Π_(n=1) ^∞ ((4n^2 (10n−6)(10n−4))/((2n−1)^2 (10n−1)(10n+1))) =?


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Question Number 114758 by bobhans last updated on 21/Sep/20

$\underset{n = 1}{\prod^{\infty}} \frac{4 n^{2} (10 n - 6) (10 n - 4)}{{(2 n - 1)}^{2} (10 n - 1) (10 n + 1)} = ?$
	Answered by Olaf last updated on 21/Sep/20

	$u_{n} = \frac{4 n^{2} (10 n - 6) (10 n - 4)}{{(2 n - 1)}^{2} (10 n - 1) (10 n + 1)}$ $u_{n} = \frac{{(10 n)}^{2} (10 n - 6) (10 n - 4)}{{(10 n - 5)}^{2} (10 n - 1) (10 n + 1)}$ $u_{n} = \frac{4}{5} \frac{(10 n - 0) (10 n + 0) (10 n + 4) (10 n - 4)}{(10 n + 5) (10 n - 5) (10 n - 1) (10 n + 1)}$ $Let Π = \underset{n = 1}{\prod^{\infty}} u_{n}$ $Π = \frac{4}{5} \times \frac{\underset{n = 1}{\prod^{\infty}} (\frac{10 n + 0}{10 n + 1} \times \frac{10 n + 4}{10 n + 5})}{\underset{n = 1}{\prod^{\infty}} (\frac{10 n - 1}{10 n - 0} \times \frac{10 n - 5}{10 n - 4})}$ $work in progress . . .$
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