These days I saw a really interesting limit as I was reading more information on Napier's constant here : http://mathworld.wolfram.com/e.html. It seems a pretty young limit since it appears under the name and year "Brothers and Knox 1998". It's also new for me and I'd like to know more approaching ways for it.
$\lim_{n\to\infty} \frac{{(n+1)}^{n+1}}{n^n} - \frac{{n}^{n}}{{(n-1)}^{n-1}} =e$
A possible way to go is to use the first converse of the Stolz–Cesàro theorem, but since $\lim_{n\to\infty}\frac{b_{n}}{b_{n+1}}=1$ we can at most check the given result because the theorem may work
or not in this case. For the case when $\lim_{n\to\infty}\frac{b_{n}}{b_{n+1}}=1$ it is required more research in
order to be sure that we can safely apply the first converse theorem. I'll make an update
as soon as things are clarified such that I may turn it into a rigorous proof.
$\lim_{n\to\infty} \frac{{(n+1)}^{n+1}}{n^n} - \frac{{n}^{n}}{{(n-1)}^{n-1}}=$ $\lim_{n\to\infty} f(n+1) - f(n)=$ By the first converse of the Stolz–Cesàro theorem we have $\lim_{n\to\infty} \frac{f(n+1) - f(n)}{n+1-n}=$ $\lim_{n\to\infty} \frac{f(n)}{n}=$ $\lim_{n\to\infty} \frac{n^n}{(n-1)^{n-1}}\cdot \frac{1}{n}=$ $\lim_{n\to\infty} \frac{n^{n-1}}{(n-1)^{n-1}}=\left(1+\frac{1}{n-1}\right)^{n-1}=e.$
What else can we do here? Thanks.