Property of $a_{n+1} = a_n - \frac{1}{a_n}$

Question

For a $a_n$ defined recursively by $a_{n+1} = a_n - \frac{1}{a_n}$,$a_0 = k >0$. Prove that if the first $n$ such that $a_n \leq 0$, then $n \in O(k^2)$.

I ran a computer simulation, and it seems true.

Moreover, these questions appear similar to this question: Closed form for the sequence defined by $a_0=1$ and $a_{n+1} = a_n + a_n^{-1}$ .

However, I have no idea how to prove that. Do you have any ideas?

Answer 1

Show the following for $ a_0 = k$. If you're stuck, explain what you've tried.

1. $a_{k-1} > k-1 $
2. $a_{k-1 + k-2} > k-2 $
3. $a_{k-1 + k-2 + k-3 } > k-3$
4. $ a_{k-1 + k-2 + k-2 + \ldots + 1 } > 0 $
5. $a_k < k-1 $
6. $a_{k + k-1} < k-2 $
7. $a_{k + k-1 + k-2 } < k-3 $
8. $a_{k + k-1 + k-2 + \ldots + 1 } < 0 $
9. Hence $ \frac{k^2 - k } { 2 } < n_k \leq \frac{ k^2 + k}{2}$.

Hence, the result follows.