Recall that, for $k\geq 0$ the Bernoulli numbers $B_k$ are defined as the coefficients in the Taylor expansion: \begin{eqnarray} \label{ber} \frac{t}{e^t-1}=\sum_{k\geq 0} B_k \frac{t^k}{k!}. \end{eqnarray}Just to name a few: $$B_0=1,\quad B_1=-\frac{1}{2},\quad B_2=\frac{1}{6},\quad B_3=0,\quad B_4=-\frac{1}{30},\ B_5=0,\ldots,\ B_{10}=\frac{5}{66},\ldots $$

Proof. From the right hand side of () we extract the term corresponding to $k=1$ \begin{eqnarray} \label{ber2} \frac{t}{e^t-1}=-\frac{t}{2}+\sum_{k\geq 0,\ k\neq 1} B_k \frac{t^k}{k!}. \end{eqnarray}Thus: \begin{eqnarray} \label{ber21} \frac{t}{e^t-1}+\frac{t}{2}=\sum_{k\geq 0,\ k\neq 1} B_k \frac{t^k}{k!} \end{eqnarray}and the left hand side can be rewritten as: \begin{eqnarray} \label{ber3} \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}}. \end{eqnarray}Hence, if one replaces $t$ by $-t$ then () is unchanged. Since () is the left hand side of (), 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 $B_k=(-1)^kB_k$ for $k\neq 1$ We conclude that if $k\geq 3$ and $k$ is odd, $B_k=0$