Convergence of $\sum_{n=1}^{\infty}f(n)$ $\iff $ convergence of $\int_1^{\infty}$f(x)dx

Question

f is a real valued $C^1$ function on [0,$\infty$]. Suppose that $\int_1^{\infty}|f'(x)|dx$ converges. Show that Convergence of $\sum_{n=1}^{\infty}f(n)$ $\iff $ convergence of $\int_1^{\infty}$f(x)dx.

My thought is like this: $\int_1^{\infty}|f'(x)|dx$ converge $\Rightarrow$ $\sum_{n=1}^{\infty}\int_n^{n+1}|f'(x)|dx $ converge $\Rightarrow$ $\lim_{n \to \infty} \int_n^{n+1}|f'(x)|dx = 0$ $\Rightarrow$ $\lim_{x \to \infty}f'(x) = 0$

If $\sum_{n=1}^{\infty}f(n)$ converge, then $\lim_{n \to \infty}f(n) = 0$ $\Rightarrow$ $\forall \epsilon>0$, $\exists n_0 >0$, such that n>$n_0$ $\Rightarrow$ $|f(n)|<\epsilon$.

Then I don't know what to continue.

Answer 1

My hint:

$$\left|\int_k^{k+1}f(x)\,dx-f(k)\right| \leq \max_{k<x<k+1} |f(x)-f(k)| \leq \int_k^{k+1} |f'(x)|\,dx $$

Answer 2

Hint (based on mlu's comment). Let $f:[1,\infty)\to[0,\infty)$ be decreasing then

$$0\leq \lim_{n\to\infty}\left(\sum_{k=1}^nf(k)-\int_1^{n+1}f(x)~\mathrm{d} x\right)\leq f(1).$$