calculus - A limit problem about $a_{n+1}=a_n+frac{n}{a_n}$

Let $a_{n+1}=a_n+\frac{n}{a_n}$ and $a_1>0$. Prove $\lim\limits_{n\to \infty} n(a_n-n)$ exists.

In my view, maybe we can use
$${a_{n + 1}} = {a_n} + \frac{n}{{{a_n}}} \Rightarrow {a_{n + 1}} - \left( {n + 1} \right) = \left( {{a_n} - n} \right)\left( {1 - \frac{1}{{{a_n}}}} \right).$$
And then
$${a_n} - n = \left( {{a_1} - 1} \right)\prod\limits_{k = 1}^{n - 1} {\left( {1 - \frac{1}{{{a_k}}}} \right)} .$$
By Stolz formula, we have
$$\begin{align*} &\mathop {\lim }\limits_{n \to \infty } n\left( {{a_n} - n} \right) = \mathop {\lim }\limits_{n \to \infty } \frac{n}{{\frac{1}{{{a_n} - n}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{\frac{1}{{{a_{n + 1}} - \left( {n + 1} \right)}} - \frac{1}{{{a_n} - n}}}}\\ = &\mathop {\lim }\limits_{n \to \infty } \frac{{\left[ {{a_{n + 1}} - \left( {n + 1} \right)} \right]\left( {{a_n} - n} \right)}}{{{a_n} - {a_{n + 1}} + 1}} = \mathop {\lim }\limits_{n \to \infty } \frac{{\left[ {{a_{n + 1}} - \left( {n + 1} \right)} \right]\left( {{a_n} - n} \right)}}{{ - \frac{n}{{{a_n}}} + 1}}\\ = &\mathop {\lim }\limits_{n \to \infty } {a_n}\left[ {{a_{n + 1}} - \left( {n + 1} \right)} \right]. \end{align*}$$
And how can we continue?