\}-\frac23x^3+\ln2\ x^2-\frac12\operatorname{Li}_3(e^{-2x})-x\operatorname{Li}_2(e^{-2x})\\ &=\frac{x^3}{3}+x^2\ln(1-e^{-2x})-\frac12\operatorname{Li}_3(e^{-2x})-x\operatorname{Li}_2(e^{-2x})\\ \end{align}

Thus \begin{align} S&=4\left[\frac{x^3}{3}+x^2\ln(1-e^{-2x})-\frac12\operatorname{Li}_3(e^{-2x})-x\operatorname{Li}_2(e^{-2x})\right]_0^{\frac{\ln2}{2}}\\ &=4\left[\frac12\zeta(3)-\frac5{24}\ln^32-\frac12\operatorname{Li}_3\left(\frac12\right)-\frac{\ln2}{2}\operatorname{Li}_2\left(\frac12\right)\right]\\ &=4\left[\frac1{16}\zeta(3)-\frac1{24}\ln^32\right]\\ &\boxed{=\frac1{4}\zeta(3)-\frac1{6}\ln^32} \end{align}

Note that we used $$\operatorname{Li}_2\left(\frac12\right)=\frac12\zeta(2)-\frac12\ln^22$$ and $$\operatorname{Li}_3\left(\frac12\right)=\frac78\zeta(3)-\frac12\ln2\zeta(2)+\frac16\ln^32$$

One possible way is to use $$S=\sum_{k=1}^\infty \frac{x^k}{k^3 \binom{2 k}{k}}=\frac{x}{2} \, _4F_3\left(1,1,1,1;\frac{3}{2},2,2;\frac{x}{4}\right)$$

which is

$$S=2 \text{Li}_3\left(-\frac{x}{2}-\frac{1}{2} i \sqrt{(4-x) x}+1\right)+4 i \text{Li}_2\left(-\frac{x}{2}-\frac{1}{2} i \sqrt{(4-x) x}+1\right) \csc ^{-1}\left(\frac{2}{\sqrt{x}}\right)+\frac{4}{3} i \csc ^{-1}\left(\frac{2}{\sqrt{x}}\right)^3+4 \log \left(\frac{1}{2} \left(x+i \sqrt{(4-x) x}\right)\right) \csc ^{-1}\left(\frac{2}{\sqrt{x}}\right)^2-2 \zeta (3)$$ Computing for $$x=-\frac 12$$, this leads before any simplification to $$2 \text{Li}_3(2)-4 \text{Li}_2(2) \sinh ^{-1}\left(\frac{1}{2 \sqrt{2}}\right)-2 \zeta (3)-4 i \pi \sinh ^{-1}\left(\frac{1}{2 \sqrt{2}}\right)^2+\frac{4}{3} \sinh ^{-1}\left(\frac{1}{2 \sqrt{2}}\right)^3$$ which simplifies to $$\frac{\log ^3(2)}{6}-\frac{\zeta (3)}{4}$$

I found a few other $$x=4\implies S=\pi ^2 \log (2)-\frac{7 \zeta (3)}{2}$$ $$x=2\implies S=\pi C-\frac{35 \zeta (3)}{16}+\frac{1}{8} \pi ^2 \log (2)$$ $$x=-1\implies S=-\frac{2 \zeta (3)}{5}$$

• its interesting to see different solutions. thanks for the efforts. – Ali Shadhar Jul 27 '19 at 22:20

Similarly, one may obtain the following equality, used by Apery to prove the irrationality of $$\zeta(3)$$: $$\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^3 \binom{2n}{n}}=\frac25\sum_{n=1}^\infty \frac{1}{n^3}$$

Using $$\arcsin^2 \sqrt{-z}=-\operatorname{arcsinh}^2z$$ we get:$$S=\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^32^k {2k\choose k}}=-2\int_0^{-1}\frac{\arcsin^2\left(\sqrt{\frac x8}\right)}{x} dx\overset{x=-t}=2\int_0^1 \frac{\operatorname{arcsinh}^2\left(\sqrt{\frac t8}\right)}{t}dt$$ Furthermore, we let $$\operatorname{arcsinh}\sqrt{\frac t8}=y$$, which yields $$S=4\int_0^{\ln{\sqrt 2}} y^2 \coth y dy\overset{y=\ln x}=4\int_1^{\sqrt 2}\ln^2 x\ \frac{x^2+1}{x^2-1}\frac{dx}{x}$$$$=4\int_1^{\sqrt 2} \frac{(2x)\ln^2 x}{x^2-1}dx-4\int_1^{\sqrt 2}\frac{\ln^2 x}{x}dx\overset{x^2=t}=\int_1^2 \frac{\ln^2 t}{t-1}dt-\frac{\ln^3 2}{6}$$ $$\overset{t-1=x}=\int_0^1 \frac{\ln^2(1+x)}{x}dx-\frac{\ln^3 2}{6}=\boxed{\frac{\zeta(3)}{4}-\frac{\ln^3 2}{6}}$$ See here for the last integral, or just let $$m=1,n=0,q=1,p=0$$ in the following relation: $$\small \int_0^1 \frac{[m\ln(1+x)+n\ln(1-x)][q\ln(1+x)+p\ln(1-x)]}{x}dx=\left(\frac{mq}{4}-\frac{5}{8}(mp+nq)+2np\right)\zeta(3)$$

• nice work Zacky. I missed that beautiful sub $y=\ln x$. I wonder why you changed your name. – Ali Shadhar Jul 27 '19 at 22:18
• Thank you! Indeed, that substitution does some great job.// Well, I got bored :D – Zacky Jul 27 '19 at 22:33
• haha i agree.. change is always good. – Ali Shadhar Jul 27 '19 at 22:34