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.
Best Answer
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 $$\lim x_n=\frac12e$$