Let $x_0 = 5,$ and $x_{n+1} = x_n + \frac {1} {x_n}$ for n = $0, 1, 2, . . ..$ Show that $45 < x_{1000} < 45.1.$
First noticing that $x_{n+1}-x_n= \frac {1}{x_n},$ we get $\sum \limits_{n=1}^{1000} \frac{1}{x_n} = x_{1000} - 5$ with some simple telescoping.
Now we want to find some inequalities for $\sum \limits_{n=1}^{1000} \frac{1}{x_n},$ which I tried to do with more telescoping, but I've had no luck finding a good way to express $\frac{1}{x_n}$ between two bounds.
