# A new formula for Apery's constant and other zeta(s)?

I recently found these Plouffe-like formulas using Mathematica's LatticeReduce. Has anybody seen/can prove these are indeed true?

\begin{aligned}\frac{3}{2}\,\zeta(3) &= \frac{\pi^3}{24}\sqrt{2}-2\sum_{k=1}^\infty \frac{1}{k^3(e^{\pi k\sqrt{2}}-1)}-\sum_{k=1}^\infty\frac{1}{k^3(e^{2\pi k\sqrt{2}}-1)}\\ \frac{3}{2}\,\zeta(5) &= \frac{\pi^5}{270}\sqrt{2}-4\sum_{k=1}^\infty \frac{1}{k^5(e^{\pi k\sqrt{2}}-1)}+\sum_{k=1}^\infty \frac{1}{k^5(e^{2\pi k\sqrt{2}}-1)}\\ \frac{9}{2}\,\zeta(7) &= \frac{41\pi^7}{37800}\sqrt{2}-8\sum_{k=1}^\infty\frac{1}{k^7(e^{\pi k\sqrt{2}}-1)}-\sum_{k=1}^\infty\frac{1}{k^7(e^{2\pi k\sqrt{2}}-1)} \end{aligned}

And so on for other $\zeta(2n+1)$. The background for these are in my blog.

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 These are indeed very similar to the Plouffe ones. Really the number of formulas surrounding Riemann Zeta are already so mindboggling that formulas that are so close to highly publicized ones have a hard time getting attention. – Peter Sheldrick Jun 11 '12 at 17:52

Your identities for $\zeta(4n-1)$ follow from the results immediately following his Corollary 2 $$q^{k-1}\sum\frac{1}{n^k(e^{2\pi pn/q}-1)}+p^{k-1}\sum\frac{1}{n^k(e^{2\pi qn/p}-1)} = q^{k-1}I_k(2\pi p/q)$$ with $p=\sqrt{2},q=1$, and using the expression for $I_k(x)$ from the middle of page 7 (I confirmed for $k=3$ and expect it will also match for $k=7$).
He uses additional machinery to establish the identities in Plouffe's form for $k=4n+1$, and it doesn't immediately admit generalization to your form, but maybe it can be tweaked.
Yes, these Plouffe-like identities usually fall into two classes: for $\zeta(4n-1)$ and $\zeta(4n+1)$. I had hoped the general family using $\sqrt{2}$ was new, but even if it is not, I'm glad I found them on my own. – Tito Piezas III Jun 25 '12 at 18:58