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

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

$B_{2 n + 1} = 0$

$\blacksquare$


Sources

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