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

lim_(n→∞) Π_(k=1) ^n (1−(1/(k+1)))=?

$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\prod}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{k}+\mathrm{1}}\right)=? \\ $$

Commented by Dwaipayan Shikari last updated on 26/Sep/20

First term (1−(1/1))=0 Product will be 0

$$\mathrm{First}\:\mathrm{term}\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}}\right)=\mathrm{0} \\ $$$$\mathrm{Product}\:\mathrm{will}\:\mathrm{be}\:\mathrm{0} \\ $$

Commented by Aziztisffola last updated on 26/Sep/20

yes sir k=1 not 0 , I rectify.

$$\mathrm{yes}\:\mathrm{sir}\:\mathrm{k}=\mathrm{1}\:\mathrm{not}\:\mathrm{0}\:,\:\mathrm{I}\:\mathrm{rectify}. \\ $$

Answered by TANMAY PANACEA last updated on 26/Sep/20

when k=0 (1−(1/(0+1)))=0 so i think Π_(k=1) ^n should be (1−(1/(1+1)))(1−(1/(2+1)))(1−(1/(3+1)))...(1−(1/(n−1+1)))(1−(1/(n+1))) =(1/2)×(2/3)×(3/4)×..×((n−1)/n)×(n/(n+1))=(1/(n+1)) lim_(n→∞) (1/(n+1))=0

$${when}\:{k}=\mathrm{0} \\ $$$$\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{0}+\mathrm{1}}\right)=\mathrm{0} \\ $$$${so}\:{i}\:{think}\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\:{should}\:{be} \\ $$$$\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+\mathrm{1}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}+\mathrm{1}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}+\mathrm{1}}\right)...\left(\mathrm{1}−\frac{\mathrm{1}}{{n}−\mathrm{1}+\mathrm{1}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{3}}{\mathrm{4}}×..×\frac{{n}−\mathrm{1}}{{n}}×\frac{{n}}{{n}+\mathrm{1}}=\frac{\mathrm{1}}{{n}+\mathrm{1}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{{n}+\mathrm{1}}=\mathrm{0} \\ $$

Commented by Aziztisffola last updated on 26/Sep/20

yes sir k=1.

$$\mathrm{yes}\:\mathrm{sir}\:\mathrm{k}=\mathrm{1}. \\ $$

Answered by Dwaipayan Shikari last updated on 26/Sep/20

lim_(n→∞) Π_(k=1) ^n (1−(1/(k+1)))=y Π^∞ (k/(k+1))=(1/2).(2/3).(3/4).(4/5).......((n−1)/n).(n/(n+1))=lim_(n→∞) (1/(n+1))=0

$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\prod}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{k}+\mathrm{1}}\right)=\mathrm{y} \\ $$$$\overset{\infty} {\prod}\frac{\mathrm{k}}{\mathrm{k}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{2}}{\mathrm{3}}.\frac{\mathrm{3}}{\mathrm{4}}.\frac{\mathrm{4}}{\mathrm{5}}.......\frac{\mathrm{n}−\mathrm{1}}{\mathrm{n}}.\frac{\mathrm{n}}{\mathrm{n}+\mathrm{1}}=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}=\mathrm{0} \\ $$

Commented by TANMAY PANACEA last updated on 26/Sep/20

tumi kothai thako...kolkata

$${tumi}\:{kothai}\:{thako}...{kolkata} \\ $$

Commented by Dwaipayan Shikari last updated on 26/Sep/20

Ha sir

$$\mathrm{Ha}\:\mathrm{sir} \\ $$

Commented by TANMAY PANACEA last updated on 26/Sep/20

i am 49 years ...service...stay at nagpur...home town barrackpire

$${i}\:{am}\:\mathrm{49}\:{years}\:...{service}...{stay}\:{at}\:{nagpur}...{home}\:{town}\:{barrackpire} \\ $$

Commented by Aziztisffola last updated on 26/Sep/20

That′s it.

$$\mathrm{That}'\mathrm{s}\:\mathrm{it}. \\ $$

Answered by Bird last updated on 27/Sep/20

let A_n =Π_(k=1) ^n (1−(1/(k+1))) ⇒ A_n =Π_(k=1) ^n (k/(k+1)) =(1/2).(2/3).(3/4)....((n−1)/n).(n/(n+1)) =(1/(n+1)) ⇒ lim_(n→+∞) A_n =0

$${let}\:{A}_{{n}} =\prod_{{k}=\mathrm{1}} ^{{n}} \left(\mathrm{1}−\frac{\mathrm{1}}{{k}+\mathrm{1}}\right)\:\Rightarrow \\ $$$${A}_{{n}} =\prod_{{k}=\mathrm{1}} ^{{n}} \frac{{k}}{{k}+\mathrm{1}}\:=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{2}}{\mathrm{3}}.\frac{\mathrm{3}}{\mathrm{4}}....\frac{{n}−\mathrm{1}}{{n}}.\frac{{n}}{{n}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{{n}+\mathrm{1}}\:\Rightarrow\:{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} =\mathrm{0} \\ $$

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