An unusual limit involving $e$ From MathWorld I have the following quote:
$e$ is given by the unusual limit $$\lim_{n \to \infty}\left(\frac{(n + 1)^{n + 1}}{n^{n}} - \frac{n^{n}}{(n - 1)^{n - 1}}\right) = e\tag{1}$$ Now if we put $$a_{n} = \frac{(n + 1)^{n + 1}}{n^{n}}$$ then the above result says that $$\lim_{n \to \infty}(a_{n} - a_{n - 1}) = e$$ Now we know that
Theorem: If $a_{n} - a_{n - 1} \to L$ then $a_{n}/n \to L$.
Using the example here we see that $$\frac{a_{n}}{n} = \frac{n + 1}{n}\left(1 + \frac{1}{n}\right)^{n} \to e$$ I am thinking of some partial converse to the standard theorem above which can get us from limit of $a_{n}/n$ to limit of $a_{n} - a_{n - 1}$.
Another option to prove $(1)$ is to make use of real variable theory. Thus we can put $x = 1/n$ and deal with function $f(x) = \left(1 + \dfrac{1}{x}\right)(1 + x)^{1/x}$ and use Taylor expansion $$(1 + x)^{1/x} = e - \frac{ex}{2} + \frac{11e}{24}x^{2} + \cdots$$ and $$g(x) = \left(\frac{1}{x} - 1\right)(1 - x)^{-1/x}$$ and calculate the limit of $f(x) - g(x)$ as $x \to 0$. This way we see that $$f(x) = \frac{e}{2} + \frac{e}{x} + o(1), g(x) = -\frac{e}{2} + \frac{e}{x} + o(1)$$ and clearly $f(x) - g(x) \to e$ as $x \to 0$.
Is there a proof without using real variable theory and just dealing with theorems on sequences which establishes the result $(1)$?
 A: For $n > 1$, we can write
$$a_n - a_{n-1} = \biggl(1+\frac{1}{n}\biggr)^n\cdot \biggl(n+1 - (n-1)\frac{n^{2n}}{(n^2-1)^n}\biggr).\tag{1}$$
Bernoulli's inequality says on the one hand that
$$\frac{n^{2n}}{(n^2-1)^n} = \biggl(1 + \frac{1}{n^2-1}\biggr)^n \geqslant 1 + \frac{n}{n^2-1},$$
so
$$n+1 - (n-1)\frac{n^{2n}}{(n^2-1)^n} \leqslant 2 - \frac{n}{n+1} = 1+ \frac{1}{n+1},$$
and on the other hand it says
$$\frac{n^{2n}}{(n^2-1)^n} = \frac{1}{\Bigl(1 - \frac{1}{n^2}\Bigr)^n} \leqslant \frac{1}{1-\frac{1}{n}} = \frac{n}{n-1},$$
whence
$$n+1 - (n-1)\frac{n^{2n}}{(n^2-1)^n} \geqslant (n+1) - n = 1.$$
Since the first factor in $(1)$ converges to $e$, and the second to $1$, as we just saw, it follows that $a_n - a_{n-1} \to e.$
A: Given that:
$$\begin{eqnarray*} a_n = \frac{(n+1)^{n+1}}{n^n} &=& (n+1)\left(1+\frac{1}{n}\right)^n=n\left(1+\frac{1}{n}\right)^{n+1}\end{eqnarray*} $$
we have $ ne\leq a_n\leq (n+1)e$ by the Bernoulli inequality, hence $b_n=a_n-a_{n-1}$ is non-negative and bounded by $2e$. If we manage to prove that $\{b_n\}$ is a decreasing sequence, then $\{b_n\}$ is a converging sequence (to its infimum). If $b_n$ is converging, then:
$$ \lim_{n\to +\infty}b_n = \lim_{n\to +\infty}\frac{b_1+b_2+\ldots+b_n}{n}=\lim_{n\to +\infty}\frac{a_n-a_0}{n}=e$$
by Cesàro theorem. So you only need to prove that $\{a_n\}$ is a concave sequence to prove your claim.
For instance, that follows from:
$$\frac{d^2 a_n}{dn^2} = \frac{(n+1)^n}{n^{n+1}}\left[-1+n(n+1)\log^2\left(1+\frac{1}{n}\right)\right].$$
