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




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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