TITLE: Why are the Bernoulli numbers of even index nonzero?
QUESTION [7 upvotes]: 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

REPLY [3 votes]: 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 $B_{2n}\neq 0$.