From Wikipedia
\begin{eqnarray*}
B_{2 k} &=&
(-1)^{k+1} \frac{2 \, (2 \, k)!}{ (2 \, \pi)^{2 k}}
\sum_{n=1}^{\infty} \frac{1}{n^{2k}}
= (-1)^{k+1} \frac{2 \, (2 \, k)!}{ (2 \, \pi)^{2 k}}
+ \mathcal{O} \left ( \left ( \frac{1}{2} \right)^{2k} \right )
\end{eqnarray*}
Then:
\begin{eqnarray}
\lim_{k \to \infty}
\left | \frac{B_{2k}}{2 \ (2k)!} \right |^{\frac{1}{2k}} = \frac{1}{2 \pi}
\label{toseries}.
\end{eqnarray}
Since $\lim_{k \to \infty} (1/2)^{1/2k}=1$ we can also write
\begin{eqnarray}
\lim_{k \to \infty}
\left | \frac{B_{2k}}{(2k)!} \right |^{\frac{1}{2k}} = \frac{1}{2 \pi}
\label{toseries2}.
\end{eqnarray}
Let us call the $k$ term of the series:
\begin{eqnarray*}
a_k = \left . \frac{B_{2k}}{(2k)!} f^{(2k-1)}(x) \; \right |_{x=a}^b
\end{eqnarray*}
then the radius of convergence of the series is found from the limit
$\lim_{k \to \infty} |a_{k+1}|/ |a_{k}|$. In this case we have
\begin{eqnarray*}
R =
\lim_{k \to \infty}
\left |
\frac{ \left . B_{2k +2} f^{(2k+1)}(x) \; \right |_{x=a}^b (2k)!}
{\left . B_{2k} f^{(2k-1)} \right |_{x=a}^b (2k+2)!} \right |
= \lim_{k \to \infty} \left | \frac{ \left .f^{(2k+1)} \; \right |_{x=a}^b}
{\left . f^{(2k-1)} \;\right |_{x=a}^b} \right |.
\end{eqnarray*}
We need $R < 1$ for convergence.
That is, we need to have the condition:
\begin{eqnarray}
\lim_{k \to \infty} \left | \frac{ \left .f^{(2k+1)} \; \right |_{x=a}^b}
{\left . f^{(2k-1)} \;\right |_{x=a}^b} \right | < 1
\end{eqnarray}
for convergence of the series.
