convergence of integral vs convergence of infinite series Problem:
Let $f\in C^{1}([0,\infty ))$ such that: $\int_{1}^{\infty }\left | f^{'}(x) \right |dx$ converges. The question is to prove the following: 
$\left ( \sum_{n=1}^{\infty }f(n) \right )$ converges $\Leftrightarrow \left ( \int_{1}^{\infty }f(x)dx \right )$ converges
I don't know how to prove it. For the direction: $\Leftarrow $ I was trying to use the definition of Rieamann integrals as an infinite sum where the mesh goes to zero, and somehow try to prove that $\left ( \sum_{n=1}^{\infty }f(n) \right )$ converges. 
Any solution or ideas for this problem?
 A: Suppose first that the sum converges. By the fundamental theorem of calculus, for each $x\in [n,n+1]$, 
$$f(x) = f(n) + \int_{n} ^x f'(t) dt$$
and therefore
$$|f(x)| \leq |f(n)| + \int_n ^x |f'(t)| dt\;.$$
Integrating and summing over all $n$, we get
$$\begin{align*}
\int_1 ^\infty |f(x)| dx &= \sum_{n=1} ^\infty \int_n ^{n+1} |f(x)| dx\\
&\leq \sum_{n=1} ^\infty |f(n)| + \sum_{n=1} ^\infty \int_n ^{n+1}\int_n ^x |f'(t)| dt dx \;.\tag{1}\end{align*}$$ 
We have 
$$\begin{align*}
\int_n ^{n+1}\int_n ^x |f'(t)| dtdx &= \int_{n} ^{n+1} \int_t ^{n+1} |f'(t)| dxdt \\
&= \int_{n} ^{n+1} (n+1 - t)|f'(t)| dt \\
&\leq  \int_{n} ^{n+1} |f'(t)| dt\;.\end{align*}$$
So, the final sum in $(1)$ is at most $$\sum_{n=1} ^\infty \int_n ^{n+1}|f'(t)| dt = \int_1 ^\infty |f'(t)| dt\;.$$ Thus
$$\int_1 ^\infty |f(x)| dx  \leq \sum_{n=1} ^\infty|f(n)| + \int_1 ^\infty |f'(t)| dt < \infty\;.$$ 
For the converse, we use a similar argument, beginning from the equation
$$f(n) = f(x) -  \int_{n} ^x f'(t) dt$$
for each $x\in [n, n+1]$. 
