Compute $\lim\limits_{n\to\infty}n^2\left(\left(1+\frac1{n+1}\right)^{n+1}-\left(1+\frac1n\right)^n\right)$

A problem I saw recently is to compute $$\lim_{n\to\infty} n^2 \left(\left(1+\dfrac{1}{n+1}\right)^{n+1}-\left(1+\dfrac{1}{n}\right)^n\right)$$

I thought it would be fun to give this a go but it completely stumped me. I am not aware of any of the standard techniques used to compute $\infty\times (e-e)$ - type limits, bar possibly L'Hôpital's Rule. But I doubt that would be a good idea here, at best it would be ugly. I would appreciate solutions using any methods, the more elegant the better of course. Though I welcome any relevant hints as well if full solutions are discouraged.

Only necessary ingredient: $\log(1+x)=x-\frac12x^2+o(x^2)$ when $x\to0$; in particular, $\frac1x\log(1+x)\to1$ when $x\to0$.

Rewrite this as $$x_n=n^2\left(f\left(\frac1{n+1}\right)-f\left(\frac1n\right)\right)$$ where $$f(x)=\left(1+x\right)^{1/x}$$ Now, $f(x)=\exp\left(\frac1x\log\left(1+x\right)\right)$ hence $$f'(x)=g(x)\frac1{x^2(1+x)}f(x)$$ with $$g(x)=x-(1+x)\log(1+x)$$ When $x\to0$, $\log(1+x)=x-\frac12x^2+o(x^2)$ hence $$g(x)=x-(1+x)\left(x-\frac12x^2+o(x^2)\right)\sim-\frac12x^2$$ Since $f(x)\to e$ when $x\to0$, this yields $$f'(x)\sim-\frac12x^2\cdot\frac1{x^2}\cdot e=-\frac12e$$ hence $$f\left(\frac1{n+1}\right)-f\left(\frac1n\right)\sim-\frac12e\left(\frac1{n+1}-\frac1n\right)\sim\frac12e\frac1{n^2}$$ and finally $$\lim x_n=\frac12e$$

Let consider the sequence: \begin{align} x_n &=n\log\left(1+\frac 1n\right)\\ &\sim n\left(\frac 1n-\frac 1{n^2}+o\left(\frac 1{n^2}\right)\right)\\ &\sim 1-\frac 1{2n}+\frac 1{3n^2}+o\left(\frac 1{n^2}\right) \end{align} Then \begin{align} x_{n+1}-x_n &=\left(1-\frac 1{2(n+1)}+\frac 1{3(n+1)^2}\right)-\left(1-\frac 1{2n}+\frac 1{3n^2}\right)+o\left(\frac 1{n^2}\right)\\ &=\frac 1{2n(n+1)}+\frac 1{3(n+1)^2}-\frac 1{3n^2}+o\left(\frac 1{n^2}\right)\\ &=\frac 1{2n^2}+o\left(\frac 1{n^2}\right)\\ &\sim\frac 1{2n^2} \end{align} consequently \begin{align} n^2 \left(\left(1+\dfrac{1}{n+1}\right)^{n+1}-\left(1+\dfrac{1}{n}\right)^n\right) &=n^2(e^{x_{n+1}}-e^{x_n})\\ &=n^2e^{x_n}(e^{x_{n+1}-x_n}-1)\\ &\sim n^2e^{x_n}(x_{n+1}-x_n)\\ &\sim \frac{e^{x_n}}2\\ &\to\frac e2 \end{align}

Too long for a comment.

Using the same approach as in Fabio Lucchini's answer, you could even get more than the limit itself at the price of one single extra term in the expansions. $$x_n=n\log\left(1+\frac 1n\right)=1-\frac{1}{2 n}+\frac{1}{3 n^2}-\frac{1}{4 n^3}+O\left(\frac{1}{n^4}\right)$$ $$e^{x_n}=e-\frac{e}{2 n}+\frac{11 e}{24 n^2}-\frac{7 e}{16 n^3}+O\left(\frac{1}{n^4}\right)$$ Doing the same and continuing with Taylor series (or long division) $$x_{n+1}=(n+1)\log\left(1+\frac 1{n+1}\right)=1-\frac{1}{2 n}+\frac{5}{6 n^2}-\frac{17}{12 n^3}+O\left(\frac{1}{n^4}\right)$$ $$e^{x_{n+1}}=e-\frac{e}{2 n}+\frac{23 e}{24 n^2}-\frac{89 e}{48 n^3}+O\left(\frac{1}{n^4}\right)$$ $$e^{x_{n+1}}-e^{x_{n}}=\frac{e}{2 n^2}-\frac{17 e}{12 n^3}+O\left(\frac{1}{n^4}\right)$$

$$n^2\left(e^{x_{n+1}}-e^{x_{n}} \right)=\frac{e}{2}-\frac{17 e}{12 n}+O\left(\frac{1}{n^2}\right)$$ which shows th limpt and also how it is approached.

Try even with a small number such as $n=10$. The exact result would be $\frac{29833762621839898789}{28531167061100000000}$ which is $\approx 1.04566$ while the above truncated expression would give $\frac{43 e}{120}\approx 0.97405$.