Proving a result in infinite products: $\prod (1+a_n)$ converges (to a non zero element) iff the series $\sum a_n$ converges

Question

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.

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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.

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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.

Answer 1

You're on the right track. You've said that for small $x$, $\ln(1+x) < 2x$. That gave you an upper bound.

Can you make a similar claim, such as "for small enough $x$, $\ln(1 + x) > x/2$", for example? If so, you can reverse all your inequalities and do the lower bound proof.

I'm not saying that the bound I've suggested above is correct...merely that it might be, and you should try to prove it or find something similar that works.

Answer 2

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