I left the Maclaurin expansion of the function $f(x)=x/(e^x-1)$
$$\frac{x}{e^x-1}=\sum_{k=1}^{\infty} B_k\frac{x^k}{k!}=B_0\frac{x^0}{0!}+B_1\frac{x^1}{1!}+\sum_{k=2}^{\infty}B_k\frac{x^k}{k!}$$ $$ \frac{x}{e^x-1}=1-\frac12x+\sum_{k=2}^{\infty}B_k\frac{x^k}{k!}$$ $$ \frac{x}{e^x-1}+\frac12x=1+\sum_{k=2}^{\infty}B_k\frac{x^k}{k!} $$ $$\frac{x}{e^x-1}+\frac12x=x\left( \frac{1}{e^x-1}+\frac12 \right)=x\left( \frac{2+e^x-1}{2(e^x-1)} \right)=\frac x2\left ( \frac{e^x+1}{e^x-1} \right )$$ $$\frac x2\left ( \frac{e^x+1}{e^x-1} \right )\cdot\frac{e^{-x/2}}{e^{-x/2}}=\frac x2\cdot \left( \frac{e^{x/2}+e^{-x/2}}{e^{x/2}-e^{-x/2}} \right) $$
knowing that $$\sinh(t)=\frac{e^t-e^{-t}}{2}\text{ and also, }\cosh(t)=\frac{e^t+e^{-t}}{2}\Longrightarrow$$ $$\tanh(t)=\frac{\sinh(t)}{\cosh(t)}=\frac{e^t-e^{-t}}{e^t+e^{-t}}\Longrightarrow \coth(t)=\frac{e^t+e^{-t}}{e^t-e^{-t}} $$
Therefore $$\frac x2\cdot \left( \frac{e^{x/2}+e^{-x/2}}{e^{x/2}-e^{-x/2}} \right)=\frac{x}{2}\coth\left( \frac{x}{2}\right) $$
$$ \frac x2\coth\left(\frac x2\right)=1+\sum_{k=1}^{\infty} B_k\frac{x^k}{k!} $$
In a text I found here, we have the function $\frac x2\coth\left(\frac x2\right)$ is an even function (OK!), this text concludes that $B_{2k+1}=0$.
And that I did not understand, because the fact of the function be even assures me that the Bernoulli numbers are zero whenever the index is odd?
Assume that $f(z)$ is an even analytic function in a neighbourhood of zero, $$ f(z) = \sum_{k\geq 0} a_k z^{k}. $$ We have: $$ 0=f(z)-f(-z) = 2\,\sum_{k\geq 0} a_{2k+1}\, z^{2k+1} $$ hence for every $k\geq 0$, $a_{2k+1}=0$.