# the odd Bernoulli numbers are zero

Recall that, for $k\geq 0$, the Bernoulli numbers $B_{k}$ are defined as the coefficients in the Taylor expansion:

 $\displaystyle\frac{t}{e^{t}-1}=\sum_{k\geq 0}B_{k}\frac{t^{k}}{k!}.$ (1)

Just to name a few:

 $B_{0}=1,\quad B_{1}=-\frac{1}{2},\quad B_{2}=\frac{1}{6},\quad B_{3}=0,\quad B% _{4}=-\frac{1}{30},\ B_{5}=0,\ldots,\ B_{10}=\frac{5}{66},\ldots$
###### Lemma.

If $k\geq 3$ is odd then $B_{k}=0$.

###### Proof.

From the right hand side of (1) we extract the term corresponding to $k=1$:

 $\displaystyle\frac{t}{e^{t}-1}=-\frac{t}{2}+\sum_{k\geq 0,\ k\neq 1}B_{k}\frac% {t^{k}}{k!}.$ (2)

Thus:

 $\displaystyle\frac{t}{e^{t}-1}+\frac{t}{2}=\sum_{k\geq 0,\ k\neq 1}B_{k}\frac{% t^{k}}{k!}$ (3)

and the left hand side can be rewritten as:

 $\displaystyle\frac{t}{e^{t}-1}+\frac{t}{2}=\frac{2t+t(e^{t}-1)}{2(e^{t}-1)}=% \frac{t}{2}\cdot\frac{e^{t}+1}{e^{t}-1}=\frac{t}{2}\cdot\frac{e^{t/2}+e^{-t/2}% }{e^{t/2}-e^{-t/2}}.$ (4)

Hence, if one replaces $t$ by $-t$ then (4) is unchanged. Since (4) is the left hand side of (3), the quantity

 $\sum_{k\geq 0,\ k\neq 1}B_{k}\frac{t^{k}}{k!}$

is also unchanged when $t$ is exchanged by $-t$, and so we must have $B_{k}=(-1)^{k}B_{k}$ for $k\neq 1$. We conclude that if $k\geq 3$ and $k$ is odd, $B_{k}=0$. ∎

Title the odd Bernoulli numbers are zero TheOddBernoulliNumbersAreZero 2013-03-22 15:12:04 2013-03-22 15:12:04 alozano (2414) alozano (2414) 5 alozano (2414) Theorem msc 11B68 KummersCongruence CongruenceOfClausenAndVonStaudt