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Show that $\sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}$

I have to show this using results I have found earlier. I started with $$0 = \frac{512}{10}+\sum_{n=1}^{\infty} 2048(\pi^2n^2-6)\frac{(-1)^n}{π^4n^4}$$ which was obtained from a fourier series.

And have got to $$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^4 }=\frac{ -7\pi^4}{720} $$

But I have no idea how to proceed with getting rid of the $(-1)^n.$

Any help/tips would be greatly appreciated. Thanks in advance.


marked as duplicate by Dietrich Burde, Sahiba Arora, Yagna Patel, mechanodroid, Misha Lavrov Dec 9 '17 at 1:01

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  • $\begingroup$ Is that all results you have found earlier? If no, can you tell us more about the problem? $\endgroup$ – Shashi Dec 8 '17 at 21:57

$$\frac{ -7\pi^4}{720}=\sum_{n=1}^{\infty} \frac{(-1)^n}{n^4 }=-\sum_{n=1}^{\infty} \frac1{n^4 }+2\sum_{n=1}^{\infty} \frac1{(2n)^4 }=-\left(1-\frac2{2^4}\right)\sum_{n=1}^{\infty} \frac1{n^4 }=-\frac78\sum_{n=1}^{\infty} \frac1{n^4 }$$ $$\sum_{n=1}^{\infty} \frac1{n^4 }=\frac{\pi^4}{90}$$


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