I have a question regarding a slight modification of the integral test.
Suppose that $a_k = f(k)$ for some continuous function $f:[1,\infty) \rightarrow [0,\infty)$, which satisfies $f(k)\rightarrow 0$ as $k\rightarrow \infty$. If $\sum_{k\geq 1} a_k$ converges, then $\int_{1}^{\infty} f(k)\,{\rm d}k$ converges.
My question is whether or not it's true obviously; I feel that it's not true because it doesn't say that $f(k)$ has to be a decreasing function like it does in the integral test, it only says that it has to go to 0. So, if this were to be true, why would this not be a useful test?
In contrast to what I said earlier, I've been trying to find a counterexample for a little while but can not seem to come up with one. If any of you can give me a hint for a counterexample or a hint on how to prove this statement, it would be greatly appreciated!
Let $$ f(x) = \frac{|\sin{\pi x}|}{x}. $$ Then $a_k = f(k) = 0$, so the series converges trivially, but it's not hard to see that the integral diverges, since $$ \int_{k}^{k+1} \frac{|\sin{\pi x}|}{x} \ge \int_{k}^{k+1} \frac{|\sin{\pi x}|}{k+1} = \frac{2}{(k+1)\pi}, $$ and $\sum_{k=0}^\infty \frac{2}{(k+1)\pi}$ is a divergent (harmonic) series.