Can the limit of averages of $f(1),f(2),\dots, f(n)$ be expressed as an integral?

Question

If $\int_0^1 f(x) dx$ exists then, of course, $$ \lim_{n\to\infty} \frac{f(\frac{1}{n})+f(\frac{2}{n})+\ldots+f(\frac{n}{n})}{n} = \int_0^1 f(x) dx. $$ I would like to know is there a similar formula for $$ \lim_{n\to\infty} \frac{f(1)+f(2)+\ldots+f(n)}{n} $$ under some assumptions on $f$.

Answer 1

It looks like a Cesaro sommation (but Cesaro cared more about series than functions)

https://en.wikipedia.org/wiki/Ces%C3%A0ro_summation

Answer 2

If $f(x)$ is monotonic increasing, then $f(n) < \int_n^{n+1} f(x) dx < f(n+1) $.

Therefore $\frac1{n}\sum_{k=0}^{n-1} f(k) < \int_0^n f(x) dx < \frac1{n}\sum_{k=1}^{n} f(k) $ or $\frac1{n}(f(0)-f(n)) < \int_0^n f(x) dx-\frac1{n}\sum_{k=1}^{n} f(k) < 0 $.

Therefore, if $f(x)$ is monotonic increasing and $\dfrac{f(n)}{n} \to 0 $, then $\lim_{n \to \infty} \int_0^n f(x) dx-\frac1{n}\sum_{k=1}^{n} f(k) \to 0 $.

This is a sufficient condition.

Other sufficient conditions can be similarly derived.

You might see what you can derive if $f(x)$ is monotonic decreasing.

Have fun.