TITLE: Given the sequence $a_1=1$,$a_{n+1}=1+\frac{n}{a_n}$, does the sequence $a_n+n-a_n^2$ converges?
QUESTION [10 upvotes]: Given the sequence $a_1=1$,$a_{n+1}=1+\frac{n}{a_n}$, does the sequence $a_n+n-a_n^2$ converges? If it converges, find the limit.

Now, I proved
$\frac{1+\sqrt{1+4(n-1)}}{2}\leq a_n\leq \frac{1+\sqrt{1+4n}}{2}$,
and
$0\leq p_n=a_n+n-a_n^2\leq 1,p_{n+1}+\frac{n}{a_n^2}p_n=1$,which implies $\lim\limits_{n\rightarrow+\infty}p_{n+1}=\lim\limits_{n\rightarrow\infty}p_{n-1}$,
but I cannot prove $\lim\limits_{n\rightarrow+\infty}p_{n+1}=\lim\limits_{n\rightarrow\infty}p_{n}$ which means the squence $a_n+n-a_n^2$ limit equals to ${1\over2}$
so the question is: how can I prove $\lim\limits_{n\rightarrow+\infty}p_{n+1}=\lim\limits_{n\rightarrow\infty}p_{n}$ ?

REPLY [2 votes]: Link with the number of involutions in the symmetric goup $S_n$.
Let $I_0=1$. Let, for $n\geqslant 1$, $I_n$ be the number of involutions in the symmetric group $S_n$ (i.e $\ \sigma \in S_n \ | \ \sigma^2=I$). We can show that $\ I_1=1 \ $ , and $\ \forall n \in \mathbb N^{\star} \ , \ I_{n+1}=I_n+nI_{n-1}$. And, by induction, $\ \forall n\in \mathbb N \ , \ a_n=\frac{I_{n+1}}{I_n}\  $ .
L Moser and M Wyman, On solutions of x^d=1 in symmetric groups, Canad. J. Math., 7 (1955)  This paper can be found here . In this article, it is shown that exists $(\beta_n)_{n\in\mathbb N}$ such that $\ \forall p \in \mathbb N \ , \ a_n =\sqrt{n} +\displaystyle \sum_{k=0}^p \dfrac{\beta_k}{n^{k/2}}+{\rm o}\left(\dfrac{1}{n^{p/2}}\right)\ $. For example $\ a_n=\sqrt{n}+\dfrac{1}{2}-\dfrac{1}{8\sqrt{n}} +{\rm o}\left(\dfrac{1}{\sqrt{n}}\right)$. With that  result, it's easy to show that $\ \lim (a_n+n-a_n^2)=\dfrac{1}{2}$ .
If $\ b_0 \in \mathbb R_+^{\star}\ $ and $\ \forall n \in \mathbb N \ , \ b_{n+1}=1+\frac{n}{b_n} \ $, then, $(b_n)_{n\in\mathbb N}$ has the same asymptotic expansion: $\ \forall p \in \mathbb N \ , \ a_n =\sqrt{n} +\displaystyle \sum_{k=0}^p \dfrac{\beta_k}{n^{k/2}}+{\rm o}\left(\dfrac{1}{n^{p/2}}\right)\ $. (the same coefficients $\beta_k$ )