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?
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I had a look there but no one uses the same proof that Euler did which what i'm interested in – jake walsh Oct 30 '16 at 15:07
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The answer with score 25. That is the closest to Euler's method. – Parcly Taxel Oct 30 '16 at 15:08
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Yes it's very close and a nice proof but I' trying to find how Euler did it using only his original proof. – jake walsh Oct 30 '16 at 15:09
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See 1st comment there for a derivation from $\frac{\pi^2}{\sin^2(\pi z)} = \sum_{n=-\infty}^\infty \frac{1}{(z-n)^2}$ – reuns Oct 30 '16 at 15:44
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The answer with score 14 is also the method that Euler used for Basel problem! – user91500 Oct 30 '16 at 15:46
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1Someone 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 $z \in \mathbb{C}$ – reuns Oct 30 '16 at 15:49
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@user1952009 Sh... believe in the great Euler XD. And theres that other squeeze theorem proof Euler gave too. – Simply Beautiful Art Oct 31 '16 at 23:31
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@SimpleArt hmm ? – reuns Oct 31 '16 at 23:48
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@user1952009 Just reminding that there is a second proof that uses the squeeze theorem instead. – Simply Beautiful Art Oct 31 '16 at 23:49
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@user1952009 Rigorous proof concerning the Basel problem – Simply Beautiful Art Oct 31 '16 at 23:53
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@SimpleArt I don't see how it is related to what I wrote, which is not about proving $\zeta(2) = \pi^2/6$, but $k! 2 (2^{2k}-1)\zeta(2k) = f^{(k)}(1/2)$ from $f(z) = \frac{\pi^2}{\sin(\pi z)^2} = \sum_{n=-\infty}^\infty \frac{1}{(z-n)^2}$. So I ask if Euler could prove $\frac{\pi^2}{\sin(\pi z)^2} = \sum_{n=-\infty}^\infty \frac{1}{(z-n)^2}$ rigorously, and without making some complicated digressions about (his) product formula for $\sin(x), x \in \mathbb{R}$ – reuns Oct 31 '16 at 23:59