lim_(n→∞) n^2 [(1+(1/(n+1)))^(n+1) −(1+(1/n))^n ]=?


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Question Number 212544 by MrGaster last updated on 17/Oct/24

$\lim_{n \to \infty} n^{2} [{(1 + \frac{1}{n + 1})}^{n + 1} - {(1 + \frac{1}{n})}^{n}] = ?$
	Answered by lepuissantcedricjunior last updated on 17/Oct/24
	$lim_(n→+∞) n^2 [(1+(1/(n+1)))^(n+1) −(1+(1/n))^n ] =lim_(n→+∞) {n^2 (1+(1/(n+1)))^(n+1) −n^2 (1+(1/n))^n } posons : { (((1/(n+1))=a)),(((1/n)=b)) :}=> { ((n=(1/a)−1)),((n=(1/b))) :}=> { ((n→+∞;a→0)),((n→+∞;b→0)) :} =lim_((a;b)→(0;0)) {((1/a)−1)^2 e^((ln(1+a))/a) −(1/b^2 )e^((ln(1+b))/b) } =lim_((a;b)→(0;0)) {e^(((ln(1+a))/a)+2ln(1−a)−2lna) −e^(((ln(1+b))/b)+2lnb) } =lim_((a;b)→(0;0)) (1−(2/a))e^((ln(1+a))/a) =−∞×1=−∞ =>lim_(n→+∞) n^2 [(1+(1/(n+1)))^(n+1) −(1+(1/n))^n ]=−∞$
	$\lim_{n \to + \infty} n^{2} [{(1 + \frac{1}{n + 1})}^{n + 1} - {(1 + \frac{1}{n})}^{n}]$ $= \lim_{n \to + \infty} {n^{2} {(1 + \frac{1}{n + 1})}^{n + 1} - n^{2} {(1 + \frac{1}{n})}^{n}}$ $p o s o n s : {\begin{cases} \frac{1}{n + 1} = a \\ \frac{1}{n} = b \end{cases} => {\begin{cases} n = \frac{1}{a} - 1 \\ n = \frac{1}{b} \end{cases} => {\begin{cases} n \to + \infty; a \to 0 \\ n \to + \infty; b \to 0 \end{cases}$ $= \lim_{(a; b) \to (0; 0)} {{(\frac{1}{a} - 1)}^{2} e^{\frac{l n (1 + a)}{a}} - \frac{1}{b^{2}} e^{\frac{l n (1 + b)}{b}}}$ $= \lim_{(a; b) \to (0; 0)} {e^{\frac{l n (1 + a)}{a} + 2 l n (1 - a) - 2 l n a} - e^{\frac{l n (1 + b)}{b} + 2 l n b}}$ $= \lim_{(a; b) \to (0; 0)} (1 - \frac{2}{a}) e^{\frac{l n (1 + a)}{a}} = - \infty \times 1 = - \infty$ $=> \lim_{n \to + \infty} n^{2} [{(1 + \frac{1}{n + 1})}^{n + 1} - {(1 + \frac{1}{n})}^{n}] = - \infty$
	Commented by Ghisom last updated on 17/Oct/24

	$wrong$ ${(1 + \frac{1}{n + 1})}^{n + 1} - {(1 + \frac{1}{n})}^{n} > 0 \forall n \in Z^{+}$ $\Rightarrow the limit cannot be < 0$
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