Question Number 115575 by Aziztisffola last updated on 26/Sep/20
$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\prod}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{k}+\mathrm{1}}\right)=? \\ $$
Commented by Dwaipayan Shikari last updated on 26/Sep/20
$$\mathrm{First}\:\mathrm{term}\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}}\right)=\mathrm{0} \\ $$$$\mathrm{Product}\:\mathrm{will}\:\mathrm{be}\:\mathrm{0} \\ $$
Commented by Aziztisffola last updated on 26/Sep/20
$$\mathrm{yes}\:\mathrm{sir}\:\mathrm{k}=\mathrm{1}\:\mathrm{not}\:\mathrm{0}\:,\:\mathrm{I}\:\mathrm{rectify}. \\ $$
Answered by TANMAY PANACEA last updated on 26/Sep/20
$${when}\:{k}=\mathrm{0} \\ $$$$\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{0}+\mathrm{1}}\right)=\mathrm{0} \\ $$$${so}\:{i}\:{think}\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\:{should}\:{be} \\ $$$$\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+\mathrm{1}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}+\mathrm{1}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}+\mathrm{1}}\right)…\left(\mathrm{1}−\frac{\mathrm{1}}{{n}−\mathrm{1}+\mathrm{1}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{3}}{\mathrm{4}}×..×\frac{{n}−\mathrm{1}}{{n}}×\frac{{n}}{{n}+\mathrm{1}}=\frac{\mathrm{1}}{{n}+\mathrm{1}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{{n}+\mathrm{1}}=\mathrm{0} \\ $$
Commented by Aziztisffola last updated on 26/Sep/20
$$\mathrm{yes}\:\mathrm{sir}\:\mathrm{k}=\mathrm{1}. \\ $$
Answered by Dwaipayan Shikari last updated on 26/Sep/20
$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\prod}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{k}+\mathrm{1}}\right)=\mathrm{y} \\ $$$$\overset{\infty} {\prod}\frac{\mathrm{k}}{\mathrm{k}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{2}}{\mathrm{3}}.\frac{\mathrm{3}}{\mathrm{4}}.\frac{\mathrm{4}}{\mathrm{5}}…….\frac{\mathrm{n}−\mathrm{1}}{\mathrm{n}}.\frac{\mathrm{n}}{\mathrm{n}+\mathrm{1}}=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}=\mathrm{0} \\ $$
Commented by TANMAY PANACEA last updated on 26/Sep/20
$${tumi}\:{kothai}\:{thako}…{kolkata} \\ $$
Commented by Dwaipayan Shikari last updated on 26/Sep/20
$$\mathrm{Ha}\:\mathrm{sir} \\ $$
Commented by TANMAY PANACEA last updated on 26/Sep/20
$${i}\:{am}\:\mathrm{49}\:{years}\:…{service}…{stay}\:{at}\:{nagpur}…{home}\:{town}\:{barrackpire} \\ $$
Commented by Aziztisffola last updated on 26/Sep/20
$$\mathrm{That}'\mathrm{s}\:\mathrm{it}. \\ $$
Answered by Bird last updated on 27/Sep/20
$${let}\:{A}_{{n}} =\prod_{{k}=\mathrm{1}} ^{{n}} \left(\mathrm{1}−\frac{\mathrm{1}}{{k}+\mathrm{1}}\right)\:\Rightarrow \\ $$$${A}_{{n}} =\prod_{{k}=\mathrm{1}} ^{{n}} \frac{{k}}{{k}+\mathrm{1}}\:=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{2}}{\mathrm{3}}.\frac{\mathrm{3}}{\mathrm{4}}….\frac{{n}−\mathrm{1}}{{n}}.\frac{{n}}{{n}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{{n}+\mathrm{1}}\:\Rightarrow\:{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} =\mathrm{0} \\ $$