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The proof using Fourier series: Reference: Stein Shakarchi "Fourier Analysis, the introduction" p 97 Exercise 4 The key ingredient of this proof is the following identity $$\sum_{n=1}^{\infty} \frac{1}{n^2-\alpha^2}=\frac{1}{2\alpha^2}-\frac{\pi}{2\alpha\tan{\alpha\pi}}$$ which can be proved by using Fourier series of $\cos(\alpha x)$. This allows an expression of $\zeta(2n)$ by Bernoulli numbers.
The proof using the functional equation of $\zeta(s)$: Reference: E.Titchmarsh "The Theory of the Riemann zeta function" p18 (2.4 Second method)
Chapter 2 of this book is entirely devoted to the proofs of the functional equation $$\pi^{-s/2}\Gamma(s)\zeta(s)=\pi^{-(1-s)/2}\Gamma(1-s)\zeta(1-s).$$ $\pi^{-s/2}\Gamma(s/2)\zeta(s)=\pi^{-(1-s)/2}\Gamma((1-s)/2)\zeta(1-s).$$Section 2.4 is one of the proofs, it uses te residue theorem of complex analysis, and derives the formula$$\zeta(1-2m)=\frac{(-1)^mB_{2m}}{2m}$$for m=1,2,\cdots The formula for \zeta(2n) is now followd by the functional equation. The result is:$$2\zeta(2n)=(-1)^{n+1}\frac{(2\pi)^{2n}}{(2n)!}B_{2n}$$where$$\frac{z}{e^z-1}=\sum_{n=0}^{\infty} \frac{B_n}{n!}z^n.$$2 corrected minor errors These are the proofs that I have seen: The proof using Fourier series: Reference: Stein Shakarchi "Fourier Analysis, the introduction" p 97 Exercise 4 The key ingredient of this proof is the following identity$$\sum_{n=1}^{\infty} \frac{1}{n^2-\alpha^2}=\frac{1}{2\alpha^2}-\frac{\pi}{2\alpha\tan{\alpha\pi}}$$which can be proved by using Fourier series of \cos(\alpha\pi). \cos(\alpha x). This allows an expression of \zeta(2n) by Bernoulli numbers. The proof using the functional equation of \zeta(s): Reference: E.Titchmarsh "The Theory of the Riemann zeta function" p18 (2.4 Second method) Chapter 2 of this book is entirely devoted to the proofs of the functional equation$$\pi^{-s/2}\Gamma(s)\zeta(s)=\pi^{-(1-s)/2}\Gamma(1-s)\zeta(1-s).$$Section 2.4 is one of the proofs, it uses te residue theorem of complex analysis, and derives the formula$$\zeta(1-2m)=\frac{(-1)^mB_{2m}}{2m}$$for m=1,2,\cdots The formula for \zeta(2n) is now followd by the functional equation. The result is:$$2\zeta(2n)=(-1)^{n+1}\frac{(2\pi)^{2n}}{(2n)!}B_{2n}$$where$$\frac{z}{e^z-1}=\sum_{n=0}^{\infty} \frac{B_n}{n!}z^n.$$1 These are the proofs that I have seen: The proof using Fourier series: Reference: Stein Shakarchi "Fourier Analysis, the introduction" p 97 Exercise 4 The key ingredient of this proof is the following identity$$\sum_{n=1}^{\infty} \frac{1}{n^2-\alpha^2}=\frac{1}{2\alpha^2}-\frac{\pi}{2\alpha\tan{\alpha\pi}}$$which can be proved by using Fourier series of \cos(\alpha\pi). This allows an expression of \zeta(2n) by Bernoulli numbers. The proof using the functional equation of \zeta(s): Reference: E.Titchmarsh "The Theory of the Riemann zeta function" p18 (2.4 Second method) Chapter 2 of this book is entirely devoted to the proofs of the functional equation$$\pi^{-s/2}\Gamma(s)\zeta(s)=\pi^{-(1-s)/2}\Gamma(1-s)\zeta(1-s).$$Section 2.4 is one of the proofs, it uses te residue theorem of complex analysis, and derives the formula$$\zeta(1-2m)=\frac{(-1)^mB_{2m}}{2m}$$for m=1,2,\cdots The formula for \zeta(2n) is now followd by the functional equation. The result is:$$2\zeta(2n)=(-1)^{n+1}\frac{(2\pi)^{2n}}{(2n)!}B_{2n}$$where$$\frac{z}{e^z-1}=\sum_{n=0}^{\infty} \frac{B_n}{n!}z^n.$\$