# 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:

 $\ds \frac x {e^x - 1}$ $=$ $\ds \frac x 2 \paren {\frac 2 {e^x - 1} }$ multiplying top and bottom by $2$ $\ds$ $=$ $\ds \frac x 2 \paren {\frac {e^x - e^x + 2} {e^x - 1} }$ adding zero $\ds$ $=$ $\ds \frac x 2 \paren {\frac {\paren {e^x + 1} - \paren {e^x - 1} } {e^x - 1} }$ $\ds$ $=$ $\ds \frac x 2 \paren {\frac {e^x + 1} {e^x - 1} - 1}$ $\ds$ $=$ $\ds -\frac x 2 + \frac x 2 \paren {\frac {e^x + 1} {e^x - 1} }$

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

 $\ds \map f {-x}$ $=$ $\ds \frac {-x} 2 \paren {\dfrac {e^{-x} + 1} {e^{-x} - 1} }$ $\ds$ $=$ $\ds -\frac {-x} 2 \paren {\dfrac {1 + e^x} {1 - e^x} }$ multiplying top and bottom by $e^x$ $\ds$ $=$ $\ds -\frac {-x} 2 \paren {\dfrac {e^x + 1} {-\paren {e^x - 1} } }$ $\ds$ $=$ $\ds \frac x 2 \paren {\frac {e^x + 1} {e^x - 1} }$ $\ds$ $=$ $\ds \map f x$

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

 $\ds \frac x {e^x - 1}$ $=$ $\ds \sum_{n \mathop = 0}^\infty \frac {B_n x^n} {n!}$ $\ds \frac x {e^x - 1}$ $=$ $\ds 1 - \dfrac 1 2 x + \sum_{n \mathop = 2}^\infty \frac {B_n x^n} {n!}$ $\ds \frac x {e^x - 1} + \dfrac 1 2 x$ $=$ $\ds 1 + \sum_{n \mathop = 2}^\infty \frac {B_n x^n} {n!}$ adding $\dfrac 1 2 x$ to both sides $\ds \frac {2x + x\paren {e^x - 1 } } {2\paren {e^x - 1 } }$ $=$ $\ds 1 + \sum_{n \mathop = 2}^\infty \frac {B_n x^n} {n!}$ $\ds \dfrac x 2 \paren {\dfrac {e^x + 1} {e^x - 1} }$ $=$ $\ds 1 + \sum_{n \mathop = 2}^\infty \frac {B_n x^n} {n!}$ $\ds \map f x$ $=$ $\ds 1 + \sum_{n \mathop = 2}^\infty \frac {B_n} {n!} x^n$

We now have:

 $\ds \map f x$ $=$ $\ds \map f {-x}$ $\map f x$ established to be an even function above $\ds 1 + \sum_{n \mathop = 2}^\infty \frac {B_n} {n!} \paren x^n$ $=$ $\ds 1 + \sum_{n \mathop = 2}^\infty \frac {B_n} {n!} \paren {-x}^n$

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

 $\ds \frac {B_{2 k + 1} } { \paren {2 k + 1}!} \paren x^{2 k + 1}$ $=$ $\ds \frac {B_{2 k + 1} } { \paren {2 k + 1}!} \paren {-x}^{2 k + 1}$ $\ds B_{2 k + 1} \paren x$ $=$ $\ds B_{2 k + 1} \paren {-x}$ multiplying both sides by $\dfrac {\paren {2 k + 1}!} {x^2}$ $\ds B_{2 k + 1}$ $=$ $\ds -B_{2 k + 1}$ dividing both sides by $x$ $\ds 2 B_{2 k + 1}$ $=$ $\ds 0$ $\ds B_{2 k + 1}$ $=$ $\ds 0$

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

$B_{2 n + 1} = 0$

$\blacksquare$