Why are the Bernoulli numbers of even index nonzero?

Question

The Bernoulli numbers are defined through: $\frac{z}{e^z-1}=\sum_{i=0}^{\infty}B_i \frac{z^i}{i!} $.

One can show easily that $B_i=0$ for all $i\geq3$ odd. Is there a similar way to show that $B_i\neq 0$ for all even $i$?

Best regards

Answer 1

This can be shown using the formula $$ \zeta(2n)=\frac{(-1)^{n+1}B_{2n}(2\pi)^{2n}}{2(2n)!}, $$ where $\zeta(s)$ is the Riemann zeta function. The series representation of $\zeta$ makes it clear that $\zeta(2n)\neq 0$, which implies $B_{2n}\neq 0$.