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

Question Number 80792 by john santu last updated on 06/Feb/20

\underset{n = 1}{\prod^{\infty}} [\frac{2 n}{2 n - 1} . \frac{2 n}{2 n + 1}] = ?

Answered by mind is power last updated on 06/Feb/20

l n (\prod_{n ⩾ 1} \frac{2 n . 2 n}{(2 n - 1) (2 n + 1)}) = \sum_{n ⩾ 1_{}} l n (\frac{1}{(1 - \frac{1}{2 n}) (1 + \frac{1}{2 n})})

= - \sum_{n ⩾ 1} l n (1 - \frac{1}{4 n^{2}}) \sim \frac{1}{4 n^{2}} \Rightarrow o u r s u m e x i s t \Rightarrow p r o d u c t e x i s t

\underset{n = 1}{\prod^{+ \infty}} \frac{4 n^{2}}{(2 n - 1) (2 n + 1)} = \frac{1}{Π \frac{(2 n - 1) (2 n + 1)}{4 n^{2}}}

= \frac{1}{\prod_{n ⩾ 1} (1 - \frac{1}{4 n^{2}})}

w e h a v e e u l e r f o r m u l a \Rightarrow s i n (x) = x \prod_{k ⩾ 1} (1 - \frac{x^{2}}{k^{2} . π^{2}})

x = \frac{π}{2} \Rightarrow 1 = \frac{π}{2} \prod_{k ⩾ 1} (1 - \frac{1}{4 k^{2}}) \Rightarrow \frac{1}{\prod_{n ⩾ 1} (1 - \frac{1}{4 n^{2}})} = \frac{π}{2}

\Rightarrow \underset{n = 1}{\prod^{+ \infty}} \frac{2 n}{2 n - 1} . \frac{2 n}{2 n + 1} = \frac{π}{2}

Commented by jagoll last updated on 07/Feb/20

t h a n k y o u s i r

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