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 of the Bernoulli numbers:


 * $\displaystyle \frac x {e^x - 1} = \sum_{n \mathop = 0}^\infty \frac{B_n x^n} {n!}$

or equivalently:


 * $\displaystyle \frac x {e^x - 1} + \frac x 2 = 1 + \sum_{n \mathop = 2}^\infty \frac{B_n x^n} {n!}$

It remains to show that this is an even function.