We assume that $\sum |a_n|^{2}$ converges, then I want to conclude that $\prod (1+a_n)$ converges to a non zero element $\iff$ the series $\sum a_n$ converges.
My attempt
If $\prod (1+a_n)$ converges to a non zero element then we can write
$$\prod (1+a_n)= \prod \exp(\ln(1+a_n))= \exp(\sum \ln (1+a_n)) \le \exp(\sum |\ln (1+a_n)|)$$
Then using that $\sum |a_n|^{2}$ converges we can choose $n$ such that $|a_n|^2 < \frac{1}{4}$ and we know that $|ln(1+z)|\le 2 |z|$ if $|z|< \frac{1}{2}$ using the series expansion, we get that
$$\exp(\sum |\ln (1+a_n)|) \le \exp(\sum 2|a_n|)$$
For the converse I want to use the result that says that if $\sum |a_n|$ converges then $\prod (1+a_n)$ converges so since we have that $\sum |a_n|^{2}$ converges then $\sum |a_n|$ converges and $\prod (1+a_n)$ does too.
Questions
But from here I don't know if I am right, how to conclude and solve the converse part to say that we have a non zero limit, and another thing Can someone provide explicit examples of a sequence of complex numbers $\{a_n\}$ such that $\sum a_n$ converges but $\prod (1+a_n)$ diverges and the other way around (This is $\prod (1+a_n)$ converges but $\sum a_n$ diverges )?
Thanks a lot in advance.
Best Answer
If $\sum |a_n|^2$ converges, then $|a_n| \to 0$ and there is a positive integer $N$ such that $|a_n| < 1/2$ for all $n > N$.
We then have
$$| \log(1 + a_n) - a_n| = \left|\sum_{k=2}^\infty (-1)^{k+1}\frac{a_n^k}{k} \right| \leqslant \sum_{k=2}^\infty \frac{|a_n|^k}{k} \leqslant \frac1{2}|a_n|^2 \sum_{k=0}^\infty |a_n|^k \\ = \frac{|a_n|^2}{2(1 - |a_n|)} < |a_n|^2 $$
Hence, the series $\sum (\log(1+a_n) - a_n)$ is absolutely convergent. Therefore, $\sum a_n$ and $\sum \log(1+a_n)$ converge or diverge together , which implies that $\sum a_n$ and $\prod (1+a_n)$ converge or diverge together