Question-144607

Question Number 144607 by aliibrahim1 last updated on 26/Jun/21

Answered by mathmax by abdo last updated on 27/Jun/21

A_{n} = \prod_{k = 2}^{n} e {(1 - \frac{1}{k^{2}})}^{k^{2}} \Rightarrow \prod_{n = 2}^{\infty} (\dots .) = \lim_{n ⌣ \infty} A_{n}

A_{n} = e^{n - 1} \prod_{k = 2}^{n} {(1 - \frac{1}{k^{2}})}^{k^{2}} \Rightarrow \log (A_{n}) = n - 1 + \sum_{k = 2}^{n} k^{2} \log (1 - \frac{1}{k^{2}})

\log^{^{'}} (1 - x) = - \frac{1}{1 - x} = - (1 + x + x^{2} + o (x^{3})) \Rightarrow

\log (1 - x) \sim - x - \frac{x^{2}}{2} \Rightarrow \log (1 - \frac{1}{k^{2}}) \sim - \frac{1}{k^{2}} - \frac{1}{{2 k}^{4}} \Rightarrow

k^{2} \log (1 - \frac{1}{k^{2}}) \sim - 1 - \frac{1}{{2 k}^{2}} \Rightarrow \sum_{k = 2}^{n} k^{2} \log (\dots) \sim - (n - 1)

- \frac{1}{2} \sum_{k = 2}^{n} \frac{1}{k^{2}} \Rightarrow \log (A_{n}) \sim - \frac{1}{2} \sum_{k = 2}^{n} \frac{1}{k^{2}}

= - \frac{1}{2} (ξ_{n} (2) - 1) = \frac{1}{2} - \frac{ξ_{n} (2)}{2} \Rightarrow

\lim_{n \to + \infty} \log (A_{n}) = \frac{1}{2} - \frac{π^{2}}{12} \Rightarrow \lim_{n \to + \infty} A_{n} = e^{\frac{1}{2} - \frac{π^{2}}{12}}

Commented by aliibrahim1 last updated on 27/Jun/21

t h x s i r

Commented by mathmax by abdo last updated on 28/Jun/21

you are welcome