Question Number 61834 by aliesam last updated on 09/Jun/19
Commented by maxmathsup by imad last updated on 09/Jun/19
![let A_n = {(1/p) Σ_(k=1) ^p (1+(k/p))^(1/n) }^n ⇒ln(A_n ) =n ln((1/p)Σ_(k=1) ^p (1+(k/p))^(1/n) ) we have lim_(p→+∞) (1/p)Σ_(k=1) ^p (1+(k/p))^(1/n) =∫_0 ^1 (1+x)^(1/n) dx =[(1/(1+(1/n)))(1+x)^((1/n)+1) ]_0 ^1 =(n/(n+1)) {2^((n+1)/n) −1} ⇒ln(A_n ) = n ln{(n/(n+1))( 2^(1+(1/n)) −1)} we have 2^(1+(1/n)) −1 =2 e^((ln(2))/n) −1 ∼2{ 1 +((ln(2))/n)} −1 =1+(2/n)ln(2)⇒ (n/(n+1))(2^(1+(1/n)) −1) ∼(n/(n+1))(1+(2/n)ln(2)) =(n/(n+1)) +(2/(n+1))ln(2) ⇒ nln{(n/(n+1))(2^(1+(1/n)) −1)} ∼ nln((n/(n+1)) +(2/(n+1))ln(2)) but nln((n/(n+1)) +((2ln(2))/(n+1))) =n ln(1−(1/(n+1)) +((2ln(2))/(n+1))) ∼n(−(1/(n+1)) +((2ln(2))/(n+1))) =(n/(n+1)){−1+2ln(2)}→2ln(2)−1 ⇒ lim_(n→+∞) A_n =e^(2ln(2)−1) =(4/e) ⇒ lim_(n→+∞) X_n =(4/e) .](https://www.tinkutara.com/question/Q61836.png)
$${let}\:{A}_{{n}} =\:\left\{\frac{\mathrm{1}}{{p}}\:\sum_{{k}=\mathrm{1}} ^{{p}} \:\left(\mathrm{1}+\frac{{k}}{{p}}\right)^{\frac{\mathrm{1}}{{n}}} \right\}^{{n}} \:\Rightarrow{ln}\left({A}_{{n}} \right)\:={n}\:{ln}\left(\frac{\mathrm{1}}{{p}}\sum_{{k}=\mathrm{1}} ^{{p}} \:\left(\mathrm{1}+\frac{{k}}{{p}}\right)^{\frac{\mathrm{1}}{{n}}} \right)\:{we}\:{have} \\ $$$${lim}_{{p}\rightarrow+\infty} \:\:\frac{\mathrm{1}}{{p}}\sum_{{k}=\mathrm{1}} ^{{p}} \left(\mathrm{1}+\frac{{k}}{{p}}\right)^{\frac{\mathrm{1}}{{n}}} \:\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{{n}}} {dx}\:\:=\left[\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{{n}}}\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{{n}}+\mathrm{1}} \right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{{n}}{{n}+\mathrm{1}}\:\left\{\mathrm{2}^{\frac{{n}+\mathrm{1}}{{n}}} −\mathrm{1}\right\}\:\Rightarrow{ln}\left({A}_{{n}} \right)\:=\:{n}\:{ln}\left\{\frac{{n}}{{n}+\mathrm{1}}\left(\:\mathrm{2}^{\mathrm{1}+\frac{\mathrm{1}}{{n}}} \:−\mathrm{1}\right)\right\}\:\:\:{we}\:{have} \\ $$$$\mathrm{2}^{\mathrm{1}+\frac{\mathrm{1}}{{n}}} −\mathrm{1}\:=\mathrm{2}\:{e}^{\frac{{ln}\left(\mathrm{2}\right)}{{n}}} −\mathrm{1}\:\sim\mathrm{2}\left\{\:\mathrm{1}\:+\frac{{ln}\left(\mathrm{2}\right)}{{n}}\right\}\:−\mathrm{1}\:\:=\mathrm{1}+\frac{\mathrm{2}}{{n}}{ln}\left(\mathrm{2}\right)\Rightarrow \\ $$$$\frac{{n}}{{n}+\mathrm{1}}\left(\mathrm{2}^{\mathrm{1}+\frac{\mathrm{1}}{{n}}} −\mathrm{1}\right)\:\sim\frac{{n}}{{n}+\mathrm{1}}\left(\mathrm{1}+\frac{\mathrm{2}}{{n}}{ln}\left(\mathrm{2}\right)\right)\:=\frac{{n}}{{n}+\mathrm{1}}\:+\frac{\mathrm{2}}{{n}+\mathrm{1}}{ln}\left(\mathrm{2}\right)\:\Rightarrow \\ $$$${nln}\left\{\frac{{n}}{{n}+\mathrm{1}}\left(\mathrm{2}^{\mathrm{1}+\frac{\mathrm{1}}{{n}}} −\mathrm{1}\right)\right\}\:\sim\:{nln}\left(\frac{{n}}{{n}+\mathrm{1}}\:+\frac{\mathrm{2}}{{n}+\mathrm{1}}{ln}\left(\mathrm{2}\right)\right)\:{but} \\ $$$${nln}\left(\frac{{n}}{{n}+\mathrm{1}}\:+\frac{\mathrm{2}{ln}\left(\mathrm{2}\right)}{{n}+\mathrm{1}}\right)\:={n}\:{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\:+\frac{\mathrm{2}{ln}\left(\mathrm{2}\right)}{{n}+\mathrm{1}}\right)\:\sim{n}\left(−\frac{\mathrm{1}}{{n}+\mathrm{1}}\:+\frac{\mathrm{2}{ln}\left(\mathrm{2}\right)}{{n}+\mathrm{1}}\right) \\ $$$$=\frac{{n}}{{n}+\mathrm{1}}\left\{−\mathrm{1}+\mathrm{2}{ln}\left(\mathrm{2}\right)\right\}\rightarrow\mathrm{2}{ln}\left(\mathrm{2}\right)−\mathrm{1}\:\:\Rightarrow\:{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} ={e}^{\mathrm{2}{ln}\left(\mathrm{2}\right)−\mathrm{1}} \:=\frac{\mathrm{4}}{{e}} \\ $$$$\Rightarrow\:{lim}_{{n}\rightarrow+\infty} \:\:{X}_{{n}} =\frac{\mathrm{4}}{{e}}\:. \\ $$
Commented by aliesam last updated on 09/Jun/19
$${brilliant}\:{sol}\:{thank}\:{you}\:{sir}\:{you}\:{are}\:{graet} \\ $$
Commented by maxmathsup by imad last updated on 09/Jun/19
$${you}\:{are}\:{welcome}\:{sir}\:{issam}. \\ $$