# Condition for monotonic decrease of $a_n = \sum_{k=1}^n f\left(\cfrac{k}{n}\right)\cfrac1{n}$

When $f(x)$ is continuous over $[0,1]$ and differentiable over $(0,1)$, what is the condition that the $a_n$ would monotonically decrease for all $n$ $(n\ge1)$? $$a_n = \sum_{k=1}^n f\left(\cfrac{k}{n}\right)\cfrac1{n}$$

Similarly, what is the condition that the $b_n$ would monotonically increase for all $n$? $$b_n = \sum_{k=0}^{n-1} f\left(\cfrac{k}{n}\right)\cfrac1{n}$$ This is a question on behavior of an approximation error in numerical evaluation of an integral, as $$Error_n = a_n - \int_0^1f(x) dx$$ My guess is that when $f(x)$ is increasing and is a convex (upward or downward) then $a_n$ and/or $b_n$ would be decreasing/increasing monotonically, but I am not sure whether it is sufficient or necessary. The reason for the reasoning of convex is because there exists an obvious counter-example where $f(x)$ itself is a stair-case shape.

• I don't think that $b_n \le \int_0^1f(x) dx\le a_n$ is generally true, unless you assume that $f$ is increasing. – Martin R Feb 21 '16 at 20:43
• @MartinR Yeah, you're right. Let me edit that comment. – Kay K. Feb 21 '16 at 20:59