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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 $\operatorname{Li}_3\left(\frac12\right)=\frac78\zeta(3)-\frac12\ln2\zeta(2)+\frac16\ln^32$

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3 Answers 3

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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: $$\small \int_0^1 \frac{[m\ln(1+x)+n\ln(1-x)][q\ln(1+x)+p\ln(1-x)]}{x}dx=\left(\frac{mq}{4}-\frac{5}{8}(mp+nq)+2np\right)\zeta(3)$$

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    $\begingroup$ nice work Zacky. I missed that beautiful sub $y=\ln x$. I wonder why you changed your name. $\endgroup$ Jul 27, 2019 at 22:18
  • $\begingroup$ Thank you! Indeed, that substitution does some great job.// Well, I got bored :D $\endgroup$
    – Zacky
    Jul 27, 2019 at 22:33
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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 $$x=4\implies S=\pi ^2 \log (2)-\frac{7 \zeta (3)}{2}$$ $$x=2\implies S=\pi C-\frac{35 \zeta (3)}{16}+\frac{1}{8} \pi ^2 \log (2)$$ $$x=-1\implies S=-\frac{2 \zeta (3)}{5}$$

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  • $\begingroup$ its interesting to see different solutions. thanks for the efforts. $\endgroup$ Jul 27, 2019 at 22:20
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I just came across this question (a little late) and I noted one approach that was not presented, I thought it was worthy to share (If you dislike it I can just delete). It will be a little long because it´s a systematic approach to deal with this type of Series.

First, we need two preliminary results:

First preliminary result :

$$\ln\left(2 \sinh\left(\frac{x}{2}\right)\right)=\frac{x}{2}-\sum_{k=1}^{\infty}\frac{e^{-kx}}{k} \tag{1}$$

Proof:

$$ \begin{aligned} \ln\left( \sinh(x)\right)&=\ln\left( \frac{1}{2} \left( e^{x}-e^{-x} \right)\right)\\ &=-\ln 2+\ln\left( e^{x}-e^{-x} \right)\\ &=-\ln 2+\ln\left( \frac{e^{-x}}{e^{-x}} \left( e^{x}-e^{-x} \right)\right)\\ &=-\ln 2+x+\ln\left( 1-e^{-2x} \right)\\ &=-\ln 2+x-\sum_{n=1}^{\infty}\frac{e^{-2nx}}{n} \qquad \blacksquare \end{aligned} $$

Letting $x \to \frac{x}{2}$ completes the proof


Second preliminary result:

We have

$$\frac{\operatorname{arcsinh}\left(\frac{x}{2}\right)}{\sqrt{1+\left(\frac{x}{2}\right)^2}}=\sum_{n=1}^\infty \frac{(-1)^{n-1}x^{2n-1}}{\binom{2n}{n}n} \tag{2}$$

Letting $x \to \sqrt{a}x$ we obtain

$$\frac{\sqrt{a}\operatorname{arcsinh}\left(\frac{\sqrt{a} x}{2}\right)}{\sqrt{1+\left(\frac{ \sqrt{a} x}{2}\right)^2}}=\sum_{n=1}^\infty \frac{(-1)^{n-1}a^nx^{2n-1}}{\binom{2n}{n}n} \tag{3}$$


Claim:

$$\sum_{n=1}^\infty\frac{ (-1)^{n-1}a^n}{\binom{2n}{n}n^k}=\frac{(-2)^{k-2}}{(k-2)!}\int_0^{2\operatorname{arcsinh}\left(\frac{\sqrt{a} }{2} \right)}x\ln^{k-2}\left(\frac{2}{\sqrt{a}}\sinh\left(\frac{x}{2}\right) \right)\,dx \tag{4}$$

