# Odd Bernoulli Numbers Vanish/Proof 2

## 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:

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

We have:

 $\ds \frac x {e^x - 1}$ $=$ $\ds \frac x 2 \left({\frac 2 {e^x - 1} }\right)$ $\ds$ $=$ $\ds \frac x 2 \left({\frac {e^x - e^x + 2} {e^x - 1} }\right)$ $\ds$ $=$ $\ds \frac x 2 \left({\frac {\left({e^x + 1}\right) - \left({e^x - 1}\right)} {e^x - 1} }\right)$ $\ds$ $=$ $\ds \frac x 2 \left({\frac {e^x + 1} {e^x - 1} - 1}\right)$ $\ds$ $=$ $\ds -\frac x 2 + \frac x 2 \left({\frac {e^x + 1} {e^x - 1} }\right)$

Take $f \left({x}\right) := \dfrac x 2 \left({\dfrac {e^x + 1} {e^x - 1} }\right)$, and note that:

 $\ds f \left({-x}\right)$ $=$ $\ds \frac {-x} 2 \left({\dfrac {e^{-x} + 1} {e^{-x} - 1} }\right)$ $\ds$ $=$ $\ds -\frac {-x} 2 \left({\dfrac {1 + e^x} {1 - e^x} }\right)$ multiplying top and bottom by $e^x$ $\ds$ $=$ $\ds -\frac {-x} 2 \left({\dfrac {e^x + 1} {-\left({e^x - 1}\right)} }\right)$ $\ds$ $=$ $\ds \frac x 2 \left({\frac {e^x + 1} {e^x - 1} }\right)$ $\ds$ $=$ $\ds f \left({x}\right)$

and so $f \left({x}\right) := \dfrac x 2 \left({\dfrac {e^x + 1} {e^x - 1} }\right)$ is an even function.

Thus $f \left({x}\right)$ can be seen to take the form:

$\displaystyle f \left({x}\right) = 1 + \sum_{n \mathop = 2}^\infty \frac {B_n} {n!} x^n$

and so:

$\displaystyle f \left({x}\right) = 1 + \sum_{n \mathop = 2}^\infty \frac {B_n} {n!} x^{-n}$

That is:

$\forall n \in \N: n > 1: \dfrac {B_n} {n!} x^n = \dfrac {B_n} {n!} x^{-n}$

and so for odd $n$ where $n > 1$ it follows that $B_n = 0$

$\blacksquare$