Let-be-P-the-set-of-prime-numbers-and-A-P-0-1-Prove-that-n-A-n-n-2-1-2-pi-3-

Question Number 117838 by snipers237 last updated on 13/Oct/20

L e t b e P t h e s e t o f p r i m e n u m b e r s a n d

A = P \cup {0, 1}

P r o v e t h a t \prod_{n \notin A} \frac{n}{\sqrt{n^{2} - 1}} = \frac{2}{π} \sqrt{3}

Answered by mindispower last updated on 14/Oct/20

l e t \prod_{n ⩾ 2} \frac{n}{\sqrt{n^{2} - 1}} = I

I^{2} = \prod_{n ⩾ 2} \frac{n}{(n - 1)} . \frac{n}{(n + 1)} = \frac{2.2}{1.3} . \frac{3.3}{2.4} . \frac{4.4}{3.5} \dots \dots

I^{2} = \lim_{N \to \infty} . \underset{n = 2}{\prod^{N}} (\frac{n}{n - 1} . \frac{n}{n + 1}) = \lim_{N \to \infty} \frac{2 N}{N + 1}

I^{2} \to 2 \Rightarrow I \to \sqrt{2}, s i n c e I ⩾ 0 a n d x \to \sqrt{x} c o n t i n u s

l e t P b e e s e t o f r i m e s n u m b e r s

\prod_{p \in P} \frac{n^{2}}{n^{2} - 1} = \prod_{p \in P} \frac{1}{1 - \frac{1}{p^{2}}}

= \prod_{p \in P} (\sum_{s ⩾ 0} \frac{1}{p^{2 s}}) = \sum_{n ⩾ 1} \frac{1}{n^{2}} = ζ (2) = \frac{π^{2}}{6}

A = P \cup {0, 1}

\prod_{n \notin A} \frac{n^{2}}{n^{2} - 1} = \prod_{n ⩾ 2} \frac{n^{2}}{n^{2} - 1} / (\prod_{n \in P} \frac{n^{2}}{n^{2} - 1}) = \frac{2}{\frac{π^{2}}{6}}

= \frac{12}{π^{2}}

\prod_{n \notin A} \frac{n}{\sqrt{n^{2} - 1}} = \sqrt{\frac{12}{π^{2}}} = \frac{2 \sqrt{3}}{π}