Odd Bernoulli Numbers Vanish

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Theorem

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

Then:

$B_{2n + 1} = 0$

for $n \ge 1$.


Proof 1

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 be shown that this is an even function.


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

$\blacksquare$


Proof 2

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 $B_n = 0$

$\blacksquare$