Apery's original proof that $$ \zeta(3) \equiv \sum_1^\infty \frac1{n^3} $$ is irrational starts from an alternating series $$ \zeta(3) = \frac52\sum_1^\infty\frac{(-1)^{n-1}}{n^3\binom{2n}{n}} $$ There must be a way to see that those two series are equal, but I have tried various telescoping and other techniques and I just can't see it. Help me out:
Show (preferably by reasonably elementary means) that$$\sum_1^\infty \frac1{n^3}=\frac52\sum_1^\infty\frac{(-1)^{n-1}}{n^3\binom{2n}{n}}$$
On
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:
$$ \begin{aligned} \sum_{n=1}^\infty\frac{ (-1)^{n-1}a^n}{\binom{2n}{n}n^k}&=\frac{(-1)^{k-1}a^n}{(k-1)!}\sum_{n=1}^\infty\frac{ (-1)^{n-1}}{\binom{2n}{n}}\int_0^1 \ln^{k-1}(x) x^{n-1}\,dx\\ &=\frac{2(-1)^{k-1}}{(k-1)!}\sum_{n=1}^\infty\frac{ (-1)^{n-1}a^n}{\binom{2n}{n}}\int_0^1 \ln^{k-1}\left(x^2\right) x^{2n-1}\,dx & \left(x \to x^2\right)\\ &=\frac{2(-2)^{k-1}}{(k-1)!}\sum_{n=1}^\infty\frac{ (-1)^{n-1}a^n}{\binom{2n}{n}}\int_0^1 \ln^{k-1}\left(x\right) x^{2n-1}\,dx \\ &=\frac{2(-2)^{k-1}}{(k-1)!}\sum_{n=1}^\infty\frac{ (-1)^{n-1}a^n}{\binom{2n}{n}}\left(\frac{x^{2n}\ln^{k-1}(x)}{2n}\Bigg|_0^1-\frac{(k-1)}{2n}\int_0^1 \ln^{k-2}\left(x\right) x^{2n-1}\,dx \right)\\ &=-\frac{(-2)^{k-1}}{(k-2)!}\sum_{n=1}^\infty\frac{ (-1)^{n-1}a^n}{\binom{2n}{n}n}\int_0^1 \ln^{k-2}\left(x\right) x^{2n-1}\,dx \\ &=-\frac{(-2)^{k-1}}{(k-2)!}\int_0^1 \ln^{k-2}\left(x\right)\left(\sum_{n=1}^\infty\frac{(-1)^{n-1} a^n x^{2n-1}}{\binom{2n}{n}n} \right) \,dx \\ &=-\frac{(-2)^{k-1}}{(k-2)!}\int_0^1 \ln^{k-2}\left(x\right)\left(\frac{\sqrt{a} \operatorname{arcsinh}\left(\frac{\sqrt{a} x}{2} \right)}{\sqrt{1+\left( \frac{\sqrt{a} x}{2}\right)^2}} \right) \,dx & \left( \text{by eq. (3)}\right)\\ &=\frac{(-2)^{k-2}a}{(k-2)!}\int_0^{\frac{2}{\sqrt{a}}\operatorname{arcsinh}\left(\frac{\sqrt{a} }{2} \right)} \frac{x\ln^{k-2}\left(\frac{2}{\sqrt{a}}\sinh\left(\frac{\sqrt{a}x}{2}\right)\right) \cosh\left(\frac{\sqrt{a}x}{2} \right)}{\sqrt{1-\sin^2\left( \frac{\sqrt{a}x}{2}\right)}} \,dx & \left( \frac{\sqrt{a}x}{2} \to \sinh\left(\frac{\sqrt{a} x}{2} \right)\right)\\ &=\frac{(-2)^{k-2}}{(k-2)!}\int_0^{2\operatorname{arcsinh}\left(\frac{\sqrt{a} }{2} \right)}x\ln^{k-2}\left(\frac{2}{\sqrt{a}}\sinh\left(\frac{x}{2}\right) \right)\,dx & \left( \sqrt{a}x \to x\right)\\ \end{aligned} $$
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
$\mathrm{Li}_{2}\left(\frac{1}{\phi^{2}}\right) =\frac{\pi^{2}}{15}-\ln ^{2} \phi$
$\operatorname{Li}_{3}\left(\frac{1}{\phi^{2}}\right)=\frac45\zeta(3)+\frac{2\ln ^{3}(\phi)}{3}-\frac{2\pi^{2} \ln (\phi)}{15}$
We may notice that: $$\frac{1}{n^3}=\frac{1}{(n-1)n(n+1)}+\frac{(-1)}{(n-1)n^3(n+1)}$$ $$\frac{1}{(n-1)n^3(n+1)}=\frac{1}{(n-2)(n-1)n(n+1)(n+2)}+\frac{-2^2}{(n-2)(n-1)n^3(n+1)(n+2)}$$ Continuing on telescoping we get that: $$\frac{1}{n^3} = \frac{(-1)^m m!^2}{(n-m)\ldots n^3 \ldots (n+m)}+\sum_{j=1}^{m} \frac{(-1)^{j-1} (j-1)!^2}{(n-j)\ldots(n+j)}$$ So by setting $m=n-1$: $$\frac{1}{n^3} = \frac{(-1)^{n-1} (n-1)!^2}{n^2 (2n-1)!}+\sum_{j=1}^{n-1} \frac{(-1)^{j-1} (j-1)!^2}{(n-j)\ldots(n+j)}$$ The terms of the last series can be managed through partial fraction decomposition: $$\frac{1}{(n-j)\ldots(n+j)} = \frac{1}{(2j)!(n-j)} - \frac{1}{(2j-1)! 1! (n-j+1)} + \frac{1}{(2j-2)! 2! (n-j+2)} -\ldots$$ $$\frac{(n-j-1)!}{(n+j)!}=\sum_{k=0}^{2j}\frac{(-1)^k}{(2j-k)\,k!\,(n-j+k)}=\frac{1}{(2j)!}\sum_{k=0}^{2j}\frac{(-1)^k{\binom{2j}{k}}}{n-j+k}$$ and since: $$\sum_{n>j}\sum_{k=0}^{2j}\frac{(-1)^k{\binom{2j}{k}}}{n-j+k}=\sum_{h=1}^{2j}\frac{(-1)^{h-1}{\binom{2j-1}{h-1}}}{h}=\int_{0}^{1}(1-x)^{2j-1}\,dx=\frac{1}{2j}$$ we get: $$\zeta(3)=\sum_{n=1}^{+\infty}\frac{(-1)^{n-1} n!^2}{n^4 (2n-1)!}+\sum_{j=1}^{+\infty}\sum_{n>j}\frac{(-1)^{j-1} (j-1)!^2}{(n-j)\ldots(n+j)}$$ $$\zeta(3)=\sum_{n=1}^{+\infty}\frac{(-1)^{n-1} n!^2}{n^4 (2n-1)!}+\sum_{j=1}^{+\infty}\frac{(-1)^{j-1}j!^2}{2j^3\,(2j)!}=\color{red}{\frac{5}{2}\sum_{n=1}^{+\infty}\frac{(-1)^{n-1}}{n^3{\binom{2n}{n}}}}$$ as wanted.