Consider the sequence $(u_n)$ defined by: $u_1>0; u_{n+1}=u_n+\frac{n}{u_n}$ and $n=1,2,....$
Prove the convergency and find the limit of sequence $\left(\frac{u_n}{n}\right)$
I have tried finding the limit by proving $(\frac{u_n}{n})$ is monotonic and bounded sequence. By induction, I knew that $\frac{u_{n+1}}{n+1}< \frac{u_n}{n}$ when $n\geq 2$. It isn't correct with $n=1$. I'm so confused because I don't know how to continue.
Does anyone have any ideas? Thank in advance!
The following is inspired by A limit problem about $a_{n+1}=a_n+\frac{n}{a_n}$ and by Sequence limit, but not identical.
$$ u_{n+1}^2 = u_n^2 + 2n + \frac{n^2}{u_n^2} \ge u_n^2 + 2n $$ implies that $$ \tag{*} u_n^2 \ge n(n-1) $$ for all $n$. Then $$ \frac{u_{n+1}}{n} - \frac{u_n}{n-1} = \frac{u_n}{n} + \frac{1}{u_n} - \frac{u_n}{n-1} = \frac{1}{u_n} - \frac{u_n}{(n-1)n} \le 0 $$ for $n \ge 2$, so that the sequence $u_n/(n-1)$ is decreasing and therefore convergent. It follows that $$ L = \lim_{n \to \infty} \frac{u_n}{n}= \lim_{n \to \infty} \frac{u_n}{n-1} $$ exists as well, and $(*)$ implies that $L \ge 1$.
It remains to determine the value of $L$. Here we can use the Stolz–Cesàro theorem: $$ L = \lim_{n \to \infty} \frac{u_n}{n} = \lim_{n \to \infty} \frac{u_{n+1}-u_n}{(n+1)-n} = \lim_{n \to \infty} \frac{n}{u_n} = \frac 1 L $$ and therefore $L=1$.