When do Riemann sums converge monotonically? Let $f : [0, 1] \to \mathbb{R}$ be a 'well behaved' function and define $c_n$ by the Riemann sums
$$
c_n := \frac{1}{n} \sum_{k = 1}^n f\left(\frac{k - \theta}{n}\right)
$$
for some given $\theta \in [0, 1]$. Then obviously $c_n$ will converge to the integral of $f$. Under which conditions on $f$ will this convergence be monotone? In particular, will the convergence be monotone if $f$ is twice differentiable with $f > 0$, $f' < 0$ and $f'' > 0$?
 A: The answer to your final question is no for arbitrary $\theta$.  Specifically, let $f(x) = 1/\sqrt(x)$ for $\epsilon < x <= 1$ and define $f$ however you like for $0 \le x < \epsilon$ << .01, say.  (This can be done in a way that maintains the concavity and twice-differentiability of $f$.)  With $\theta = 3/4$, the sequence of Riemann sums begins (with $n=1$):
2., 2.04667, 2.056, 2.0579, 2.05757, 2.05647, 2.05511, 2.05368, ...,
peaking at $n=4$.  (The sequence is monotonic after that.)
The problem occurs because there is a section at the end of the graph of $f$, in this case near zero, that grows quickly enough to appreciably alter the very first term of the sum once $n$ grows large enough; this section is not sampled by the sum for smaller $n$.  By modifying $f$ appropriately within disjoint intervals converging to 0, you should be able to construct an example whose sequence of Riemann sums wiggles arbitrarily many times.  What this hand-waving argument does not settle is whether this process can be carried out ad infinitum: the concern is that the resulting $f$ might not be continuous at 0, which conceivably is part of your concept of "well-behaved" (although it is not necessary for the theory of Riemann integration, which permits some points of discontinuity).
A: Well, if $f$ is decreasing and convex, it is true that $c_n$ increases monotonically, if $\theta<0$. 
The proof is fairly straightforward, left as an exercise to the reader.
See the question on MO.
