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Show that $\displaystyle\prod_{p\equiv1(3)}\frac p{p-1}=\infty$ where $p$ is a prime number.

I know that $\displaystyle\prod_p\frac p{p-1}=\infty$. However, how can I show the above one?

Is there any inequality that implies $\displaystyle\prod_p\frac p{p-1}<\displaystyle\prod_{p\equiv1(3)} \frac p{p-1}$?

Otherwise, can you help me to solve it?

Parcly Taxel
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abcdmath
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    I think some strong form of Dirichlet's theorem (on primes in arithmetic progressions) is necessary here. I don't know whether this statement can be proven without it. – Wojowu Jul 16 '18 at 10:18
  • Yah..I think so, Dirichlet's L-series is required for some sort of contradiction. – abcdmath Jul 16 '18 at 10:24
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    You don't use it to get a contradiction. You can show that e.g. the sum $\sum_{p\equiv 1\pmod 3}\frac{1}{p}$ diverges and go from there. – Wojowu Jul 16 '18 at 10:26

1 Answers1

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We need two parts. The first part is $$\prod_{p\equiv1\bmod3}\frac p{p-1}=\prod_{p\equiv1\bmod6}\left(1+\frac1{p-1}\right)\ge\prod_{p\equiv1\bmod6}\left(1+\frac1p\right)\ge1+\sum_{p\equiv1\bmod6}\frac1p$$ The second part is proving the last infinite sum diverges. For one such proof, see here.

Therefore the original infinite product diverges.

Parcly Taxel
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