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Question Number 34226 by abdo imad last updated on 03/May/18

let u_n = (n+1)^((n+1)/n) −n^(n/(n−1)) find lim_(n→+∞) u_n

letun=(n+1)n+1n−nnn−1findlimn→+∞un

Commented by abdo mathsup 649 cc last updated on 05/May/18

u_n = n^((n+1)/n) (1+(1/n))^((n+1)/n) −n^(n/(n−1)) but n^((n+1)/n) =e^((1+(1/n))ln(n)) ∼n (n→+∞) n^(n/(n−1)) = e^((n/(n+1))ln(n)) ∼ n ⇒ u_n ∼ n((1+(1/n))^(1+(1/n)) −1) (1+(1/n))^(1+(1/n)) = e^((1+(1/n))ln(1+(1/n))) ∼ e^((1+(1/n))((1/n) −(1/(2n^2 )))) = e^((1/n) −(1/(2n^2 )) +(1/n^2 )−(1/(2n^3 ))) ∼ 1 +(1/n) +(1/(2n^2 )) −(1/(2n^3 )) ⇒ u_n ∼n( (1/n) +(1/(2n^2 )) −(1/(2n^3 ))) ⇒u_n ∼ 1 +(1/(2n)) −(1/(2n^2 )) ⇒ lim_(n→+∞) u_n =1 .

un=nn+1n(1+1n)n+1n−nnn−1butnn+1n=e(1+1n)ln(n)∼n(n→+∞)nnn−1=enn+1ln(n)∼n⇒un∼n((1+1n)1+1n−1)(1+1n)1+1n=e(1+1n)ln(1+1n)∼e(1+1n)(1n−12n2)=e1n−12n2+1n2−12n3∼1+1n+12n2−12n3⇒un∼n(1n+12n2−12n3)⇒un∼1+12n−12n2⇒limn→+∞un=1.

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