Some infinite products related to prime numbers. - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T12:07:29Z http://mathoverflow.net/feeds/question/70542 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/70542/some-infinite-products-related-to-prime-numbers Some infinite products related to prime numbers. Mahmood Alaghmandan 2011-07-17T08:40:15Z 2011-07-17T23:09:59Z <p>Let $P$ be the set of all odd prime numbers. I am looking for all $s\in(1,\infty)$ for them</p> <p>$A=\prod_{p\in P} (1+\frac{1}{(p-1)^s})^{p-1}$</p> <p>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.</p> <p>If I use natural logarithm I will get:</p> <p>$\ln(A)=\sum_{p\in P} (p-1) \ln(1+ \frac{1}{(p-1)^s})$</p> <p>whcih I am not sure is useful of not!</p> http://mathoverflow.net/questions/70542/some-infinite-products-related-to-prime-numbers/70544#70544 Answer by Geoff Robinson for Some infinite products related to prime numbers. Geoff Robinson 2011-07-17T10:30:51Z 2011-07-17T11:01:56Z <p>The observation regarding the logarithm shows that the product exists if $s >2$, since $\ln(1+x) &lt; x$ 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 $1 &lt; s \leq 2$, 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).</p>