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.
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| \(\ds \frac x {e^x - 1} + \frac x 2\) | \(=\) | \(\ds \frac {x \left({1 + e^x}\right)} {2\left({e^x - 1}\right)}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {-x \left({e^{-x} + 1}\right)} {2\left({e^{-x} - 1}\right)}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {-x \left({2 + e^{-x} - 1}\right)} {2\left({e^{-x} - 1}\right)}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {-x} {e^{-x} - 1} + \frac {-x} 2\) |
$\blacksquare$
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 $B_n = 0$
$\blacksquare$