I came across the following results. $$\zeta(2)=\sum_{n=1}^\infty\dfrac{1}{n^2}=\dfrac{\pi^2}{6}$$ $$\zeta(4)=\sum_{n=1}^\infty\dfrac{1}{n^4}=\dfrac{\pi^4}{90}$$ $$\zeta(8)=\sum_{n=1}^\infty\dfrac{1}{n^8}=\dfrac{\pi^8}{9450}$$ Are there any good papers/journals or books which rigoursly prove the above results?
Evaluating Riemann Zeta function( reference request)
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11 minute of Googling with the key words "zeta function even integers" lead to many answer, such as [this](https://proofwiki.org/wiki/Riemann_Zeta_Function_at_Even_Integers). – 2017-02-20
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0In Gamelin's complex analysis book and in Newman's-Bak's book you can find proofs of the 1st equality by using the residue theorem. An open problem is the evaluation of the series $ \sum_{n=1}^{\infty}1/n^3$. There us a variety of other proofs for the 1st one. One way is using trigonometric series. As about the others, what you are looking for is what we know as Bernoulli numbers. You can check it in the web. – 2017-02-20
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0Most comprehensive book about Riemann zeta function - Titchmarch's "The Theory of the Riemann Zeta-Function". – 2017-02-20
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1Here are 14 proofs of the $\zeta(2)$ result: https://www.uam.es/personal_pdi/ciencias/cillerue/Curso/zeta2.pdf – 2017-02-20
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0As companion of previous, Mikael Passare, *How to compute* $\sum 1/n^2$ *by solving triangles*, currently there is a version (2007) of this paper in arXiv, that is arXiv:math/0701039 – 2017-02-20
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Let $B_n$'s are Bernoulli numbers defined by $$\frac{z}{e^z-1}+\dfrac12z=1+\sum_{n=2}^\infty\frac{B_n}{n!}z^n$$ for $n>1$, then for $k>1$ $$\zeta(2k)=(-1)^{k+1}\frac{(2\pi)^{2k}B_{2k}}{2(2k)!}$$ Some of Bernoulli numbers are $$B_0=1,-\frac12,\frac16,0,-\frac{1}{30},0,\frac{1}{42},0,-\frac{1}{30},0,\frac{5}{66},0,\cdots$$ The best Reference I know is
Apostol Tom Mike, Introduction to Analytic Number Theory, Springer-Verlag, New York, (1976)(352s), pp.264-268.
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The historical way is showing that $$g(z) = \pi \text{ cotan } \pi z = \frac{1}{z}+ \sum_{n=1}^\infty (\frac{1}{z-n}+\frac{1}{z+n})$$ $$ f(z) = -g'(z) =\frac{\pi^2}{\sin^2 \pi z} = \sum_{n=-\infty}^\infty \frac{1}{(z-n)^2}$$ from which you get $$f^{2k-2}(1/2) = \frac{(2k)!}{2}\sum_{n=-\infty}^\infty \frac{1}{(1/2-n)^{2k}} =(2k)!2^{2k} (1-2^{-2k})\zeta(2k)$$
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0https://en.wikipedia.org/wiki/Basel_problem#Euler.27s_approach – 2017-02-21