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$$\lim_{n \to \infty}n \cdot \left [ \frac{\left (1+\frac{1}{n+1} \right )^{n+1}}{e}-1 \right ]$$ I was trying to calculate a limit that drove me to this case of Raabe-Duhamel's test, but I don't know how to finish it. Please give me a hint or a piece of advise.

I cannot use any of the solution below, but they are clear and good. I'm trying to prove it using squeeze theorem like this: $$\lim_{n \to \infty}n \cdot \left [ \frac{\left (1+\frac{1}{n+1} \right )^{n+1}}{e}-1 \right ]=\frac{-1}{e} \cdot\lim_{n \to \infty}n \cdot \left [e- \left (1+\frac{1}{n+1} \right )^{n+1} \right ]$$ I found this: $$\frac{e}{2n+2}<e- \left (1+\frac{1}{n} \right )^{n}<\frac{e}{2n+1}$$ Is this true? How can I prove this? Thanks for the answers.

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    $\begingroup$ Get an asymptotic expansion for $(n+1)\ln\left(1+\frac1{n+1}\right)$ $\endgroup$ – Lord Shark the Unknown Dec 6 '19 at 19:30
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    $\begingroup$ -1/2 :D such a nice limit $\endgroup$ – Evgeny Kuznetsov Dec 6 '19 at 19:38
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HINT

As suggested in the comments by Lord Shark the Unknown, we can use that

$$\left (1+\frac{1}{n+1} \right )^{n+1}=e^{(n+1)\log\left (1+\frac{1}{n+1} \right )}=e^{(n+1)\left (\frac{1}{n+1}-\frac{1}{2(n+1)^2}+o\left(\frac1{n^2}\right) \right )}$$

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    $\begingroup$ Do you mean $o(\frac1{n^2})$ instead of $O(\frac1{n^2})$? otherwise the $-\frac1{2(n+1)^2}$ would be useless $\endgroup$ – supinf Dec 6 '19 at 19:42
  • $\begingroup$ @supinf Yes of course I fix that! Thanks $\endgroup$ – user Dec 6 '19 at 19:42
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An overkilled method: \begin{align*} n\left[\dfrac{\left(1+\dfrac{1}{n+1}\right)^{n+1}}{e}-1\right]&=\dfrac{n}{n+1}\cdot(n+1)\cdot\left[\dfrac{\left(1+\dfrac{1}{n+1}\right)^{n+1}}{e}-1\right], \end{align*} so we are to look at \begin{align*} (n+1)\cdot\left[\dfrac{\left(1+\dfrac{1}{n+1}\right)^{n+1}}{e}-1\right], \end{align*} somehow it is the same looking at \begin{align*} n\cdot\left[\dfrac{\left(1+\dfrac{1}{n}\right)^{n}}{e}-1\right]=\dfrac{1}{e}\cdot n\cdot\left[\left(1+\dfrac{1}{n}\right)^{n}-e\right]. \end{align*} We note that \begin{align*} \lim_{x\rightarrow 0}(1+x)^{1/x}=e, \end{align*} so \begin{align*} n\cdot\left[\left(1+\dfrac{1}{n}\right)^{n}-e\right]=n\int_{0}^{1/n}\left((1+x)^{1/x}\right)'dx. \end{align*} Taking the derivative of the integrand, we find that \begin{align*} n\int_{0}^{1/n}\left((1+x)^{1/x}\right)'dx=n\int_{0}^{1/n}\left((1+x)^{1/x}\left(-\dfrac{1}{x^{2}}\log(1+x)+\dfrac{1}{x}\dfrac{1}{x+1}\right)\right)dx. \end{align*} By Integral Mean Value Theorem applied to the interval $[0,1/n]$, we get \begin{align*} &n\int_{0}^{1/n}\left((1+x)^{1/x}\left(-\dfrac{1}{x^{2}}\log(1+x)+\dfrac{1}{x}\dfrac{1}{x+1}\right)\right)dx\\ &=(1+\eta_{x})^{1/\eta_{x}}\left(-\dfrac{1}{\eta_{x}^{2}}\log(1+\eta_{x})+\dfrac{1}{\eta_{x}}\dfrac{1}{\eta_{x}+1}\right). \end{align*} However, it is not hard to compute that \begin{align*} \lim_{x\rightarrow 0}(1+x)^{1/x}\left(-\dfrac{1}{x^{2}}\log(1+x)+\dfrac{1}{x}\dfrac{1}{x+1}\right)=-e/2, \end{align*} so as $x\rightarrow 0$, $\eta_{x}\rightarrow 0$ and the limit is $-1/2$.

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  • $\begingroup$ Too too long. we don't have much time. $\endgroup$ – hamam_Abdallah Dec 7 '19 at 8:37
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hint

Since $$\lim_{n\to\infty}\frac{n}{n-1}=1,$$ It is the same to compute $$\lim_{n\to\infty}n(\frac{(1+\frac 1n)^{n}}{e}-1)$$

use the fact that $$n\ln(1+\frac 1n)=n(\frac 1n -\frac{1}{2n^2} +\frac{1}{n^2}\epsilon(n))$$ $$=1-\frac{1}{2n}+\frac 1n\epsilon(n).$$

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  • $\begingroup$ @hamam_Abdallah :D It is not hint it is full calculation steps, this is what the User meant! $\endgroup$ – Evgeny Kuznetsov Dec 6 '19 at 19:46

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