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

• Diverges to 0???? It's new to me, is there any definition of this! Aug 10, 2020 at 10:15
• @Subhajit Yes. An infinite product of complex numbers diverges to $0$ if the limit of the partial products is $0$.
– user239203
Aug 10, 2020 at 10:18
• @Subhajit When an infinite product is equal to $0$ this is usually called divergence, not convergence. If you think about it, this makes sense, since there might be examples where the product equals to $0$ because the first term is $0$, then the other terms are simply irrelevant.
– Mark
Aug 10, 2020 at 10:20
• @Subhajit The main reason is that you want the assertion "$\prod_n (1+a_n)$ converges if and only if $\sum_n \log (1+a_n)$ converges" to make the most sense it can.
– user239203
Aug 10, 2020 at 10:23
• @Charlie Chang I think ,it would help;math.stackexchange.com/questions/2712825/… Aug 10, 2020 at 10:27

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 $$|M|$$.