Similar to the Integral Test In this answer, it is said that

If $\int_{1}^{\infty}\vert f'(x)\rvert \,\mathrm{d}x$ converges, then
  $\sum_{k=1}^{\infty}f(k)$ and $\int_{1}^{\infty}f(x)\,\mathrm{d}x$ converge or diverge together.

In the proof, it says that

$$\sum_{k=2}^{\infty}\left | f(k)-\int_{k-1}^{k}f(x)\,\mathrm{d}x \right |<\infty$$
  implies the convergence of
  $$\sum_{k=2}^{\infty} f(k)-\int_{1}^{\infty}f(x)\,\mathrm{d}x$$

Why does this implication hold? If think some conditions of $f$ is missing here. 
Also, this idea brings me another question: Let $(a_n)$ and $(b_n)$ be two sequences of numbers. Is it true in general that if $\sum_{n=1}^{\infty}|a_n+b_n|$ converges, then $\sum_{n=1}^{\infty}a_n$ and $\sum_{n=1}^{\infty}b_n$ converge? It is clear if those sequences were positive, but what if at least of them were negative?
 A: It was shown that we have (absolute) convergence of
$$\sum_{k=2}^\infty \left|f(k) - \int_{k-1}^k f(x) \, dx \right|$$
By the comparison test, this implies convergence  of 
$$\sum_{k=2}^\infty \left[f(k) - \int_{k-1}^k f(x) \, dx \right],$$
and there exists a finite limit $L$ such that
$$L = \sum_{k=2}^\infty \left[f(k) - \int_{k-1}^k f(x) \, dx \right] = \lim_{n \to \infty}\left[ \sum_{k=2}^n f(k) - \sum_{k=2}^n \int_{k-1}^k f(x) \, dx\right] \\ =  \lim_{n \to \infty}  \left[ \underbrace{\sum_{k=2}^n f(k)}_{A_n} - \underbrace{\int_{1}^n f(x) \, dx}_{B_n} \right]$$
If either one of $A_n$ or $B_n$ converge, then the other must converge.  For example, if $A_n \to A$ as $n \to \infty$, then
$$\lim_{n\to \infty} B_n = \lim_{n\to \infty}[A_n - (A_n - B_n)] = \lim_{n\to \infty}A_n - \lim_{n\to \infty}(A_n - B_n) = A- L$$
If either one of $A_n$ and $B_n$ does not converge (diverges to $\pm \infty$ or the limit fails to exist), then the other must not converge.
For example, if the limit of $A_n$ fails to exist we have
$$\limsup_{n \to \infty} \,A_n \neq \liminf_{n \to \infty} \,A_n,$$
and 
$$\begin{align} \limsup_{n \to \infty} \,B_n &= \limsup_{n \to \infty} \,\, [A_n - (A_n - B_n)] \\ &= \limsup_{n \to \infty} \, A_n - \limsup_{n \to \infty}\,\, (A_n - B_n) \\&=  \limsup_{n \to \infty} \, A_n - L \\ &\neq \liminf_{n \to \infty} \,\, A_n - L\\ &= \liminf_{n \to \infty} \, A_n - \liminf_{n \to \infty} \,\,(A_n - B_n) \\ &=  \liminf_{n \to \infty} \,\, [A_n - (A_n - B_n)] \\ &= \liminf_{n \to \infty} \, B_n \end{align} $$
A: Note that, by the additivity of the integral and the triangle inequality,
\begin{align} \left| \sum_{k=2}^\infty f(k)- \int_0^1 f(x) \mathrm dx \right| &= \left| \sum_{k=2}^\infty f(k)- \sum_{k=2}^\infty \int_{k-1}^k f(x) \mathrm dx \right|\\
&= \left|\sum_{k=2}^\infty \left(f(k)- \int_{k-1}^k f(x) \mathrm dx\right)\right|\\
&\leq \sum_{k=2}^\infty \left|f(k)- \int_{k-1}^k f(x) \mathrm dx\right| < \infty.
\end{align}
For your second question, take $a_n = n = -b_n$. Then $ a_n + b_n = 0$, so
$ \sum |a_n+b_n| = 0, $ but $\sum a_n = -\sum b_n = \sum n$ which is divergent.
