lim_(n→∞) Π_(k=1) ^n (1−(1/(k+1)))=?


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Question Number 115575 by Aziztisffola last updated on 26/Sep/20

$\lim_{n \to \infty} \underset{k = 1}{\prod^{n}} (1 - \frac{1}{k + 1}) = ?$
Commented by Dwaipayan Shikari last updated on 26/Sep/20

$First term (1 - \frac{1}{1}) = 0$ $Product will be 0$
Commented by Aziztisffola last updated on 26/Sep/20

$yes sir k = 1 not 0, I rectify .$
	Answered by TANMAY PANACEA last updated on 26/Sep/20

	$w h e n k = 0$ $(1 - \frac{1}{0 + 1}) = 0$ $s o i t h i n k \underset{k = 1}{\prod^{n}} s h o u l d b e$ $(1 - \frac{1}{1 + 1}) (1 - \frac{1}{2 + 1}) (1 - \frac{1}{3 + 1}) . . . (1 - \frac{1}{n - 1 + 1}) (1 - \frac{1}{n + 1})$ $= \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times . . \times \frac{n - 1}{n} \times \frac{n}{n + 1} = \frac{1}{n + 1}$ $\lim_{n \to \infty} \frac{1}{n + 1} = 0$
	Commented by Aziztisffola last updated on 26/Sep/20

	$yes sir k = 1 .$
	Answered by Dwaipayan Shikari last updated on 26/Sep/20

	$\lim_{n \to \infty} \underset{k = 1}{\prod^{n}} (1 - \frac{1}{k + 1}) = y$ $\prod^{\infty} \frac{k}{k + 1} = \frac{1}{2} . \frac{2}{3} . \frac{3}{4} . \frac{4}{5} . . . . . . . \frac{n - 1}{n} . \frac{n}{n + 1} = \lim_{n \to \infty} \frac{1}{n + 1} = 0$
	Commented by TANMAY PANACEA last updated on 26/Sep/20

	$t u m i k o t h a i t h a k o . . . k o l k a t a$
	Commented by Dwaipayan Shikari last updated on 26/Sep/20

	$Ha sir$
	Commented by TANMAY PANACEA last updated on 26/Sep/20

	$i a m 49 y e a r s . . . s e r v i c e . . . s t a y a t n a g p u r . . . h o m e t o w n b a r r a c k p i r e$
	Commented by Aziztisffola last updated on 26/Sep/20

	${That}^{'} s it .$
	Answered by Bird last updated on 27/Sep/20

	$l e t A_{n} = \prod_{k = 1}^{n} (1 - \frac{1}{k + 1}) \Rightarrow$ $A_{n} = \prod_{k = 1}^{n} \frac{k}{k + 1} = \frac{1}{2} . \frac{2}{3} . \frac{3}{4} . . . . \frac{n - 1}{n} . \frac{n}{n + 1}$ $= \frac{1}{n + 1} \Rightarrow {l i m}_{n \to + \infty} A_{n} = 0$
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