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Let $P$ be the set of all odd prime numbers. I am looking for all $s\in(1,\infty)$ for them

$ A=\prod_{p\in P} (1+\frac{1}{(p-1)^s})^{p-1} $

exists (i.e. is finite). I know that it should be somehow related to Riemann zeta function but I was not sure how can I pursue the calculations.

If I use natural logarithm I will get:

$ \ln(A)=\sum_{p\in P} (p-1) \ln(1+ \frac{1}{(p-1)^s})$

whcih I am not sure is useful of not!

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Sensible question, but perhaps more appropriate for math.stackexchange.com ;) (The zeta-function is a red herring; my suggestion would have been to use the kinds of estimate suggested in Geoff Robinson's answer. After all, you only need an estimate not a formula.) –  Yemon Choi Jul 17 '11 at 17:49
    
Yes, I believe that this estimation is enough. –  Mahmood Alaghmandan Jul 17 '11 at 17:59
    
Hi Mahmood. If you think the answer below suffices, then you can click to "accept" it. That way, the question will show up on the main menu as "answered". –  Yemon Choi Jul 17 '11 at 18:19
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1 Answer 1

up vote 7 down vote accepted

The observation regarding the logarithm shows that the product exists if $s >2$, since $\ln(1+x) < x$ for $x >0$, so that the expression for $\ln(A)$ is less than $\sum_{p \in P} (p-1)^{1-s}$, which converges. However, for $1 < s \leq 2$, the product diverges, since, for a given $p$, the contribution to the product from $p$ is at least $1 + (p-1)^{1-s}$, (using the binomial theorem), so at least $\frac{p}{p-1}$. Hence the product is at least $\frac{1}{2} \left( \sum_{n=1}^{\infty} \frac{1}{n} \right)$, which diverges ( the half factor occurs since $P$ consists of only the odd primes. Since the sequence of partial sums diverges anyway, it doesn't really matter).

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