Continuous function whose series of function values converges but its improper integral doesn't converge

Question

Does a continuous function $f: [1, \infty) \to [0,\infty)$ exist, such that $\sum_{k=1}^\infty f(k)$ converges but $\int_1^\infty f(x)\,dx$ does not converge?

I need help coming up with an example. If $\sum_{k=1}^\infty f(k)$ converges, we know that $f(k) \xrightarrow[k \to \infty]{}0$. I tried to come up with a function that's constant $0$ on the natural numbers, so the convergence of the series is guaranteed. One idea was $|\sin(x \pi)|$, which also shows that $f$ itself doesn't necessarily need to converge, however I have no idea how to integrate that (Intuitively that integral shouldn't converge though). If there is an elementary example where it's really easy to see why it's true, I would appreciate it.

Answer 1

Your suggestion works. The function $f: [1, \infty) \to [0, \infty)$ given by $$ f(t) = | \sin(\pi t) | $$ is periodic with period $1$, i.e., $f(t + 1) = f(t)$ for all $t$ in the domain (and $T = 1$ is the smallest such).

It's clear that $f(k) = 0$ for all $k \in \mathbb{N}$, so $$ \sum_{k=1}^\infty f(k) = 0 < \infty, $$ and by breaking up $[1, \infty) = [1, 2] \cup [2, 3] \cup \cdots \cup [k, k+1] \cup \cdots$ $$ \begin{aligned} \int_1^\infty \! f(t) \, \mathrm{d}t &= \int_1^\infty \! |\sin(\pi t)| \, \mathrm{d}t \\ &= \sum_{k=1}^\infty \int_k^{k+1} \!\!\!\!\!\sin(\pi t) \, \mathrm{d}t \\ &= \sum_{k=1}^\infty \int_0^1 \sin(\pi t) \, \mathrm{d}t \\ &= \sum_{k=1}^\infty \Bigl.\bigl( -\tfrac1\pi \cos(\pi t) \bigr)\Bigr|_0^1 \\ &= \sum_{k=1}^\infty \tfrac2\pi, \end{aligned} $$ which clearly diverges to $\infty$.

Here's a picture of $f(t)$ (in red) and it's integral function $F(t) = \displaystyle\int_1^t f(u) \, \mathrm{d}u$ (in blue).