TITLE: Prove a recursive relation
QUESTION [4 upvotes]: Suppose one has the following recursive relation :
$$
a_{n+1} = a_n + \dfrac{1}{a_n}
$$
Where:
$$a_n > 0$$
Is there any way to find a closed form formula for something like this? I tried looking at the local factors by looking at how it develops, but it seems very chaotic. I found an inverse formula for this relation:
$$
a_{n-1} = \dfrac{a_n+ \sqrt{{a_n}^2 - 4}}{2}
$$
A bit odd is the fact that the inverse function is undefined ( in the reals) for $a_n < 2$ even though the regular function is defined for values lower than 2. Also, since the original function is increasing and monotonic, we know that the inverse is decreasing and monotonic - which means that for very large values of $a_0$ we always expect that the limit as n goes to infinity to be undefined, as it is guaranteed to go lower than 2 at some point in the sequence.
Thats all I got basically, Would love to hear if you have any ideas or know anything about these types of recursive relations, because I noticed  for example:
$$
a_{n+1} = a_n + \dfrac{1}{2a_n}
$$
has a very similar inverse:
$$
a_{n-1} = \dfrac{a_n+ \sqrt{{a_n}^2 - 2}}{2}
$$

REPLY [3 votes]: I can't provide a closed formula, but here is double inequality which show how fast $(a_n)_n$ goes to infinity.
Let $a_0>0$ and let $b_n=a_n^2-2n$. Then $b_0=a_0^2>0$ and
$$b_{n+1}=\left(a_n+\frac{1}{a_n}\right)^2-2n-2=a_n^2+\frac{1}{a_n^2}-2n=b_n+\frac{1}{b_n+2n}$$
which implies by induction that $b_n>0$ for all $n\geq 0$.
Moreover
$$b_{n+1}-b_n<\frac{1}{2n}\implies b_{n+1}-b_1<\frac{1}{2}\sum_{k=1}^n\frac{1}{k}\leq \frac{\ln(n)+1}{2}$$
and therefore, since $b_1=a_0^2+\frac{1}{a_0^2}$, we find that
$$\sqrt{2n}<a_n<\sqrt{2n+a_0^2+\frac{1}{a_0^2}+\frac{\ln(n)+1}{2}}.$$
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