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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$$\sum_1^\infty \frac1{n^3}=\frac52\sum_1^\infty\frac{(-1)^{n-1}}{n^3\binom{2n}{n}}$$

Mark Fischler
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  • Related: https://math.stackexchange.com/questions/499926/evaluating-the-series-sum-n-1-infty-frac1n3-binom2nn The same types of method (manipulations of the Taylor series of $\arcsin^2(x)$) can very likely also be applied to this sum. – Winther Jun 21 '17 at 21:18
  • hint: use the beta-integral representation of $1/\binom{2n}{n}$ – tired Jun 21 '17 at 21:18
  • Yet another textbook exercise. Will they never get old? Apery proved that when, in 1979? Being born after that is no excuse, just get his publication and read it. –  Jun 21 '17 at 21:19
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    http://pracownicy.uksw.edu.pl/mwolf/Poorten_MI_195_0.pdf , see in particular p.3. – Chappers Jun 21 '17 at 21:21
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    @ProfessorVector 1) Textbook please. 2) Some things just never get old. 3) The relevant publication is not available at the journal site (Asterisque). – Matemáticos Chibchas Jun 21 '17 at 21:26

2 Answers2

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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.

Jack D'Aurizio
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    This is a very elementary (and very tedious) derivation I got some time ago, I am not sure if Apery himself or Beuker took the same approach, but I hope it fits your needs. – Jack D'Aurizio Jun 21 '17 at 21:30
  • @tired's suggested approach of starting from the RHS and expressing $\frac{2n}{n}^{-1}$ through Euler's Beta function should lead to a shorter but less elementary proof. – Jack D'Aurizio Jun 21 '17 at 21:33
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    This is exactly the level of clarity I was looking for. Tedious, perhaps, but I would not call it very tedious. – Mark Fischler Jun 21 '17 at 22:00
  • Starting from the RHS isn't trivial too. You'll arrive to something like $\mathrm{Li}_{2}\left(,t\left[,t - 1,\right],\right)$. – Felix Marin Jun 21 '17 at 22:57
  • @FelixMarin: with tired's approach the question boils down to showing that $$ \int_{0}^{1}\frac{\text{Li}2(1-x^2)}{1-x^2},dx = \frac{1}{4}\left(\pi^2 \log(4)-7\zeta(3)\right) $$ and by integration by parts that boils down to computing $$\int{0}^{1}\frac{x\log(x)\log(1\pm x)}{1-x^2},dx $$ – Jack D'Aurizio Jun 21 '17 at 23:44
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    that belong to the realm of fashion logarithmic/dilogathmic integrals, very popular on MSE. With a probability close to $1$, to compute the value of such integrals has already been asked and answered by some guru (like you) in the use of Euler's Beta function, Feynman's trick and the dilogarithm reflection formulas :D – Jack D'Aurizio Jun 21 '17 at 23:48
  • $\Huge{\bullet\quad\bullet \brace {{\brace} \atop \smile}}$. – Felix Marin Jun 22 '17 at 01:14
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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

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

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

Ricardo770
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