$\sum\limits_{n=1}^\infty \log(1+a_n)$ converges absolutely $\iff\sum\limits_{n=1}^\infty a_n$ converges absolutely. 
$$\sum\limits_{n=1}^\infty \log(1+a_n) \text{ converges absolutely}
 \Leftrightarrow \sum_{n=1}^\infty a_n \text{ converges absolutely}.$$

How to prove this,
Suppose $$\sum_{n=1}^\infty a_n \text{ converges absolutely}.$$ Let $u_{n}=a_{n}$ and $v_{n}=\log(1+a_n)$, then $$\lim_{n\to\infty} \frac{u_{n}}{v_{n}}=1>0 \implies\sum_{n=1}^\infty \log(1+ a_n) \text{ converges absolutely}.$$ How to prove the converse part? 
 A: Hint: From the definition of $\ln'(1),$ we have
$$\lim_{u\to 0}\frac{\ln (1+u)}{u} = 1.$$
Thus there is $a>0$ such that
$$\frac{1}{2}\le \left|\frac{\ln (1+u)}{u}\right| \le \frac{3}{2}$$
for $u\in (-a,a),u\ne0.$
A: The limit comparison test says that if you have two sequences $\{a_n\}_{n=1}^\infty$ and $\{b_n\}_{n=1}^\infty$ such that $\lim_{n\to \infty}\frac{a_n}{b_n}=c$ with $0<c<\infty$. Then $\sum_{n=1}^\infty a_n<\infty$ if and only if $\sum_{n=1}^\infty b_n<\infty$.
So we have to prove that $\lim_{n\to \infty}\frac{\vert ln(1+x_n)\vert}{\vert x_n\vert}=c$ with $0<c<\infty$. To do this we observe that $\lim_{x\to 0}\frac{\vert ln(1+x)\vert}{\vert x\vert}=1$ by the L'Hopital Rule. As in any of the two cases ($\sum_{n=1}^\infty x_n$ converges absolutely or $\sum_{n=1}^\infty ln(1+x_n)$ converges absolutely) we will have that $\lim_{n\rightarrow \infty}x_n=0$, then $\lim_{n\to\infty}\frac{\vert ln(1+x_n)\vert}{\vert x_n\vert}=1$.
A: [This is essentially from @Guy Fsone's answer. However, it is rewritten in a significantly different way.]
First of all, observe that (by continuity of the exponential function and the logarithm)
$$\lim_{n\to\infty } \log(1+a_n) =0 \Leftrightarrow \lim_{n\to\infty }a_n = 0.$$


*

*If $\lim_{n\to\infty}a_n=0$ is not true, (note carefully that
this assumption is fundamentally different from
"$\lim_{n\to\infty}a_n\neq 0$"), then both $\sum_{n=1}^\infty
   |\log(1+a_n)|$ and $\sum_{n=1}^\infty |a_n|$ diverge.

*Assuming that $a_n \to 0$ as $n\to\infty$, we have $$\lim_{n\to\infty
   }\frac{|\log(1+a_n)|}{|a_n|} =\left|\lim_{h\to
   0}\frac{\log(1+h)}{h}\right| = 1\tag{1}$$ where we use the fact that
the derivative of $x\mapsto \log x$ at $x=1$ is $1$ and the
continuity of the absolute value function. Without loss of generality, 
we assume here that $|a_n|>0$ for all $n$. 
It follows from (1) that there exists $n_0$ such that for $n>n_0$
$$\left|\frac{|\log(1+a_n)|}{|a_n|} -1\right|<1/2 $$ which implies
that  $$ \frac12 |a_n|< |\log(1+a_n)|<\frac32|a_n|~~~\forall ~~~n>n_0
   $$ and thus $$ \frac12 \sum_{n>n_0}|a_n|<
   \sum_{n>n_0}|\log(1+a_n)|<\frac32 \sum_{n>n_0}|a_n|.$$
This estimate completes the proof.
A: My way:
$$\sum_{n=1}^\infty \log(a_n+1) = \log\prod_{n=1}^\infty (a_n+1)$$
since the series converges absolutely so does the product. Hence $a_n +1 \rightarrow 1$ and so $a_n \rightarrow 0$.
Now the nontrivial step: $a_n \rightarrow 0$, and in proximity of $0$ you have $\log(1+x) = x + o(x^2)$ so dividing by $x$ for any $\epsilon > 0$ exists $n$ such that
$$\frac{|\log(a_n+1)|}{|a_n|} \leq 1+ \epsilon$$
or in other words
$$(1-\epsilon)|a_n| \leq |\log(a_n+1)| \leq (1+\epsilon)|a_n|$$
Apply the sum to the last inequality and you'll prove the proposition both ways.
