- 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.