<s>[INST] 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..
 [/INST] 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 $k = 1/6$.
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