Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

Question Number 36925 by maxmathsup by imad last updated on 07/Jun/18

find lim_(n→+∞) Π_(k=1) ^n (1+(k/n^2 )).

$${find}\:{lim}_{{n}\rightarrow+\infty} \:\prod_{{k}=\mathrm{1}} ^{{n}} \:\:\left(\mathrm{1}+\frac{{k}}{{n}^{\mathrm{2}} }\right). \\ $$

Commented by math khazana by abdo last updated on 26/Jun/18

let A_n =Π_(k=1) ^n (1+(k/n^2 )) we have ln(A_n )=Σ_(k=1) ^n ln(1+(k/n^2 )) but ln^′ (1+u)=(1/(1+u))=Σ_(n=0) ^∞ (−1)^n u^n ⇒ ln(1+u)=Σ_(n=0) ^∞ (((−1)^n u^(n+1) )/(n+1)) =Σ_(n=1) ^∞ (((−1)^(n−1) u^n )/n) =u −(u^2 /2) +(u^3 /3) −...⇒ u−(u^2 /2)≤ln(1+u)≤u ⇒ (k/n^2 ) −(k^2 /(2n^4 )) ≤ ln(1+(k/n^2 ))≤ (k/n^2 ) ⇒ Σ_(k=1) ^n (k/n^2 ) −Σ_(k=1) ^n (k^2 /(2n^4 )) ≤ ln(A_n )≤ Σ_(k=1) ^n (k/n^2 ) ⇒ (1/n^2 ) ((n(n+1))/2) −(1/(2n^4 )) ((n(n+1)(2n+1))/6) ≤ln(A_n ) ≤ ((n(n+1))/(2n^2 )) ⇒ lim_(n→+∞) ln(A_n )=(1/2) ⇒ lim_(n→+∞) A_n =(√e) .

$${let}\:{A}_{{n}} =\prod_{{k}=\mathrm{1}} ^{{n}} \left(\mathrm{1}+\frac{{k}}{{n}^{\mathrm{2}} }\right)\:{we}\:{have}\: \\ $$$${ln}\left({A}_{{n}} \right)=\sum_{{k}=\mathrm{1}} ^{{n}} \:{ln}\left(\mathrm{1}+\frac{{k}}{{n}^{\mathrm{2}} }\right)\:{but} \\ $$$${ln}^{'} \left(\mathrm{1}+{u}\right)=\frac{\mathrm{1}}{\mathrm{1}+{u}}=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} {u}^{{n}} \:\Rightarrow \\ $$$${ln}\left(\mathrm{1}+{u}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} {u}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:{u}^{{n}} }{{n}} \\ $$$$={u}\:−\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\:+\frac{{u}^{\mathrm{3}} }{\mathrm{3}}\:−...\Rightarrow\:{u}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\leqslant{ln}\left(\mathrm{1}+{u}\right)\leqslant{u}\:\Rightarrow \\ $$$$\frac{{k}}{{n}^{\mathrm{2}} }\:−\frac{{k}^{\mathrm{2}} }{\mathrm{2}{n}^{\mathrm{4}} }\:\leqslant\:{ln}\left(\mathrm{1}+\frac{{k}}{{n}^{\mathrm{2}} }\right)\leqslant\:\frac{{k}}{{n}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \frac{{k}}{{n}^{\mathrm{2}} }\:−\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{{k}^{\mathrm{2}} }{\mathrm{2}{n}^{\mathrm{4}} }\:\leqslant\:{ln}\left({A}_{{n}} \right)\leqslant\:\sum_{{k}=\mathrm{1}} ^{{n}} \frac{{k}}{{n}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\:\:−\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{4}} }\:\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}\:\leqslant{ln}\left({A}_{{n}} \right) \\ $$$$\leqslant\:\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}{n}^{\mathrm{2}} }\:\Rightarrow\:{lim}_{{n}\rightarrow+\infty} {ln}\left({A}_{{n}} \right)=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} =\sqrt{{e}}\:. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com