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..
 A: 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, $a_{200}<\sqrt{\frac{13}{6}\cdot 200} \approx 20.81666 <24$.
A: I've decided to submit my answer.
It's a little different from Oleg567's,
but the results are the same.
I first try to estimate the growth of $a_n$.
From the desired bounds,
it looks like it might be of order
$\sqrt{n}$.
Suppose
$a_n
=c\sqrt{n}
$.
Then
$c\sqrt{n}+1/(c\sqrt{n})
=c\sqrt{n+1}
=c\sqrt{n}\sqrt{1+1/n}
\approx c\sqrt{n}(1+1/(2n)) 
= c\sqrt{n}+c/(2\sqrt{n}) 
$
or
$1/(c\sqrt{n})
\approx c/(2\sqrt{n})
$
or
$1/c \approx c/2$
or
$c \approx \sqrt{2}$.
It therefore looks like
$a_n \approx \sqrt{2n}$.
Suppose $a_n > \sqrt{2n}$.
Then
$a_{n+1}
> \sqrt{2n}+1/\sqrt{2n}
$.
To show
$a_{n+1} > \sqrt{2(n+1)}$,
we need
$\sqrt{2n}+1/\sqrt{2n}
> \sqrt{2(n+1)}$.
Squaring both sides,
this is
$2n+2+1/(2n)
> 2(n+1)
$
which is true.
Setting $n=200$,
$a_{200} > \sqrt{400} = 20$.
Suppose
$a_n \le \sqrt{(2+k)n}$ for some $k$.
Then, since
$a_n^2 > 2n$,
$a_{n+1}^2
=a_n^2+2+1/a_n^2
< (2+k)n+2+1/(2n)
$
and for this to be
$\le (2+k)(n+1)$
we need
$(2+k)n+2+1/(2n)
\le (2+k)(n+1)$
or
$2+1/(2n) \le 2+k$
or
$1/(2n)\le k$.
This will certainly be true for
large enough $n$.
We have
$a_2 = 2$, $a_3 = 5/2$,
and $a_4 = 5/2+2/5 = 29/10$.
For $n=3$,
we want
$25/4 \le 3(2+k)=6+3k$
or
$1/4 \le 3k$
or $k \ge 1/12$.
Since we also want
$k \ge 1/(2n) = 1/6$,
we can use $k = 1/6$.
