My friend's trivia league had this math question:
$$\lim_{n \to \infty} \left[\frac{(n+1)^{n + 1}}{n^n} - \frac{n^{n}}{(n-1)^{n-1}} \right]$$
After computing a few values, one could guess the answer is $e$ = 2.718...But how can we prove that is the limit?
Someone offered up a hand-wavy proof like this:
\begin{align} \lim_{n \to \infty} \left[\frac{(n+1)^{n + 1}}{n^n} - \frac{n^{n}}{(n-1)^{n-1}} \right] & = \lim_{n \to \infty} \left[\frac{(n+1)(n+1)^{n}}{n^n} - \frac{n \cdot n^{n-1}}{(n-1)^{n-1}} \right] \\ &= \lim_{n \to \infty} \left[(n+1)\frac{(n+1)^{n}}{n^n} - n\frac{n^{n-1}}{(n-1)^{n-1}} \right] \\ &= \lim_{n \to \infty} \left[(n+1)\left(1 + \frac{1}{n} \right)^n - n\left(\frac{n - 1 + 1}{n-1} \right)^{{n-1}} \right] \\ &= \lim_{n \to \infty} \left[(n+1)\left(1 + \frac{1}{n} \right)^n - n\left(1 + \frac{1}{n-1} \right)^{n-1} \right] \\ &= \lim_{n \to \infty} \left[(n+1)e - n \cdot e \right] \\ &= \lim_{n \to \infty} e \\ &= e \end{align}
The part about substituting $e$ is hand-wavy since technically this is an indeterminate form of $\infty - \infty$. And using $e$ as as upper bound did not lead me to an easy proof either.
Is there a way to rigorously prove the limit? I tried a few approaches: (a) sandwiching the limit--I could prove $e$ was a lower bound, but I could not find a suitable upper bound converging to $e$, (b) using L'Hopital's rule with no luck, (c) using the mean value theorem--but that also got complicated.
So this is a pretty tough problem to ask at trivia! Is there a way to prove this limit formally?
Sources
Trivia question: http://learnedleague.com/question.php?72&16&4
Thread on trivia: http://learnedleague.com/viewtopic.php?f=10&t=7961&hilit=euler
Hand-wavy proof: http://imgur.com/rIXghhw
Idea for mean value theorem: http://www.pharout.com/trickylimitproblem.pdf