Calculatig $\zeta (0)$ from $\zeta (2n)=\frac{(-1)^{n+1}\mathrm B_{2n}(2\pi )^{2n}}{2(2n)!}$

Question

MathWorld and Wikipedia state that $$\zeta (2n)=\frac{(-1)^{n+1}\mathrm B_{2n}(2\pi )^{2n}}{2(2n)!}$$ is valid for positive integers $n$. But the identity works for $0$ as well. Is it rigorous to use the identity to prove that $\zeta (0)=-\frac12$?

Answer 1

Of course it depends how you proved $\zeta (2n)=\frac{(-1)^{n+1}\mathrm B_{2n}(2\pi )^{2n}}{2(2n)!}, n\ge 1$.

If it is applying the residue theorem to $\lim_{N \to \infty}\int_{|z| = N} \frac{z^{-2n}}{e^z-1}dz$ then no it doesn't apply to $n=0$.

Same if it is from the Fourier series or $\pi s\cot\pi s=s\sum_k \frac1{s+k}=1-2\sum_{n=1}^\infty \frac{(-1)^{n+1}\mathrm B_{2n}(2\pi )^{2n}}{2(2n)!}s^{2n}$.

If it is after showing the functional equation and $\Gamma(s)\zeta(s)= (-1)^n\frac{B_n}{n!}\frac{1}{s+n-1}+O(1)$ then yes it is valid for $n=0$.