TITLE: A young limit $\lim_{n\to\infty} \frac{{(n+1)}^{n+1}}{n^n} - \frac{{n}^{n}}{{(n-1)}^{n-1}} =e$
QUESTION [7 upvotes]: 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.

REPLY [4 votes]: Brothers and Knox paper can be found here.
Now here is another approach. We can write
$$
\frac{{(n+1)}^{n+1}}{n^n} - \frac{{n}^{n}}{{(n-1)}^{n-1}}=\left(1+\frac {1}{n}\right)^n+n\left[\left(1+\frac{1}{n}\right)^n-\left(1+\frac{1}{n-1}\right)^{n-1} \right].
$$
Since 
$$
\left(1+\frac{1}{x}\right)^x=e \bigg[1-\frac{1}{2x}+\frac{11}{24x^2}-\frac{7}{16x^3}+\frac{2447}{5760x^4}-\frac{959}{2304x^5}+\frac{238\,043}{580\,608x^6}-\frac{67223}{165\,888x^7}+\frac{559\,440\,199}{1\,393\,459\,200x^8}-\frac{123\,377\,159}{309\,657\,600x^9}+\frac{29\,128\,857\,391}{73\,574\,645\,760x^{10}}-\frac{5\,267\,725\,147}{13\,377\,208\,320x^{11}}+O\left(\frac{1}{x^{12}} \right) \bigg]
$$
we get
$$
\frac{{(n+1)}^{n+1}}{n^n} - \frac{{n}^{n}}{{(n-1)}^{n-1}}=e+\frac{e}{24n^2}+\frac{11e}{640n^4}+\frac{5525e}{580\,608n^6}+\frac{1\,212\,281e}{199\,065\,600n^8}+\frac{772\,193e}{181\,665\,792n^{10}}+O\left(\frac{1}{n^{11}}\right).
$$
The denominators of the coefficients $24, 640, 580\,608,199\,065\,600$ are the same as OEIS sequence, but $181\,665\,792$ is different.
What is interesting (at least for me), that the odd powers are missing and the coefficients are positive numbers.
One more thing.
If $\lim a_n=a$ and $\lim n(a_n-a_{n-1})=0$ then $\lim (n+1)a_n-na_{n-1}=a$. The special case $a_n=\left(1+\frac{1}{n}\right)^n$ is the original statement.