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I have been recently finding the values for the even positive integers of the zeta function using fourier series, and it is well know that these are all of the form $\frac{\pi^{2n}}{a_{2n}}$ and so I thought about whether or not the series below would converge: $$S=\sum_{n=1}^\infty\frac{\zeta(2n)}{\pi^{2n}}=\frac{1-\cot(1)}{2}$$ As you can see it does converge and has an interesting value however I am not sure how to prove it. I do know that: $$\zeta(2n)=(-1)^{n+1}\frac{B_{2n}(2\pi)^{2n}}{2(2n)!}$$ and so we could say: $$S=\sum_{n=1}^\infty\frac{(-1)^{n+1}B_{2n}2^{2n-1}}{(2n)!}$$ How can I manipulate this? Thanks

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Notice

$${\sum_{n=1}^{\infty}\frac{\zeta(2n)}{\pi^{2n}}=\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\left(\frac{1}{k^2}\right)^n\frac{1}{\pi^{2n}}}$$

Interchanging sums

$${\Rightarrow \sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\left(\frac{1}{(k\pi)^2}\right)^n}$$

This is the sum of an infinite number of Geometric series!

$${=\sum_{k=1}^{\infty}\left(\left(\frac{1}{1-\frac{1}{(k\pi)^2}}\right)-1\right)}$$

Simplifying this, you get

$${\Rightarrow \sum_{k=1}^{\infty}\frac{1}{(k\pi)^2 - 1}}$$

Now we can factor out the ${\pi^2}$ to get

$${=\frac{1}{\pi^2}\sum_{k=1}^{\infty}\frac{1}{k^2 - \left(\frac{1}{\pi^2}\right)}}$$

Fortunately, this is a sum we can deal with very nicely using the formula

$${\sum_{n=0}^{\infty}\frac{1}{n^2 + a^2} = \frac{1+a\pi\coth(a\pi)}{2a^2}}$$

(source: https://en.wikipedia.org/wiki/List_of_mathematical_series . Can be derived using Fourier series!)

Can you take it from here?

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    $\begingroup$ Very nice. +1 from me. $\endgroup$ – K.defaoite Jul 9 at 16:44
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    $\begingroup$ great answer, thanks! $\endgroup$ – Henry Lee Jul 9 at 19:52
  • $\begingroup$ No problem at all! Thanks! $\endgroup$ – Riemann'sPointyNose Jul 9 at 20:08

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