The title does not quite capture the essence of the difficulty, please allow me to be more explicit here.

I thought of this question when I was trying out an open problem by Ovidiu Furdui(See problem 1.32 b of Furdui).

The problem reads(I changed some notations to make it more concise):

Suppose $f \in C([0,1])$, define $x_{n}=\sum_{k=1}^{n-1} f(\frac{k}{n})$, here n is a non-zero natural number. For $n \ge2$, define $y_{n}=x_{n}-x_{n-1}$. The question is , does the sequence $y_{n}$ converge under this assumption?

Obviously, if $y_{n}$ were to converge at all, then it has to converge to $\int_{0}^{1}f(x)dx$. This is a direct consequence of the Stolz's theorem(https://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem) applied to the regular partition of riemann sum of $f$.

What is interesting is that this sequence actually converges if we suppose $f$ is $C^1$. Furdui himself saw this as well. I will only give sketches.

Observe $\frac{k}{n+1} \le \frac{k}{n} \le \frac{k+1}{n+1}$ for the k we care about.

By Mean Value theorem, $f(\frac{k}{n+1})-f(\frac{k}{n})=-\frac{k}{n(n+1)}f^\prime(\theta_{n,k})$ ,where $\theta_{n,k} \in (\frac{k}{n+1},\frac{k}{n})$.

Take difference of $y_{n}-f(1)$ and $\frac{1}{n}$ of riemann sum of $-xf^\prime(x)$ with partition defined by $\frac{k}{n+1}$, by uniform continuity of $f^\prime$ on $[0,1]$ (which we have by compactness and continuity), we see this difference converges to 0. Integrate by parts gives the desired result.

Now, we see there is no way the above approach is going to work if $f$ is merely continuous. In fact, straightforward imitation will not work even if $f$ is assumed to be differentiable.

However, numerical tests on several bad functions, including Cantor function, show the result may remain valid under the continuity hypothesis. This suggests some approximation arguments may help.

We use $x_{n}(f)$ to mean the sequence defined by a fixed $f$. I think I tried to approximate $f$ uniformly by a sequence of polynomial $P_i$. I wanted to somehow make use of the results on the $C^1$ case. Here is how triangle inequality leads to a dead end:

$|y_{n}(f)-\int_{0}^{1}f(x)dx|=|(y_{n}(f)-y_{n}(P_i))+(\int_{0}^{1}P_i(x)dx-\int_{0}^{1}f(x)dx)+(y_{n}(P_i)-\int_{0}^{1}P_i(x)dx)| \le 2n|f-P_i|_{\infty }+|y_{n}(P_i)-\int_{0}^{1}P_i(x)dx|$.

Now we are out of luck: to make the first term on the right small for a fixed n, we need to raise $i$. The second term, which depends on i, is small possibly only if we raise n. Now this loop just goes on and on.

What would be a dream on true is, if $|y_{n}(P_i)-\int_{0}^{1}P_i dx|$ has some bound for all $n \ge n_i$ , then the same bound is satisfied(for all $n \ge n_i$) if we replace i by some $i^\prime \ge i$ . The result will follow immediately. Of course, I don't see why this will be true, particularly after analyzing carefully what made the $C^1$ case work:uniform continuity of derivative was used in a crucial way. Even if the derivatives of all the polynomials in the sequence were equicontinuous(which is rarely the case), one still has no control on the rate of convergence of the riemann sums of $-xP_i^\prime$ with the specific partition we chose.

Triangle inequality is not so useful for our purpose here, since it can at best give $O(n)$ bound on $y_{n}(f)$, which we know is in fact $o(n)$. I think one has to analyze carefully what these $y_{n}$ do to $f$ viewed as operator on $C([0,1])$. The obstacle with sum of difference of this form is that there is no good way to control the contribution from the increase of terms. Perhaps one also needs to choose carefully the sequence of polynomials converging to $f$.

Any suggestions of methods, or references which may contain useful tools, or possible counter-examples will be welcome. I already looked up some results on rate of convergence of polynomial, as well as estimate of riemann sum, nothing there seemed to be directly applicable to this particular situation. What may be useful in my opinion is a linear operator view point, though I am not quite sure which direction I should be looking into.

One reason why one would care about the convergence(this is why I care): It uses the same data as one uses to compute riemann sum, but simpler arithmetic operations.

Source of problem:

"Limits, Series, and Fractional Part Integrals: Problems in Mathematical Analysis", Ovidiu Furdiu.