TITLE: Limit $(n - a_n)$ of sequence $a_{n+1} = \sqrt{n^2 - a_n}$
QUESTION [5 upvotes]: Consider the sequence $\{a_n\}_{n=1}^{\infty}$ defined recursively by
  $$a_{n+1} = \sqrt{n^2 - a_n}$$ with $a_1 = 1$. Compute $$\lim_{n\to\infty} (n-a_n)$$

I am having trouble with this. I am not even sure how to show the limit exists. I think if we know the limit exists, it is just algebra, but I'm not sure.

REPLY [3 votes]: Here is a briefer answer which illustrates that the difficulty of the problem lies in bounding the growth of $a_n$. 
All we need is $a_n \sim n$, in the sense that $$\lim_{n\to\infty} \frac{a_n}{n} = 1.$$
This would follow if we knew e.g. that $n - a_n$ is bounded. JimmyK's sharp result that $0 \le n - a_n \le 2$ is more than enough! 
So, indeed, $a_n \sim n$. From here, using difference of squares, 
$$(n+1) - a_{n+1} = \frac{(n+1)^2 - (n^2 - a_n)}{(n+1) + a_{n+1}} = \frac{2n + 1 + a_n}{n + 1 + a_{n+1}} = \frac{2 + \frac{1}{n} + \frac{a_n}{n}}{1 + \frac{1}{n} + \frac{a_{n+1}}{n}} \to \frac{2 + 0 + 1}{1 + 0 + 1} = \frac{3}{2}.$$