Let $f:(0,\infty)\to [0,\infty)$ be a continuously differentiable function.
Assume that $\int_0^{\infty} f(x)dx<\infty$ and that $\int_0^{\infty}|f'(x)|dx<\infty$.
Claim: $\sum_{n=1}^{\infty}f(n)$ converges.
I have tried a few things. My gut tells me to try applying the integral test but I need a monotone decreasing function to compare the series with.
My "bright" idea is to define a step function $g(x)=\sup_{x\geq n}f(x)$ on the interval $[n,n+1)$. Then $g(x)$ is monotone decreasing and bounds the function $\phi(x):=\sum_{n\geq 1}f(n)\chi_{[n,n+1)}(x)$ from above.
So $g(x)\geq \phi(x)\implies \int g dx \geq \int \phi dx=\sum_{n\geq 1} f(n)$ and if I can show $g(x)$ is finite then I am done.
It should be clear by the assumptions that $\lim_{x\to\infty}f(x)=0$ and that the function $f(x)$ is bounded. Does there exist a constant $M>0$ such that $\int Mf(x)dx \geq \int g(x)dx$? Because this would be fantastic.
Thank you for your help!