I got interested in the recursion $$ a_{n+1} = a_n + \frac1{a_n} $$ in response to a question on this site (which I can no longer locate).
I thought this would be a relatively easy one to solve as an explicit function of $n$. For instance, the closely related recursion $$ b_{n+1} = \frac12 \left( b_n + \frac1{b_n} \right) $$ is the sequence of guesses in Newton's algorithm for $\sqrt{1}$, given a starting guess $b_0$, and that turns out to be $$ b_{2k} = \tanh\left( 2^{2k} x \right)\\ b_{2k+1} = \frac{1}{ \tanh\left( 2^{2k+1} x \right)} $$ with $x = \tanh^{-1} b_0$.
But the recursion for $a$ is a tougher nut to crack. Although I'd llike to have in in explicit form, that might not be practical (I tried various things, including Jacobi elliptic functions, but I nevery quite get the right identities).
This question asks to prove something about the asymptotic behavior of $a_n$ for the case of arbitrary $a_0>0$, namely that
$$ \lim_{n\to\infty} \frac{H_n^4}{a_n} = \lim_{n\to\infty} \frac{a_n} {H_n^5}=0 $$ where $H_n$ are the harmonic numbers $$H_n \equiv \sum_{m=1}^n \frac1m$$
$a_{n+1}^2 = a_{n}^2 + 2 + \frac{1}{a_n^2}$ gives $a_n\geq \sqrt{2n-1}$ as well as $$ a_{n+1}^2\leq a_n^2+2+\frac{1}{2n-1}$$ from which $a_n\leq \sqrt{2n+O(\log n)}$.
The given limits are simple to compute given these bounds, but $$ \lim_{n\to +\infty}\frac{H_n^4}{a_n} = 0,\qquad \lim_{n\to +\infty}\frac{a_n}{H_n^5}=\color{red}{+\infty}$$ since $H_n=\log(n)+O(1)$.
Thanks to Clement C., here it is a plot of $a_n$ versus $\sqrt{2n}$: