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.
Best Answer
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.