If $a_{n+1}=a_n+\frac1{a_n}$, then $a_n/n$ converges to $0$ Let $a_{n+1}=a_n+\dfrac1{a_n}$, with $a_n=1$.
Prove $\lim \limits_{n\to \infty }\left(\dfrac{a_n}{n}\right)=0$.
Now I already know that it is  monotonically increasing and that $a_n\to \infty$ as $n\to \infty$.
I thought of using Cauchy here, but I don't know how exactly.
NOTE: Stolz–Cesàro theorem is forbidden in this question.
Could anyone help me with this, please?

If you know that $a_n \to \infty,$ you can apply Stolz theorem to obtain 
$$\lim_{n \to \infty} \frac{a_n}{n} = \lim_{n \to \infty} \frac{a_{n+1} - a_n}{(n+1)-n} = \lim_{n \to \infty} \frac{1}{a_n} = 0.$$
Edit: I'm adding solution which doesn't use Stolz theorem. 
Let $\varepsilon > 0$. Since $a_n \to \infty$, there is such $N$ that $a_n \geqslant \frac{2}{\varepsilon}$ for $n \geqslant N$. Thus 
$$a_{N+k+1} = a_{N+1} + \frac{1}{a_{N+1}} + \ldots + \frac{1}{a_{N+k}} \leqslant a_{N+1} + k \cdot \frac{1}{a_{N+1}} \leqslant a_{N+1} + k \cdot \frac{\varepsilon}{2}$$
so 
$$\frac{a_{N+k+1}}{N+k+1} \leqslant \frac{a_{N+1}}{N+k+1} + \frac{k}{N+k+1} \cdot \frac{\varepsilon}{2} \leqslant \frac{a_{N+1}}{N+k+1} + \frac{\varepsilon}{2}$$
For large $k$ we have $\frac{a_{N+1}}{N+k+1} < \frac{\varepsilon}{2}$ and therefore $a_{N+k+1} < \varepsilon$.
