If $x_0 = 5$ and $x_{n+1} = x_n + \frac {1}{x_n},$ show that

$45<x_{1000}<45.1$

This problem is taken from the list submitted for the $1975$ Canadian Mathematics Olympiad (but not used on the actual exam).

SOURCE :CRUX(Page Number 3 ; Question Number 162)

I tried writing out the first few terms :

$x_1 = 5+ \frac{1}{5} $

$x_2 = \big(5+\frac{1}{5}\big) + \big(5+\frac{1}{5}\big)^{-1} = \frac{x_0^2 + 1}{x_0} + \frac{x_0}{x_0^2 + 1} = \frac{(x_0^2 + 1)^2+x_0^2}{x_0(x_0^2+1)}$

$x_3 = \frac{(x_0^2 + 1)^2+x_0^2}{x_0(x_0^2+1)} + \frac{x_0(x_0^2+1)}{(x_0^2 + 1)^2+x_0^2} = Messy$

I tried a lot but could not find any general formula for the $n$th term. Does there even exist any?

Also it is clear that $\big(x_n + \frac{1}{x_n}\big)$ is an increasing function. So I think sequence diverges, but how can the $1000th$ term be calculated or aprroximated?

Any help would be gratefully acknowledged :).