Proof:

$$ \begin{aligned} \sum_{n=1}^\infty\frac{ (-1)^{n-1}a^n}{\binom{2n}{n}n^k}&=\frac{(-1)^{k-1}a^n}{(k-1)!}\sum_{n=1}^\infty\frac{ (-1)^{n-1}}{\binom{2n}{n}}\int_0^1 \ln^{k-1}(x) x^{n-1}\,dx\\ &=\frac{2(-1)^{k-1}}{(k-1)!}\sum_{n=1}^\infty\frac{ (-1)^{n-1}a^n}{\binom{2n}{n}}\int_0^1 \ln^{k-1}\left(x^2\right) x^{2n-1}\,dx & \left(x \to x^2\right)\\ &=\frac{2(-2)^{k-1}}{(k-1)!}\sum_{n=1}^\infty\frac{ (-1)^{n-1}a^n}{\binom{2n}{n}}\int_0^1 \ln^{k-1}\left(x\right) x^{2n-1}\,dx \\ &=\frac{2(-2)^{k-1}}{(k-1)!}\sum_{n=1}^\infty\frac{ (-1)^{n-1}a^n}{\binom{2n}{n}}\left(\frac{x^{2n}\ln^{k-1}(x)}{2n}\Bigg|_0^1-\frac{(k-1)}{2n}\int_0^1 \ln^{k-2}\left(x\right) x^{2n-1}\,dx \right)\\ &=-\frac{(-2)^{k-1}}{(k-2)!}\sum_{n=1}^\infty\frac{ (-1)^{n-1}a^n}{\binom{2n}{n}n}\int_0^1 \ln^{k-2}\left(x\right) x^{2n-1}\,dx \\ &=-\frac{(-2)^{k-1}}{(k-2)!}\int_0^1 \ln^{k-2}\left(x\right)\left(\sum_{n=1}^\infty\frac{(-1)^{n-1} a^n x^{2n-1}}{\binom{2n}{n}n} \right) \,dx \\ &=-\frac{(-2)^{k-1}}{(k-2)!}\int_0^1 \ln^{k-2}\left(x\right)\left(\frac{\sqrt{a} \operatorname{arcsinh}\left(\frac{\sqrt{a} x}{2} \right)}{\sqrt{1+\left( \frac{\sqrt{a} x}{2}\right)^2}} \right) \,dx & \left( \text{by eq. (3)}\right)\\ &=\frac{(-2)^{k-2}a}{(k-2)!}\int_0^{\frac{2}{\sqrt{a}}\operatorname{arcsinh}\left(\frac{\sqrt{a} }{2} \right)} \frac{x\ln^{k-2}\left(\frac{2}{\sqrt{a}}\sinh\left(\frac{\sqrt{a}x}{2}\right)\right) \cosh\left(\frac{\sqrt{a}x}{2} \right)}{\sqrt{1-\sin^2\left( \frac{\sqrt{a}x}{2}\right)}} \,dx & \left( \frac{\sqrt{a}x}{2} \to \sinh\left(\frac{\sqrt{a} x}{2} \right)\right)\\ &=\frac{(-2)^{k-2}}{(k-2)!}\int_0^{2\operatorname{arcsinh}\left(\frac{\sqrt{a} }{2} \right)}x\ln^{k-2}\left(\frac{2}{\sqrt{a}}\sinh\left(\frac{x}{2}\right) \right)\,dx & \left( \sqrt{a}x \to x\right)\\ \end{aligned} $$

Letting $a=\frac12$ and $k=3$ in $(4)$, we obtain

$$ \begin{aligned} \sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^32^n {2n\choose n}}&=-2\int_0^{\ln(2)}x\ln\left(2\sqrt{2}\sinh\left(\frac{x}{2}\right) \right)\,dx\\ &=-\ln(2)\int_0^{\ln(2)}x\,dx-2\int_0^{\ln(2)}x\ln\left(2\sinh\left(\frac{x}{2}\right) \right)\,dx\\ &=-\frac{\ln^3(2)}{2}-2\int_0^{\ln(2)}x \left(\frac{x}{2}-\sum_{k=1}^{\infty}\frac{e^{-kx}}{k}\right)\,dx\\ &=-\frac{\ln^3(2)}{2}-\frac{\ln^3(2)}{3}+2\sum_{k=1}^{\infty}\frac{1}{k}\left(-\frac{x e^{-kx}}{k}\Bigg|_0^{\ln(2)} +\frac{1}{k}\int_0^{\ln(2)}e^{-kx}\,dx\right)\\ &=-\frac{\ln^3(2)}{2}-\frac{\ln^3(2)}{3}-2\ln(2)\operatorname{Li}_2\left( \frac12\right)-2\sum_{k=1}^{\infty}\frac{1}{k^3}\left(e^{-kx}\Bigg|_0^{\ln(2)} \right)\\ &=-\frac{\ln^3(2)}{2}-\frac{\ln^3(2)}{3}-2\ln(2)\operatorname{Li}_2\left( \frac12\right)-2\operatorname{Li}_3\left( \frac12\right)+2\zeta(3)\\ &=-\frac{\ln^3(2)}{2}-\frac{\ln^3(2)}{3}-\frac{\pi^6}{6}\ln(2)+\ln^3(2)-\frac74\zeta(3)+\frac{\pi^6}{6}\ln(2)-\frac{\ln^3(2)}{3}+2\zeta(3)\\ &=\frac{\zeta(3)}{4}-\frac{\ln^3(2)}{6} \qquad \blacksquare \end{aligned} $$

