find-lim-n-n-k-1-n-1-k-n-2-

Question Number 31510 by abdo imad last updated on 09/Mar/18

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

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

Commented by abdo imad last updated on 12/Mar/18

let put A_n =(Π_(k=1) ^n (1+(k/n^2 )))^(1/n) ⇒ ln(A_n )=(1/n)Σ_(k=1) ^n ln(1+(k/n^2 )) ln(1+x)^′ = (1/(1+x))=Σ_(n=0) (−1)^n x^n ⇒ln(1+x) =Σ (−1)^n x^(n+1) /n+1 = Σ_(n=) ^∞ (−1)^(n−1) (x^n /n)=x −(x^2 /2) +(x^3 /3) +... ⇒x−(x^2 /2) ≤ln(1+x)≤x ⇒Σ_(k=1) ^n ((k/n^2 )) −(k^2 /(2n^4 ))≤ Σ_(k=1) ^n ln(1+(k/n^2 ))≤Σ_(k=1) ^n (k/n^2 ) ⇒ ((n(n+1))/(2n^2 )) −(1/(2n^4 )) ((n(n+1)(2n+1))/6) ≤ Σ_(k=1) ^n ln(1+(k/n^2 ))≤((n(n+1))/(2n^2 ))

$${let}\:{put}\:{A}_{{n}} =\left(\prod_{{k}=\mathrm{1}} ^{{n}} \left(\mathrm{1}+\frac{{k}}{{n}^{\mathrm{2}} }\right)\right)^{\frac{\mathrm{1}}{{n}}} \:\Rightarrow\:{ln}\left({A}_{{n}} \right)=\frac{\mathrm{1}}{{n}}\sum_{{k}=\mathrm{1}} ^{{n}} {ln}\left(\mathrm{1}+\frac{{k}}{{n}^{\mathrm{2}} }\right) \\ $$$${ln}\left(\mathrm{1}+{x}\right)^{'} =\:\frac{\mathrm{1}}{\mathrm{1}+{x}}=\sum_{{n}=\mathrm{0}} \:\left(−\mathrm{1}\right)^{{n}} {x}^{{n}} \:\Rightarrow{ln}\left(\mathrm{1}+{x}\right)\:=\Sigma\:\left(−\mathrm{1}\right)^{{n}} {x}^{{n}+\mathrm{1}} /{n}+\mathrm{1} \\ $$$$=\:\sum_{{n}=} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:\frac{{x}^{{n}} }{{n}}={x}\:−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\:+…\:\Rightarrow{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:\leqslant{ln}\left(\mathrm{1}+{x}\right)\leqslant{x} \\ $$$$\Rightarrow\sum_{{k}=\mathrm{1}} ^{{n}} \left(\frac{{k}}{{n}^{\mathrm{2}} }\right)\:−\frac{{k}^{\mathrm{2}} }{\mathrm{2}{n}^{\mathrm{4}} }\leqslant\:\sum_{{k}=\mathrm{1}} ^{{n}} {ln}\left(\mathrm{1}+\frac{{k}}{{n}^{\mathrm{2}} }\right)\leqslant\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{{k}}{{n}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}{n}^{\mathrm{2}} }\:−\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{4}} }\:\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}\:\leqslant\:\sum_{{k}=\mathrm{1}} ^{{n}} \:{ln}\left(\mathrm{1}+\frac{{k}}{{n}^{\mathrm{2}} }\right)\leqslant\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}{n}^{\mathrm{2}} } \\ $$

Commented by abdo imad last updated on 12/Mar/18

α_n ≤Σ_(k=1) ^n ln(1+(k/n^2 ))≤β_n ⇒(α_n /n) ≤(1/n)Σ_(k=1) ^n ln(1+(k/n^ ))≤(β_n /n) (α_n /n) ∼ (1/(2n)) −(1/(6n^2 )) →0 and (β_n /n) ∼ (1/(2n)) →0 ⇒lim_(n→) ln(A_n )=0 ⇒ lim_(n→∞) A_n =1 .

$$\alpha_{{n}} \:\leqslant\sum_{{k}=\mathrm{1}} ^{{n}} \:{ln}\left(\mathrm{1}+\frac{{k}}{{n}^{\mathrm{2}} }\right)\leqslant\beta_{{n}} \:\:\:\Rightarrow\frac{\alpha_{{n}} }{{n}}\:\leqslant\frac{\mathrm{1}}{{n}}\sum_{{k}=\mathrm{1}} ^{{n}} {ln}\left(\mathrm{1}+\frac{{k}}{{n}^{} }\right)\leqslant\frac{\beta_{{n}} }{{n}} \\ $$$$\frac{\alpha_{{n}} }{{n}}\:\sim\:\frac{\mathrm{1}}{\mathrm{2}{n}}\:−\frac{\mathrm{1}}{\mathrm{6}{n}^{\mathrm{2}} }\:\rightarrow\mathrm{0}\:{and}\:\:\frac{\beta_{{n}} }{{n}}\:\sim\:\frac{\mathrm{1}}{\mathrm{2}{n}}\:\rightarrow\mathrm{0}\:\Rightarrow{lim}_{{n}\rightarrow} {ln}\left({A}_{{n}} \right)=\mathrm{0} \\ $$$$\Rightarrow\:{lim}_{{n}\rightarrow\infty} \:{A}_{{n}} =\mathrm{1}\:. \\ $$

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