If $a_{n+1}=a_n+\frac1{a_n}$, then $a_n/n$ converges to $0$

Question

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?

Answer 1

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 $a_{N+k+1} < \varepsilon$.

Answer 2

If $a_n$ is bounded, then you are done.

Otherwise, consider the squares: $$ a_{n+1}^2 -a_n^2 = 2+\frac{1}{a_n^2} \le 3 $$ for large enough $n$. Conclude by induction that an inequality of the form $a_n^2\le C+3n$ holds.

Answer 3

You can show the result in a more straightforward manner.

By considering the associated differential equation, you can see that : $$f' = 1/f$$ Gives the solution $f(x)^2 = 2x$. This gives you the following guess : $a_n\sim \sqrt{2n} $ which is actually true.

To show that, as Care Bear suggested, consider the sequence $a_n^2$ (the square are not random exponent because of the previous explanation), since $a_n$ grows to infinity (you already showed that), you can use Cesaro lemma with $a_n^2$ to show that $a_n^2 \sim 2n$ because $a_{n+1}-a_n = 2 + 1/a_n^2$ and you are done.