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Question Number 61834 by aliesam last updated on 09/Jun/19

Commented by maxmathsup by imad last updated on 09/Jun/19
$let A_n = {(1/p) Σ_(k=1) ^p (1+(k/p))^(1/n) }^n ⇒ln(A_n ) =n ln((1/p)Σ_(k=1) ^p (1+(k/p))^(1/n) ) we have lim_(p→+∞) (1/p)Σ_(k=1) ^p (1+(k/p))^(1/n) =∫_0 ^1 (1+x)^(1/n) dx =[(1/(1+(1/n)))(1+x)^((1/n)+1) ]_0 ^1 =(n/(n+1)) {2^((n+1)/n) −1} ⇒ln(A_n ) = n ln{(n/(n+1))( 2^(1+(1/n)) −1)} we have 2^(1+(1/n)) −1 =2 e^((ln(2))/n) −1 ∼2{ 1 +((ln(2))/n)} −1 =1+(2/n)ln(2)⇒ (n/(n+1))(2^(1+(1/n)) −1) ∼(n/(n+1))(1+(2/n)ln(2)) =(n/(n+1)) +(2/(n+1))ln(2) ⇒ nln{(n/(n+1))(2^(1+(1/n)) −1)} ∼ nln((n/(n+1)) +(2/(n+1))ln(2)) but nln((n/(n+1)) +((2ln(2))/(n+1))) =n ln(1−(1/(n+1)) +((2ln(2))/(n+1))) ∼n(−(1/(n+1)) +((2ln(2))/(n+1))) =(n/(n+1)){−1+2ln(2)}→2ln(2)−1 ⇒ lim_(n→+∞) A_n =e^(2ln(2)−1) =(4/e) ⇒ lim_(n→+∞) X_n =(4/e) .$
$l e t A_{n} = {\frac{1}{p} \sum_{k = 1}^{p} {(1 + \frac{k}{p})}^{\frac{1}{n}}}^{n} \Rightarrow l n (A_{n}) = n l n (\frac{1}{p} \sum_{k = 1}^{p} {(1 + \frac{k}{p})}^{\frac{1}{n}}) w e h a v e$ ${l i m}_{p \to + \infty} \frac{1}{p} \sum_{k = 1}^{p} {(1 + \frac{k}{p})}^{\frac{1}{n}} = \int_{0}^{1} {(1 + x)}^{\frac{1}{n}} d x = {[\frac{1}{1 + \frac{1}{n}} {(1 + x)}^{\frac{1}{n} + 1}]}_{0}^{1}$ $= \frac{n}{n + 1} {2^{\frac{n + 1}{n}} - 1} \Rightarrow l n (A_{n}) = n l n {\frac{n}{n + 1} (2^{1 + \frac{1}{n}} - 1)} w e h a v e$ $2^{1 + \frac{1}{n}} - 1 = 2 e^{\frac{l n (2)}{n}} - 1 \sim 2 {1 + \frac{l n (2)}{n}} - 1 = 1 + \frac{2}{n} l n (2) \Rightarrow$ $\frac{n}{n + 1} (2^{1 + \frac{1}{n}} - 1) \sim \frac{n}{n + 1} (1 + \frac{2}{n} l n (2)) = \frac{n}{n + 1} + \frac{2}{n + 1} l n (2) \Rightarrow$ $n l n {\frac{n}{n + 1} (2^{1 + \frac{1}{n}} - 1)} \sim n l n (\frac{n}{n + 1} + \frac{2}{n + 1} l n (2)) b u t$ $n l n (\frac{n}{n + 1} + \frac{2 l n (2)}{n + 1}) = n l n (1 - \frac{1}{n + 1} + \frac{2 l n (2)}{n + 1}) \sim n (- \frac{1}{n + 1} + \frac{2 l n (2)}{n + 1})$ $= \frac{n}{n + 1} {- 1 + 2 l n (2)} \to 2 l n (2) - 1 \Rightarrow {l i m}_{n \to + \infty} A_{n} = e^{2 l n (2) - 1} = \frac{4}{e}$ $\Rightarrow {l i m}_{n \to + \infty} X_{n} = \frac{4}{e} .$
Commented by aliesam last updated on 09/Jun/19

$b r i l l i a n t s o l t h a n k y o u s i r y o u a r e g r a e t$
Commented by maxmathsup by imad last updated on 09/Jun/19

$y o u a r e w e l c o m e s i r i s s a m .$
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