The proof that if $a_{1}>1$ and $a_{n+1}=2-\frac{1}{a_{n}}$ for $n\geq 1$, then $\{a_{n}\}$ is monotone

Question

I am trying to make a proof that $\{a_{n}\}$ is monotone if

$$a_{1}>1\;\; \text{and}\;\; a_{n+1}=2-\frac{1}{a_{n}} \;\;\text{for}\;\; n\geq 1$$

Here is what I've done!

Let $a_{1}>1$ then $\frac{1}{a_{1}}<1$

Base case: When $n=1$

$$a_{2}=2-\frac{1}{a_{1}}< 2-1<2-a_{1}$$

Induction case: Assume $n=k$, then

$$a_{k+1}=2-\frac{1}{a_{k}}<2-a_{k}$$

To now prove that it holds for $n=k+1$, i.e.

$$a_{k+2}=2-\frac{1}{a_{k+1}}<2-a_{k+1}$$

My question is: Am I right? If yes, how do I continue? If no, I'll be grateful if another proof can be provided! Various proofs are welcome! Thanks!

Answer 1

$$a_{n+1}-a_n=2-\frac{1}{a_n}-a_n=\frac{2a_n-1-a_n^2}{a_n}=-\frac{(a_n-1)^2}{a_n}<0$$ since $a_n>0$ for all $n$. So your sequence is monotone decreasing.