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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.

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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
    
@MartinSleziak Can you shift it as an answer i will accept it. –  mez Mar 10 '13 at 12:05

2 Answers 2

up vote 2 down vote accepted

Per OP's request I made my comment into an answer. I am making this CW; if someone wants to add details of the solutions, feel free to do so. (Of course, if you prefer, you can post them in a separate post, so that you are rewarded by reputation for you effort.)

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.

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.

A solution is given on p.364.

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.

A solution is given on p.365.

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a) For odd i, P_i = 4 + 1/2^i and for even i, P_i = -2 + 1/2^i.

b) P_i = 1 - 1/i

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b) $\prod_{i = 2}^\infty P_i = 0$, it does not count as convergent for me. I defined convergence to be non zero. –  mez Mar 8 '13 at 14:13
    
a) it is pure wrong. sum does not converge. –  mez Mar 8 '13 at 14:17
    
I was Wong about b, I didn't notice that you excluded 0. a is correct, though: the relevant sum is 1. –  Don L. Mar 9 '13 at 7:20
    
a) cannot be correct, because each time you are plusing a 4 or a -2, if you take 4 and -2 to the very front and exclude then from the series, they do converge. 4-2+1/2+1/2+1/4+1/4... is good but not like (4+1/2)+(-2+1/2)+(4+1/4)-(-2+1/4).... this is clearly unbounded. –  mez Mar 9 '13 at 9:27

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