A young limit $\lim_{n\to\infty} \frac{{(n+1)}^{n+1}}{n^n} - \frac{{n}^{n}}{{(n-1)}^{n-1}} =e$ 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.

$$\eqalign{(n+1)^{n+1} = \exp\left((n+1) \ln(n+1)\right) &= \exp\left(((n+1) (\ln n + \frac{1}{n} - \frac{1}{2n^2} + O(n^{-3}))\right)\cr
&= \exp\left( n \ln n + \ln n + 1 + \frac{1}{2n} + O(n^{-2})\right) \cr &= n^{n+1} e \left(1 + \frac{1}{2n}+ O(n^{-2})\right)}$$
so
$$ \dfrac{(n+1)^{n+1}}{n^n} = n e + \frac{e}{2} + O(1/n)$$
Replacing $n$ by $n-1$,
$$ \dfrac{n^n}{(n-1)^{n-1}} = (n-1) e + \frac{e}{2} + O(1/(n-1)) = ne - \frac{e}{2} + O(1/n)$$
So 
$$ \dfrac{(n+1)^{n+1}}{n^n} - \dfrac{n^n}{(n-1)^{n-1}}  = e + O(1/n)$$
If you include more terms, you get
$$ \dfrac{(n+1)^{n+1}}{n^n} - \dfrac{n^n}{(n-1)^{n-1}}  = e + \dfrac{e}{24 n^2} + \dfrac{11\; e}{640 n^4} +  \ldots $$
Hmm, the denominators of the coefficients seem to be the almost the same as OEIS sequence A118051 
http://oeis.org/A118051 which comes from the asymptotics of the inverse harmonic numbers.
