Is it possible for an improper integral to converge and its series to diverge?

Question

Can an improper Riemann integral converge while its infinite series diverges?

Answer 1

I presume you're asking about $\int_0^\infty f(x)dx$ vs. $\sum_{n=0}^\infty f(n)$.

In general, the answer is "Of course not." For example, consider a continuous, nonnegative function $f$ which is zero at every integer but such that $\int_k^{k+1}f(x)dx=1$ for all integers $k$ (so $f$ has a lot of "bumps" - for example, we can take $f(x)={\pi\over 2}\vert \sin {\pi x\over 2}\vert$). Then the sum indicated is just zero, since the sum only takes into account integral values of $x$; but the integral diverges to infinity.

If we demand that $f$ be nonincreasing, though, then we do get an equivalence; see https://en.wikipedia.org/wiki/Integral_test_for_convergence.