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Question Number 104174 by Dwaipayan Shikari last updated on 19/Jul/20

Π_(n=1) ^∞ ((n/(n+1)))^2

$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\frac{\mathrm{n}}{\mathrm{n}+\mathrm{1}}\right)^{\mathrm{2}} \\ $$

Answered by OlafThorendsen last updated on 19/Jul/20

S_n = Π_(k=1) ^n ((k/(k+1)))^2 S_n = ((1/2))^2 ((2/3))^2 ((3/4))^2 ...((n/(n+1)))^2 S_n = (1/((n+1)^2 )) lim_(n→∞) S_n = 0

$$\mathrm{S}_{{n}} \:=\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\frac{{k}}{{k}+\mathrm{1}}\right)^{\mathrm{2}} \\ $$$$\mathrm{S}_{{n}} \:=\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} ...\left(\frac{{n}}{{n}+\mathrm{1}}\right)^{\mathrm{2}} \\ $$$$\mathrm{S}_{{n}} \:=\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}S}_{{n}} \:=\:\mathrm{0} \\ $$

Answered by mathmax by abdo last updated on 19/Jul/20

let S_n =Π_(k=1) ^n (k^2 /((k+1)^2 )) ⇒ln(S_n ) =Σ_(k=1) ^n ln((k^2 /((k+1)^2 ))) =2Σ_(k=1) ^n { ln(k)−ln(k+1)} =2 {ln1−ln(2)+ln(2)−ln(3)+...+ln(n)−ln(n+1)} =−2ln(n+1)→−∞ ⇒lim_(n→+∞) ln(S_n ) =0 ⇒lim_(n→+∞) S_n =0

$$\mathrm{let}\:\mathrm{S}_{\mathrm{n}} =\prod_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{k}^{\mathrm{2}} }{\left(\mathrm{k}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\mathrm{ln}\left(\mathrm{S}_{\mathrm{n}} \right)\:=\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \mathrm{ln}\left(\frac{\mathrm{k}^{\mathrm{2}} }{\left(\mathrm{k}+\mathrm{1}\right)^{\mathrm{2}} }\right)\:=\mathrm{2}\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \left\{\:\mathrm{ln}\left(\mathrm{k}\right)−\mathrm{ln}\left(\mathrm{k}+\mathrm{1}\right)\right\} \\ $$$$=\mathrm{2}\:\left\{\mathrm{ln1}−\mathrm{ln}\left(\mathrm{2}\right)+\mathrm{ln}\left(\mathrm{2}\right)−\mathrm{ln}\left(\mathrm{3}\right)+...+\mathrm{ln}\left(\mathrm{n}\right)−\mathrm{ln}\left(\mathrm{n}+\mathrm{1}\right)\right\} \\ $$$$=−\mathrm{2ln}\left(\mathrm{n}+\mathrm{1}\right)\rightarrow−\infty\:\Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{ln}\left(\mathrm{S}_{\mathrm{n}} \right)\:=\mathrm{0}\:\Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{S}_{\mathrm{n}} =\mathrm{0} \\ $$

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