Series about Euler-Maclaurin formula The Euler-Maclaurin formula says (from Concrete Mathematics section 9.5)

\[
  \sum_{a\le{}k< b}f(k)=\int_a^bf(x)dx+\left.\sum_{k=1}^m\frac{B_k}{k!}f^{(k-1)}(x)\right|_a^b+R_m
  \]
  where $\displaystyle{}R_m=(-1)^{m+1}\int_a^b\frac{B_m(\{x\})}{m!}f^{(m)}(x)dx$, integers $a\le{}b$ and $m\ge1$.

However, let $a=0$, $b=n$, and consider the series
\[
\int_0^nf(x)dx+\sum_{k=1}^\infty\frac{B_k}{k!}\left(f^{(k-1)}(n)-f^{(k-1)}(0)\right)
\]
Unfortuntely, the series usually diverges. So, I wonder whether there is some sufficient condition (easy to verify) that the series converges?
Especially when the series $\sum_{k=0}^\infty{}f(k)$ converges, or more stronger, converges absolutely, what about those conditions?
 A: We have the estimate
$$
\lim_{k\to\infty}\left(\frac{\left|B_{2k}\right|}{(2k)!}\right)^{\Large{\frac{1}{2k}}}=\frac{1}{2\pi}\tag{1}
$$
Thus, for the series
$$
\sum_{k=1}^\infty\frac{B_{2k}}{(2k)!}\left(f^{(2k-1)}(n)-f^{(2k-1)}(0)\right)\tag{2}
$$
to converge, we need
$$
\lim_{k\to\infty}\left(f^{(2k-1)}(n)-f^{(2k-1)}(0)\right)^{\Large{\frac{1}{2k}}}\le2\pi\tag{3}
$$
Since $R$, the radius of convergence of $f$, satisfies
$$
\frac1R=\limsup_{k\to\infty}\left(\frac{f^{(k)}(x)}{k!}\right)^{\Large{\frac{1}{k}}}\tag{4}
$$
$(3)$ and $(4)$ would seem to require that $\dfrac1R=0$; that is, $f$ be entire.
In fact, according to this result, if the Fourier Transform of $f$ is contained in $[-1,1]$, then
$$
\limsup_{k\to\infty}\left(f^{(k)}\right)^{\Large{\frac{1}{k}}}\le2\pi\tag{5}
$$
and therefore, the series in $(2)$ converges.
Of course, the Euler-Maclaurin Sum Formula is finite on polynomials, and so the series in $(2)$ also converges for polynomials.
A: 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. 
