Interesting Limit (11)

Note that $n^2\left[\left(1 + \tfrac{1}{n+1}\right)^{n+1} - \left(1 + \tfrac{1}{n}\right)^n\right] \; = \; n^2\left(1 + \tfrac{1}{n}\right)^n\left[\frac{(n+2)^{n+1}n^n}{(n+1)^{2n+1}} - 1 \right] \; = \; n^2\left(1 + \tfrac{1}{n}\right)^n\left[\frac{n+1}{n}\left(1 - \frac{1}{(n+1)^2}\right)^{n+1} - 1\right]$ Now $\begin{aligned} \ln\left(1 - \tfrac{1}{(n+1)^2}\right) & = \; -\tfrac{1}{(n+1)^2} + o\big(n^{-3}\big) \\ (n+1)\ln\left(1 - \tfrac{1}{(n+1)^2}\right) & = \; -\tfrac{1}{n+1} + o\big(n^{-2}\big) \\ \left(1 - \tfrac{1}{(n+1)^2}\right)^{n+1} & = \; e^{-\frac{1}{n+1} + o(n^{-2})} \; = \; 1 - \tfrac{1}{n+1} + \tfrac{1}{2(n+1)^2} + o\big(n^{-2}\big) \\ \tfrac{n+1}{n}\left(1 - \tfrac{1}{(n+1)^2}\right)^{n+1} &= \; 1 + \tfrac{1}{2n(n+1)} + o\big(n^{-2}\big) \\ n^2\left[\tfrac{n+1}{n}\left(1 - \tfrac{1}{(n+1)^2}\right)^{n+1} - 1\right] & = \; \tfrac{n}{2(n+1)} + o(1) \end{aligned}$ as $n \to \infty$ . Thus we deduce that $\lim_{n \to \infty} n^2\left[\left(1 + \tfrac{1}{n+1}\right)^{n+1} - \left(1 + \tfrac{1}{n}\right)^n\right] \; =\; \boxed{\tfrac12e}$

We apply the mean value theorem to $f(x) = \left(1+\frac{1}{x}\right)^x$ on $[n,n+1]$ to give a function $\varepsilon : (0,\infty)\to(0,1)$ such that $f(n+1)-f(n) = f'(n+\varepsilon(n))((n+1)-n) = \left(1+\frac{1}{n+\varepsilon(n)}\right)^{n+\varepsilon(n)}\left(\ln\left(1+\frac{1}{n+\varepsilon(n)}\right) - \frac{1}{n+\varepsilon(n)+1}\right)$

Now, since $0 < \varepsilon(n) < 1$ implies $\lim_{n\to\infty} \frac{n^2}{(n+\varepsilon(n))^2} = 1$ , it follows that $\begin{aligned} \lim_{n\to\infty} n^2\left(\left(1+\frac{1}{n+1}\right)^{n+1} - \left(1+\frac{1}{n}\right)^n\right) &= \lim_{n\to\infty} (n+\varepsilon(n))^2\left(1+\frac{1}{n+\varepsilon(n)}\right)^{n+\varepsilon(n)}\left(\ln\left(1+\frac{1}{n+\varepsilon(n)}\right) - \frac{1}{n+\varepsilon(n)+1}\right) \\ &= \lim_{x\to\infty} x^2\left(1+\frac{1}{x}\right)^x \left(\ln\left(1+\frac{1}{x}\right) - \frac{1}{x+1}\right) \end{aligned}$ where this last equality is true as long as the last limit exists. To evaluate it, we can use the fact $\left(1+\frac{1}{x}\right)^x\to e$ along with one application of L'Hopital's rule: $\begin{aligned} \lim_{x\to\infty} x^2\left(1+\frac{1}{x}\right)^x \left(\ln\left(1+\frac{1}{x}\right) - \frac{1}{x+1}\right) &= e \cdot \lim_{x\to\infty} \frac{\ln\left(1+\frac{1}{x}\right) - \frac{1}{x+1}}{\frac{1}{x^2}} \\ &= e \cdot \lim_{x\to\infty} \frac{\frac{1}{x+1}-\frac{1}{x} + \frac{1}{(x+1)^2}}{-\frac{2}{x^3}} \\ &= e \cdot \lim_{x\to\infty} \frac{x^2}{2(x+1)^2} \\ &= e \cdot \frac{1}{2} = \boxed{\frac{e}{2} \approx 1.35914091} \end{aligned}$

Note: In this case, $f''(x) < 0$ for all $x>0$ , so $\varepsilon$ is uniquely defined. Even for general functions $f$ satisfying the assumptions of the mean value theorem, we can always choose such a function $\varepsilon$ by the axiom of choice.