Show that if $\sum_{n=1}^{\infty} a_n$ converges conditionally, $\prod_{n=1}^{\infty} (1+a_n)$ could diverge to 0.

Question

This post follows from Show that if $\sum_{n=1}^{\infty} a_n$ converges conditionally, then $\prod_{n=1}^{\infty} (1+a_n)$ converges conditionally or diverges to 0..

There we haven't proved he following statement (particularly with my proof 2, we even seem to disprove it) that

if $\sum_{n=1}^{\infty} a_n$ converges conditionally, and $\sum_{n=1}^{\infty} {a_n}^2$ converges not, $\prod_{n=1}^{\infty} (1+a_n)$ diverges to $0$.

Any idea about ways to prove that? (Or to point out what goes wrong in my proof 2 that leads to the negation of the above statement.)

Answer 1

I see one problem in my proof 2 in Show that if $\sum_{n=1}^{\infty} a_n$ converges conditionally, then $\prod_{n=1}^{\infty} (1+a_n)$ converges conditionally or diverges to 0. is since we have 'for all p, ...<$\epsilon$', then $\epsilon$ should be independent of p, but in my proof it is not!

More specifically, $\epsilon$ there depends on x only, not p, (so p depends not on $\epsilon$) so however small $\epsilon$ is, we can choose p sufficiently large, e.g. 1/$\epsilon$-1, so that $\epsilon'+\epsilon|M|(p+1)$ can't be arbitrarily small, e.g. it being larger than $|M|$.