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Is there any way to prove that:

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

Using Euler's original proof of equating coefficients?


marked as duplicate by Adam Hughes, Parcly Taxel, user91500, Pragabhava, user223391 Oct 30 '16 at 20:55

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  • $\begingroup$ I had a look there but no one uses the same proof that Euler did which what i'm interested in $\endgroup$ – jake walsh Oct 30 '16 at 15:07
  • $\begingroup$ The answer with score 25. That is the closest to Euler's method. $\endgroup$ – Parcly Taxel Oct 30 '16 at 15:08
  • $\begingroup$ Yes it's very close and a nice proof but I' trying to find how Euler did it using only his original proof. $\endgroup$ – jake walsh Oct 30 '16 at 15:09
  • $\begingroup$ See 1st comment there for a derivation from $\frac{\pi^2}{\sin^2(\pi z)} = \sum_{n=-\infty}^\infty \frac{1}{(z-n)^2}$ $\endgroup$ – reuns Oct 30 '16 at 15:44
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    $\begingroup$ Someone knows if Euler could prove that $\pi \cot(z) = \lim_{N \to \infty}\sum_{n=-N}^N\frac{1}{z-n}$ without making some complicated digressions about $\frac{\sin(\pi x)}{ x}$ Weierstrass product ? I think he had some problems because he wasn't thinking to meromorphic functions of $z \in \mathbb{C}$ $\endgroup$ – reuns Oct 30 '16 at 15:49