Showing $\displaystyle{\zeta(3)=\frac{1}{7}\sum_{n=1}^\infty\frac{n(2n+1)\zeta(2n+1)}{2^{2n-2}}}$

Question

I want to show the following infinite series representation for $\zeta(3)$ which I stumbled across on accident. $$\zeta(3)=\frac{1}{7}\sum_{n=1}^\infty\frac{n(2n+1)\zeta(2n+1)}{2^{2n-2}}$$ Here is my attempt so far. By the series representation of $\zeta(2n+1)$, $$\begin{align*} \sum_{n=1}^\infty\frac{n(2n+1)\zeta(2n+1)}{2^{2n-2}}&=\sum_{n=1}^\infty\frac{n(2n+1)}{2^{2n-2}}\sum_{k=1}^{\infty}\frac{1}{k^{2n+1}} \\ &=\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{n(2n+1)}{2^{2n-2}k^{2n+1}} \end{align*}$$ and I am stuck. Evaluating the second sum using wolfram gives me, $$\sum_{n=1}^{\infty}\frac{n(2n+1)}{2^{2n-2}k^{2n+1}}=\frac{16k(12k^2+1)}{(4k^2-1)^3}$$ for $\vert k\vert>1/2$, turning the problem into finding the infinite sum of some polynomial fraction, which I still can't evaluate.

Looking at the wiki for $\zeta(3)$ I see Euler gave the series representation, $$\zeta(3)=\frac{\pi^2}{7}\left(1-4\sum_{k=1}^\infty\frac{\zeta(2k)}{2^{2k}(2k+1)(2k+2)}\right)$$ which some what resembles mine, so I'm guessing there's some trick I don't know of.

Any help? Thanks in advance.

Answer 1

For $|z|<1$ we have $$\sum_{n=1}^\infty\zeta(n+1)z^n=\sum_{k=1}^\infty\left(\frac1{k-z}-\frac1k\right)=-\gamma-\psi(1-z)$$ (seen here and here, respectively; the first equality is easy to get after inserting $\zeta(n+1)=\sum_{k=1}^\infty k^{-n-1}$ and summing over $n$). Extracting the even part, we obtain $$\sum_{n=1}^\infty\zeta(2n+1)z^{2n}=-\gamma-\frac12\big(\psi(1-z)+\psi(1+z)\big).$$

Now multiply by $z$ and take the second derivative: $$\sum_{n=1}^\infty 2n(2n+1)\zeta(2n+1)z^{2n-1}=\psi'(1-z)-\psi'(1+z)-\frac z2\big(\psi''(1-z)+\psi''(1+z)\big).$$

The sum in the title of the OP is obtained at $z=1/2$. For $n>0$ we have $$(-1)^{n+1}\psi^{(n)}(1-z)=n!\sum_{k=1}^\infty\frac1{(k-z)^{n+1}}$$ (taking the $n$-th derivative of the equality at the beginning), hence $$\psi'(1/2)-\psi'(3/2)=4,\\\psi''(1/2)-\psi''(3/2)=-16,\\\psi''(1/2)=-16\sum_{k=1}^\infty\frac1{(2k-1)^3}=-14\zeta(3).$$

That's sufficient.

Answer 2

A proof of the mentioned series (by Euler), for completeness. We have, for $|z|<1$, $$2\sum_{n=1}^\infty\zeta(2n)z^{2n}=\sum_{k=1}^\infty\frac{2z^2}{k^2-z^2}=1-\pi z\cot\pi z$$ (the first equality after $\zeta(2n)=\sum_{k=1}^\infty k^{-2n}$ and summing over $n$; the second one is known, or obtained as the logarithmic derivative of the infinite product for the sine function). Then $$2\sum_{n=1}^\infty\frac{\zeta(2n)z^{2n+2}}{(2n+1)(2n+2)}=\int_0^z(z-t)(1-\pi t\cot\pi t)\,dt$$ and, if we put $z=1/2$ and substitute $t=x/\pi$, we obtain $$S:=\sum_{n=1}^\infty\frac{\zeta(2n)}{2^{2n}(2n+1)(2n+2)}=\int_0^{1/2}(1-2t)(1-\pi t\cot\pi t)\,dt\\=\frac14-\frac1\pi\int_0^{\pi/2}x\cot x\,dx+\frac2{\pi^2}\int_0^{\pi/2}x^2\cot x\,dx.$$

These integrals are evaluated using integration by parts and the Fourier series $$-\log|\sin x|=\log 2+\sum_{n=1}^\infty\frac{\cos 2nx}{n}.$$

With $\int_0^{\pi/2}x\cos 2nx\,dx=\frac{(-1)^n-1}{4n^2}$ (again, using i.b.p.), we easily get $$\int_0^{\pi/2}x\cot x\,dx=\frac\pi2\log2,\qquad\int_0^{\pi/2}x^2\cot x\,dx=\frac{\pi^2}4\log2-\frac78\zeta(3),$$ hence $S=\frac14\left(1-\frac7{\pi^2}\zeta(3)\right)$ as expected.