Convergence of logs imply convergence of series Let $a_n > -1$ for all $n \in \mathbb{N}$.
$\lim \limits_N \prod_{i=1}^N (1 + a_n) \in (0, +\infty)$ $\iff$ $\sum_{n = 1}^\infty a_n$ converges.  
I am trying to show $\implies $ first. 
Suppose $\lim \limits_N \prod_{n=1}^N(1 + a_n) = b \in (0, \infty)$. Then taking log, I got: 
$$   \lim_N \sum_{n=1}^N \log\left (1 + a_n \right ) = \log(b)$$
From this, I at least know that $\lim \limits_{n \to \infty} \left [  \log(1 + a_n) \right ] = 0$ which also implies $\lim a_n = 0$. No further progress...
This question is different from other questions I encountered because of the assumption that $a_n > -1$. 
 A: The case where $a_n \geqslant 0$ is easy, since 
$$\sum_{n=1}^m a_n < \prod_{n=1}^m (1+a_n) \leqslant \exp\left(\sum_{n=1}^m a_n \right)$$
Suppose $-1 < a_n < 0$ and let $b_n = -a_n$. Using $1-x < e^{-x}$ for $0 \leqslant x <1$ we have
$$0 < P_m = \prod_{n=1}^m(1-b_n) \leqslant \exp\left(- \sum_{n=1}^m b_n \right),$$
and it follows that divergence of $\sum a_n$ to $- \infty$ implies that $P_m \to 0$, that is the product $\prod(1+a_n)$ diverges (to $0$).
On the other hand, if $\sum a_n$ and , hence, $\sum b_n$ converges, then for any $\epsilon > 0$ there exists $N$ such that $0 \leqslant \sum_{n= N}^\infty b_n < \epsilon$.
Since $P_m$ is decreasing and bounded as
$$\frac{P_m}{P_{N-1}} =\prod_{n=N}^m(1- b_n) \geqslant 1 - \sum_{n=N}^mb_n > 1 - \epsilon,$$
it follows that $P_m$ converges to a finite limit, and, thus $\prod(1+a_n)$ is convergent.
Addendum
If the $a_n$ may assume either sign (infinitely often) then it may hold that the series and product converge or diverge together under additional assumptions. It can be proved, for example, that if $\sum a_n^2$ converges, then the series and product converge or diverge together.
See here for an example where $a_n$ changes sign and the product $\prod(1+a_n)$ converges but the series $\sum a_n$ diverges.
