Check proof that if $a_{n+1} = \frac{1}{k}(a_n+\frac{k}{a_n})$ with $k>1$ and $a_1>0$, then $a_n\to\sqrt {\frac{k}{k-1}}$

Question

Let $a_n$ be a Sequence such that $a_{n+1} = \frac{1}{k}(a_n+\frac{k}{a_n}), k>1, a_1>0$, show that $\{ a_n \}$ converges to $ \sqrt {\frac{k}{k-1}}$

My attempt

case 1: $a_{n+1}-a_n \leq 0 \\ \frac{1}{k}(a_n + \frac{k}{a_n})-a_n \leq 0 \\ a_n^2+k-ka_n^2 \leq 0 \\ a_n \geq \sqrt {\frac{k}{k-1}}$

Here sequence is decreasing and bounded below by $\sqrt {\frac{k}{k-1}}$. so $a_n$ is convergent

Case 2: if $a_{n+1} - a_n \geq 0,$ i get

$a_k \leq \sqrt {\frac{k}{k-1}}$

Here sequence is increasing and bounded above.

SO given sequence is convergent. Let the convergence limit be l, then $l=1/k (l+ k/l)$ gives given limit.

can you pls tell me is this correct approach?