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Here is my question:

$$\displaystyle\lim_{n\rightarrow \infty}n^2\left[\left(1+\frac{1}{1+n}\right)^{n+1}-\left(1+\frac{1}{n}\right)^{n}\right]=?$$

Any hints will be fine. Thank you!

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HINT

Note that (to be made more rigorous)

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

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

then

$$n^2\left[\left(1+\frac{1}{1+n}\right)^{n+1} - \left(1+\frac{1}{n}\right)^{n}\right]\sim e\cdot n^2\left(\frac{1}{2n}-\frac{1}{2(1+n)}\right)=e\cdot n^2\left(\frac{2}{4n^2+4n}\right)$$

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    $\begingroup$ Upvoted. Your answer is better than mine. $\endgroup$ – mickep May 3 '18 at 15:52
  • $\begingroup$ @mickep Thanks, appreciate! $\endgroup$ – gimusi May 3 '18 at 16:05
  • $\begingroup$ @gimusi Thanks! $\endgroup$ – Yan kai May 9 '18 at 13:35
  • $\begingroup$ @Yankai You are welcome! Bye $\endgroup$ – gimusi May 9 '18 at 13:41
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Here is "any" hint:

Set $x=1/n$ and you will have $$ \lim_{x\to 0^+}\frac{1}{x^2}\biggl(\Bigl(1+\frac{x}{1+x}\Bigr)^{1+1/x}-(1+x)^{1/x}\biggr). $$ The expression $$ \Bigl(1+\frac{x}{1+x}\Bigr)^{1+1/x}-(1+x)^{1/x} $$ in the right-hand side can be expanded around $x=0$ (perhaps after defining it as $0$ at $x=0$).

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Hint: Use $$n^2\left[\left(1+\frac{1}{1+n}\right)^{n+1}-\left(1+\frac{1}{n}\right)^{n}\right]=\frac{\left(1+\frac{1}{1+n}\right)^{n+1}-\left(1+\frac{1}{n}\right)^{n}}{\frac1{n^2}}$$ and then use L'Hopital's Rule.

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