Let $f: [0,\infty)\to \Bbb R$ be a positive,decreasing monotonic function.
Prove the following statement for every a>0 providing the integral on the right side converges.
First I managed to prove that the series on the left size converges using the integral test, but now I'm having a hard time proving that the equality above is actually true.
Since the function is monotone decreasing, an upper sum for the integral would be $a\sum_{n=0}^\infty f(an)$, and a lower sum would be $a\sum_{n=1}^\infty f(an)$ The difference between these two is $af(0)$, which approaches zero.
Let $B$ be a finite number so that the integral from $0$ to $B$ is within $\epsilon$ of the integral from $0$ to $\infty$. Let $C>B$. The upper sum and lower sum are now
$$U(a,C)=a\sum_{n=0}^{\lceil C/a\rceil}f(an)\\
L(a,C)=a\sum_{n=1}^{\lfloor C/a\rfloor}f(an)$$
The lower sum is within $af(0)+af(B)$ of the upper sum, and therefore within that distance of the finite integral. The finite integral is within $\epsilon$ of the infinite integral, and will remain within $\epsilon$ no matter how large $B$ gets. So the sum $L(a,C)$ has a limit as $C\to\infty$ within $\epsilon+af(0)+af(B)$ of the infinite integral. First let $a\to0^+$ then let $\epsilon\to0^+$
We are assuming $\int_0^\infty f$ converges, so it's a nice finite positive number. Let $S_a = \sum_{n=1}^{\infty}af(na).$ Because $f$ is decreasing, we have, from the usual comparison with areas of rectangles,
$$S_a \le \int_0^\infty f \le S_a + f(0)a.$$
Thus $|S_a-\int_0^\infty f| \le f(0)a,$ which implies $\lim_{a\to 0^+} S_a = \int_0^\infty f$ as desired.
