lim-n-k-1-n-1-1-k-1-

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}) \dots (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} \dots \dots . \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 \dots 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 \dots s e r v i c e \dots s t a y a t n a g p u r \dots 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} \dots . \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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