TITLE: Given $a_{1}=1, \ a_{n+1}=a_{n}+\frac{1}{a_{n}}$, find $\lim \limits_{n\to\infty}\frac{a_{n}}{n}$
QUESTION [10 upvotes]: I started by showing that $1\leq a_{n} \leq n$ (by induction) and then $\frac{1}{n}\leq \frac{a_{n}}{n} \leq 1$ which doesn't really get me anywhere.
On a different path I showed that $a_{n} \to \infty$ but can't see how that helps me.

REPLY [12 votes]: I completely forgot about the Stolz-Cesàro theorem, from which we get:
$$\lim_{n\to \infty} \frac{a_n}{n}=\lim_{n\to\infty} \frac{a_{n+1}-a_{n}}{(n+1)-n}=\lim_{n\to \infty}\frac{\frac{1}{a_{n}}}{1}=\lim_{n\to \infty}\frac{1}{a_{n}}=0. $$ The same technique works for $\displaystyle \frac{a_{n}^2}{n}.$