Given a positive sequence $\{a_n\}$ where $a_{n+1}=a_n+\frac{n}{a_n}$, can one find an asymptotic expansion of $a_n-n$? 
Given a positive sequence $\{a_n\}$ such that
  $$a_{n+1}=a_n+\frac{n}{a_n}$$
  Can one find an symptotic expannsion of $a_n-n$?

I want one which has the term $O(1/n)$, or a stronger one.

Let $b_n=a_n-n$.
Then
$$b_{n+1}=\left(1-\frac 1{a_n}\right)b_n$$
Thus if $a_0>1$ then $a_n>1$ and $b_n>0$ for all $n\in\Bbb N$.
Consequently, $0<b_{n+1}<b_n$ hence the sequence $b_n$ has limit, say $b\geq 0$, and $a_n\sim n$ as $n\to\infty$.
\begin{align} 
\log\left(\frac{a_{n}-n}{a_2}\right)
&=\log\left(\frac{b_{n}}{b_2}\right)\\
&=\log\left(\prod_{k=2}^{n-1}\left(1-\frac 1{a_k}\right)\right)\\
&=\sum_{k=2}^{n-1}\log\left(1-\frac 1{a_k}\right)\\
&=\sum_{k=2}^{n-1}\log\left(1-\frac 1k\right)+\sum_{k=2}^{n-1}\log\left(1+\frac{b_k}{(k+b_k)(k-1)}\right)\\
&=\sum_{k=2}^{n-1}\log\left(1-\frac 1k\right)+O(1)\\
&=-\sum_{k=2}^{n-1}\frac 1k+O(1)\\
&=-\log(n)+O(1)\\
\end{align}
Consequently, $a_n=n+O(1/n)$.
