real analysis - Convergence of $sum_{k=1}^{n} f(k) - int_{1}^{n} f(x) dx$

I had asked this question sometime ago here. Now I have a question which I think is related to it.

Let $f$ be an increasing function (continuous, of course!) with $f(1)=0$.

Consider the sequence $s_{n}= ( \sum\limits_{k=1}^{n} f(k) - \int\limits_{1}^{n} f(x) dx )$.

When does $s_{n}$ converge?

Answer

Qiaochu was on the right track to use an integral-to-sum formula, but it sounds like you want the Abel-Plana summation formula:

$$\lim_{n\to\infty}\left(\sum_{k=m}^n f(k)-\int_m^n f(u)\mathrm{d}u\right)=\frac{f(m)}{2}-\int_{-\infty}^\infty \left(\frac{|t|}{\exp(|2\pi t|)-1}\right)\left(\frac{f(m+it)-f(m-it)}{2it}\right)\mathrm{d}t$$

This is used for instance to evaluate the Stieltjes constants. If the expression on the right hand side is convergent, then it is equivalent to the left hand side.

Adendum for Chandru:

Definitely $f(z)$ should be analytic, or at least analytic in the region where $\Re z\geq m$. Per Henrici's "Applied and Computational Complex Analysis", the additional conditions are

$$\lim_{t\to\infty}f(u\pm it)=0$$

uniformly in $u$, and that

$$\lim_{t\to\infty}|f(u\pm it)|\exp(\mp 2\pi t)=0$$

uniformly in $u$.

EDIT:

For those scratching their head on just how Abel-Plana and Euler-Maclaurin are connected, the identity

$$\int_{-\infty}^\infty \left(\frac{|t|}{\exp(|2\pi t|)-1}\right)|t|^{2n-2}\mathrm{d}t=\frac{|B_{2n}|}{2n}$$

might be of interest.