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?

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It diverges, yes, but the behavior of the Euler-Maclaurin series is usually asymptotic; you can truncate the series up to some point, and that truncation is a good approximation. – J. M. Jun 9 '12 at 14:08
@J.M. Yes, and O-manipulation usually works well, but I wonder when it converges. – Frank Science Jun 10 '12 at 2:12

We have the estimate $$\lim_{k\to\infty}\left(\frac{B_{2k}}{(2k)!}\right)^{\Large{\frac{1}{2k}}}=\frac{1}{2\pi}\tag{1}$$ 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.