n-2-1-1-n-2-

Question Number 63919 by raj last updated on 11/Jul/19

\underset{n = 2}{\prod^{\infty}} (1 - \frac{1}{n^{2}}) = ?

Commented by Prithwish sen last updated on 11/Jul/19

\underset{n = 2}{\prod^{n = \boldsymbol n}} (1 - \frac{1}{n^{2}}) = \frac{1.2 . 3^{2} . 4^{2} \dots \dots \dots . . n . (n + 1)}{1^{2} . 2^{2} . 3^{2} \dots \dots \dots . {(n - 1)}^{2} . n^{2}}

= {[\frac{(n + 1)!}{n!}]}^{2} . \frac{1}{2 n (n + 1)} = \frac{{(n + 1)}^{2}}{2 n (n + 1)}

= \frac{1}{2} (1 + \frac{1}{n}) \to \frac{1}{2} as n \to \infty

please check .

Commented by mathmax by abdo last updated on 11/Jul/19

l e t A_{n} = \prod_{k = 2}^{n} (1 - \frac{1}{k^{2}}) \Rightarrow \prod_{k = 2}^{\infty} (1 - \frac{1}{k^{2}}) = {l i m}_{n \to + \infty} A_{n}

b u t A_{n} = \prod_{k = 2}^{n} \frac{k^{2} - 1}{k^{2}} = \prod_{k = 2}^{n} \frac{k - 1}{k} \frac{k + 1}{k}

= \prod_{k = 2}^{n} \frac{k - 1}{k} \prod_{k = 2}^{n} \frac{k + 1}{k} = \frac{1}{2} \frac{2}{3} \dots \frac{n - 2}{n - 1} \frac{n - 1}{n} \times \frac{3}{2} . \frac{4}{3} \dots . . \frac{n}{n - 1} \frac{n + 1}{n}

= \frac{1}{n} \frac{n + 1}{2} = \frac{n + 1}{2 n} \Rightarrow {l i m}_{n \to + \infty} A_{n} = \frac{1}{2}

Commented by raj last updated on 11/Jul/19

t h a n k y o u

t h a n k y o u

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