Let $a_n$ be a real sequence such that $\sum\limits_{n\in \mathbb{N}}a_n $ is convergent. Let $b_n = 1 + a_n $. Then $\prod\limits_{n\in\mathbb{N}}b_n$ need not to converge. My question is if the above mentioned infinite product doesn't converge, then does it necessarily diverge to $0$?
It doesn't hold for complex sequences, but I am seeking an example for real case.
The answer is yes as under the hypothesis in the post (and assuming wlog $a_n \ne -1$ for any $n$ as having some $a_n=-1$ is a special case with various treatments when talking about infinite products $\Pi (1+a_n)$) the infinite product converges to a (finite) non zero number or diverges to zero depending whether $\sum a_n^2$ is finite or infinity - as $a_n \in \mathbb R$ it is clear that $\sum a_n^2$ has a limit being a sum of nonnegative numbers
The proof is fairly simple as $a_n$ convergent means $a_n \to 0$ hence $|a_n|<1/2, n \ge n_0$ and we can ignore the first $n_0$ terms for convergence purposes (here is where $a_n \ne -1$ comes into play as if some are actually $-1$ one can have different takes on what convergence of $\Pi (1+a_n)$ means - usually it's taken that the product without those finitely many $a_n =-1$ needs to be convergent in the usual sense for infinite products, so the limit of the partial products exists, is finite and non-zero).
But then $\log (1+a_n)=a_n-a_n^2/2 +c(n)a_n^2$ where $|c(n)|< 5/12 <1/2$
(by easy estimates of the Taylor series noting that $|a_n^k| \le \frac {a_n^2}{2^{k-2}}, n \ge 3$, so $\sum_{k \ge 3}\frac{|a_n|^k}{k} \le |a_n|^2\sum_{k \ge 3} \frac {1}{k 2^{k-2}} \le (|a_n|^2)(1/6+\sum_{k \ge 4} \frac {1}{ 2^{k-1}}) =1/6+1/4=5/12$)
Hence $\sum_{k=1}^n (\log (1+a_n)-a_n)$ is a sum of nonpositive terms say $-d_n$ st $ a_n^2/12 \le d_n \le 11 a_n^2 /12$, so has a limit which is either finite nonpositive or minus infinity depending whether $\sum a_n^2$ is finite or infinity, hence $\frac{\prod_{k=1}^n b_k}{e^{\sum_{k=1}^n a_k}}$ has a limit which is either in $(0,1]$ or $0$ depending whether $\sum a_n^2$ is finite or infinity and by hypothesis $e^{\sum a_n}$ has a finite (strictly) positive limit so we are done!