$\sum\limits_{n=1}^\infty \log(1+a_n)$ 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 $$\lim_{n\to\infty} \frac{u_{n}}{v_{n}}=1&gt;0 \implies\sum_{n=1}^\infty \log(1+ a_n) \text{ converges absolutely}.$$ How to prove the converse part?
Hint: From the definition of $\ln'(1),$ we have

$$\lim_{u\to 0}\frac{\ln (1+u)}{u} = 1.$$

Thus there is $a&gt;0$ such that

$$\frac{1}{2}\le \left|\frac{\ln (1+u)}{u}\right| \le \frac{3}{2}$$

for $u\in (-a,a),u\ne0.$
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 $\epsilon &gt; 0$ 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.