- Assume that is a monotonically decreasing function for all and that
Prove that the improper integral and the series

both converge or both diverge.

- Give a counterexample to the theorem in part (a) in the case that is not monotonic, i.e., find a non-monotonic function such that converges but diverges.

**Incomplete.**

How about f(x)=0 if x are integers, and f(x)=1/x otherwise?

the integral is just 0

F(x) could be sin(2*pi*x)

sin(2*pi*x) series does not converge either. The answer could be – this will make the series converge, while most probably the integral diverges, but idk how to show that.