prove-that-P-n-1-1-1-n-n-2-2-sin-pi-2-pi-m-n-

Question Number 159681 by mnjuly1970 last updated on 20/Nov/21

p r o v e t h a t :

P = \underset{n = 1}{\prod^{\infty}} (1 - \frac{1}{n (n + 2)}) \overset{?}{=} \frac{- \sqrt{2} s i n (π \sqrt{2})}{π}

m . n

Answered by mindispower last updated on 21/Nov/21

\underset{n = 1}{\prod^{m}} (1 - \frac{1}{n (n + 2)}) = \underset{n = 1}{\prod^{m}} \frac{(n + 1 - \sqrt{2}) (n + 1 + \sqrt{2})}{n (n + 2)}

= \frac{2 . Γ (m + 2 - \sqrt{2}) Γ (m + 2 + \sqrt{2})}{Γ (m + 1) Γ (m + 3) Γ (2 + \sqrt{2}) Γ (2 - \sqrt{2})}

Γ (1 + \sqrt{2}) = \sqrt{2} Γ (\sqrt{2})

\Leftrightarrow \frac{2 Γ (m + 2 - \sqrt{2}) Γ (m + 2 + \sqrt{2})}{m! . (m + 2)! Γ (\sqrt{2}) Γ (1 - \sqrt{2}) . \sqrt{2} . (1 - \sqrt{2}) (1 + \sqrt{2})}

= - \frac{\sqrt{2}}{\frac{π}{s i n (π \sqrt{2})}} . \frac{Γ (m + 2 - \sqrt{2}) Γ (m + 2 + \sqrt{2})}{m! . (m + 2)!}

Γ (z + a) \sim \sqrt{2 π .} z^{a - \frac{1}{2} + z} e^{- z}

z = m + 2

Γ (m + 2 - \sqrt{2}) Γ (m + 2 + \sqrt{2}) \sim 2 π . {(m + 2)}^{2 m + 3} e^{- 2 (m + 2)}

Γ (m + 3) Γ (m + 1) \sim 2 π {(m + 3)}^{m + \frac{5}{2}} {(m + 1)}^{m + \frac{1}{2}} e^{- 2 m - 4}

\lim_{m \to \infty} \frac{Γ (m + 2 - \sqrt{2}) Γ (m + 2 + \sqrt{2})}{m! . (m + 2)!}

= \lim_{m \to \infty} \frac{{(m + 2)}^{2 m + 3}}{{(m + 1)}^{m + \frac{1}{2}} {(m + 3)}^{m + \frac{5}{2}}}

= \lim_{m \to \infty} {(\frac{{(m + 2)}^{2}}{m^{2} + 4 m + 3})}^{m} = \lim_{m \to \infty} . {(1 + \frac{1}{m^{2} + 4 m + 3})}^{m} = 1

\Rightarrow

\prod_{n ⩾ 1} (1 - \frac{1}{n (n + 2)}) = \lim_{m \to \infty} \underset{n = 1}{\prod^{m}} (1 - \frac{1}{n (n + 2)}) = - \sqrt{2} . \frac{s i n (π \sqrt{2})}{π} . \lim_{m \to \infty} . \frac{Γ (m + 2 - \sqrt{2}) Γ (m + 2 + \sqrt{2})}{Γ (m + 1) Γ (m + 3)} = - \sqrt{2} . \frac{s i n (π \sqrt{2})}{π}