TITLE: Closed form for the sequence defined by $a_0=1$ and $a_{n+1} = a_n + a_n^{-1}$
QUESTION [30 upvotes]: Today, we had a math class, where we had to show, that $a_{100} > 14$ for
$$a_0 = 1;\qquad a_{n+1} = a_n + a_n^{-1}$$
Apart from this task, I asked myself: Is there a closed form for this sequence? Since I didn't find an answer by myself, can somebody tell me, whether such a closed form exists, and if yes what it is?

REPLY [2 votes]: Let $b_n=a_n^{-1}$, so that $b_{n+1}=(b_n+b_n^{-1})^{-1}$ is the $n$th iterate of $$f(x)=(x+x^{-1})^{-1}=x-x^3+x^5-x^7+\cdots$$ 
The correct asymptotics have been given elsewhere, but this answer is to point out that there is a general method applicable to any function $f$ analytic at 0 with a series expansion $x+a_kx^k+a_{k+1}x^{k+1}+\cdots$ with $k>1$ and $a_k<0$: I learned about it in Chapter 8 of de Bruijn's Asymptotic Methods in Analysis, where it is shown that if the starting value is small enough, the $n$th iterate of $f$ is asymptotically equivalent to $((1-k)a_kn)^{-1/(k-1)}$. In our case, $k=3$ and $a_k=-1$ so $b_n \sim (2n)^{-1/2}$ and $a_n \sim (2n)^{1/2}$.
A sketch of the proof in the case at hand is as follows.  First, show by induction that if $u_n$ is any sequence tending to zero and such that 
$$u_{n+1}=u_n-u_n^2+O(u_n^3)$$
then $u_n = n^{-1} + O(n^{-2}\log n)$. This is the tricky part of the proof.
Next make a substitution $z_n= 2b_n^2$. We have
$$ b_{n+1}=b_n(1-b_n^2+b_n^4-\cdots)$$
so
$$z_{n+1}=z_n\left(1-\frac{1}{2}z_n + \frac{1}{4}z_n^2+ \cdots\right)^2 = z_n-z_n^2 + \frac{3}{4}z_n^3 + \cdots$$
from which we get $z_n = n^{-1}+O(n^{-2}\log n)$ and the asymptotics for $b_n$ follow.
Incidentally, the sequence $b_n$ is what you get by using Newton's method to find a root of $xe^{-x^{-2}/2}$. But I couldn't find a way to make that shed any light on the asymptotics...