# Convergence/Divergence of infinite product

General condition: $(P_n)_{n\in\mathbb{N}}$ is a sequence of non-zero real numbers. If $\prod_{n=0}^\infty P_n$ exists in reals and is non-zero, then call this infinite product convergent. Otherwise divergent.

I have proven that:

If all $(P_n)_{n\in\mathbb{N}}$ are positive, then its convergence is equivalent to convergence of $\sum_{n = 0}^\infty \log(P_n)$.

If further $(P_n)_{n\in\mathbb{N}}$ are further greater or equal to 1, then the convergence of the product is equivalent to the convergence of $\sum_{n = 0}^\infty (P_n-1)$

Now I am looking for two examples where
a) $(P_n)_{n\in\mathbb{N}}$ are reals. $\sum_{n = 0}^\infty (P_n-1)$ converges but $\prod_{n=0}^\infty P_n$ diverges
b) $(P_n)_{n\in\mathbb{N}}$ are reals. $\prod_{n=0}^\infty P_n$ converges but $\sum_{n = 0}^\infty (P_n-1)$ diverges

I have spent a long time on this but failed to find any. I guess it requires complex analysis technique? (Which I don't know) Please help me out. Thank you.

• I guess you are aware that such example cannot be found for $P_n>1$. See for example here. Maybe there are some posts at MSE, where this was shown. – Martin Sleziak Mar 10 '13 at 11:51
• @MartinSleziak I'm aware of that. I also know $P_n$ should oscillate around 1 and |log(Pn)| must not decrease monotonely. – mez Mar 10 '13 at 11:54
• Such Examples are given in Problems 3.8.5 and 3.8.9 in Wieslawa J. Kaczor, Maria T. Nowak: Problems in mathematical analysis: Volume 1; Real Numbers, Sequences and Series, p.113-114. – Martin Sleziak Mar 10 '13 at 12:03

Problem 3.8.5 Set $$a_{2n-1}=\frac1{\sqrt{n}}+\frac1n,\qquad a_{2n}=-\frac1{\sqrt{n}}, n\in\mathbb N.$$ Show that the product $\prod\limits_{n=1}^\infty(1+a_n)$ converges, although the series $\sum\limits_{n=1}^\infty a_n$ diverges.
Problem 3.8.9. Prove that the product $\prod\limits_{n=1}^\infty \left(1+(-1)^{n+1}\frac1{\sqrt{n}}\right)$ diverges although the series $\sum\limits_{n=1}^\infty (-1)^{n+1}\frac1{\sqrt{n}}$ converges.