Where we used

$\operatorname{Li}_2\left(\frac12\right)=\frac12\zeta(2)-\frac12\ln^22$ and

$\operatorname{Li}_3\left(\frac12\right)=\frac78\zeta(3)-\frac12\ln2\zeta(2)+\frac16\ln^32$

To show that this approach is systematic, as a Bonus, setting $a=1$ and $k=3$ in $(4)$ we obtain

$$ \begin{aligned} \sum_{n=1}^\infty\frac{ (-1)^{n-1}}{\binom{2n}{n}n^3}&=-2\int_0^{2\ln(\phi)}x\ln\left(2 \sinh\left( \frac{x}{2}\right) \right)\,dx \\ &=-2\int_0^{2\ln(\phi)}x\left(\frac{x}{2}-\sum_{k=1}^{\infty}\frac{e^{-kx}}{k} \right)\,dx & \left( \text{by eq. (1)}\right)\\ &=-\int_0^{2\ln(\phi)}x^2\,dx+2\sum_{k=1}^{\infty}\frac{1}{k}\int_0^{2\ln(\phi)} xe^{-kx}\,dx\\ &=-\frac{8}{3}\ln^3(\phi)+2\sum_{k=1}^{\infty}\frac{1}{k}\left(-\frac{2\ln(\phi)\phi^{-2k}}{k}+\frac{1}{k}\int_0^{2\ln(\phi)} e^{-kx}\,dx \right)\\ &=-\frac{8}{3}\ln^3(\phi)-4\ln(\phi)\sum_{k=1}^{\infty}\frac{(\phi^{-2})^k}{k^2}+2\sum_{k=1}^{\infty}\frac{1}{k}\left(\frac{1}{k^2}-\frac{(\phi^{-2})^k}{k^2}\right)\\ &=-\frac{8}{3}\ln^3(\phi)-4\ln(\phi)\operatorname{Li}_2(\phi^{-2})+2\zeta(3)-2\sum_{k=1}^{\infty}\frac{(\phi^{-2})^k}{k^3}\\ &=-\frac{8}{3}\ln^3(\phi)-4\ln(\phi)\operatorname{Li}_2(\phi^{-2})-2\operatorname{Li}_3(\phi^{-2})+2\zeta(3)\\ &=-\frac{8}{3}\ln^3(\phi)-4\ln(\phi)\left( \frac{\pi^{2}}{15}-\ln ^{2} \phi\right)-2\left(\frac45\zeta(3)+\frac{2\ln ^{3}(\phi)}{3}-\frac{2\pi^{2} \ln (\phi)}{15} \right)+2\zeta(3)\\ &=-\frac{8}{3}\ln^3(\phi)+4\ln^3(\phi)-\frac{4}{3}\ln^3(\phi)-\frac85\zeta(3)+2\zeta(3)\\ &=\frac25\zeta(3) \qquad \blacksquare \end{aligned} $$

Which is the representation of $\zeta(3)$ that Apery used to prove the irrationality of $\zeta(3)$.

Note that we used

$\mathrm{Li}_{2}\left(\frac{1}{\phi^{2}}\right) =\frac{\pi^{2}}{15}-\ln ^{2} \phi$

$\operatorname{Li}_{3}\left(\frac{1}{\phi^{2}}\right)=\frac45\zeta(3)+\frac{2\ln ^{3}(\phi)}{3}-\frac{2\pi^{2} \ln (\phi)}{15}$

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