Odd Bernoulli Numbers Vanish

Theorem
Let $B_n$ denote the $n$th Bernoulli Number.

Then:


 * $B_{2n + 1} = 0$

for $n \ge 1$.

Proof
By definition, the Bernoulli numbers are given by:
 * $\ds \frac x {e^x - 1} = \sum_{n \mathop = 0}^\infty \frac {B_n x^n} {n!}$

We have:

Take $\map f x := \dfrac x 2 \paren {\dfrac {e^x + 1} {e^x - 1} }$, and note that:

and so $\map f x := \dfrac x 2 \paren {\dfrac {e^x + 1} {e^x - 1} }$ is an even function.

Rewriting the definition of Bernoulli numbers

We now have:

If we set $n$ to be an odd integer where $n > 1$, we have:

Therefore: $\forall n \in \N: n \ge 1$:


 * $B_{2 n + 1} = 0$