Proving that a sequence is between certain values at certain n I'm given that $a_1=1$, and for every $n \gt1, a_{n+1} = a_n + \frac{1}{a_{n}}$. I need to prove that $20 < a_{200} < 24$. I tried finding a limit at infinity setting both limits to $L$ ( for $a_n$ and $a_{n+1}$ but that gave me nothing useful. All I figured out was that this sequence is increasing..

Lower bound:
Let's prove that $a_n>\sqrt{2n}$ for $n\geqslant 3$, other words:
$$
(a_n)^2>2n, \qquad (n \geqslant 3).
\tag{1}
$$
$a_2 = 1+1=2$.
First, $(a_3)^2 = (2+\frac{1}{2})^2 = 4+2+\frac{1}{4} > 2\cdot 3$.

Now, using math. induction, we will show, that 
$$(a_n)^2>2n \implies (a_{n+1})^2>2(n+1).$$ 
Yes, if $(a_n)^2>2n$, then
$$
(a_{n+1})^2 = \left(a_n+\frac{1}{a_n}\right)^2 = (a_n)^2 + 2 + \left(\frac{1}{a_n}\right)^2>2n+2=2(n+1).
$$
Statement $(1)$ is proved.

So, $a_{200}>\sqrt{2\cdot 200} = 20$.

Upper bound:
Let's prove that $a_n<\sqrt{\frac{13}{6}n}$ for $n\geqslant 3$, other words:
$$
(a_n)^2 < \frac{13}{6}n, \qquad (n \geqslant 3).
\tag{2}
$$
First, $(a_3)^2 = \left(2+\frac{1}{2}\right)^2 = 4+2+\frac{1}{4}<6+\frac{1}{2} = \frac{13}{6}\cdot 3$.

Using math. induction and $(1)$, we will show, that 
$$(a_n)^2<\frac{13}{6}n \implies (a_{n+1})^2<\frac{13}{6}(n+1).$$ 
Yes, if $(a_n)^2<13n/6$, then
$$
(a_{n+1})^2 = \left(a_n+\frac{1}{a_n}\right)^2 = (a_n)^2 + 2 + \left(\frac{1}{a_n}\right)^2<
\frac{13}{6}n + 2 + \frac{1}{2n} \leqslant  \frac{13}{6}n + 2+\frac{1}{6}  = \frac{13}{6}(n+1),
$$
since $(a_n)^2>2n\geqslant 6$, when $n\geqslant 3$.
Statement $(2)$ is proved.

So, $a_{200}<\sqrt{\frac{13}{6}\cdot 200} \approx 20.81666 <24$.
