Prove $u_{n}$ is decreasing $$u_{1}=2, \quad u_{n+1}=\frac{1}{3-u_n}$$ Prove it is decreasing and convergent and calculate its limit. Is it possible to define $u_{n}$ in terms of $n$? In order to prove it is decreasing, I calculated some terms but I would like to know how to do it in a more "elaborated" way.

Assuming the claim is true for $n$: $u_n < u_{n-1}$

We'll show the claim is also right for $n+1$. Indeed:
$$u_{n+1} = \frac{1}{3-u_n} < \frac{1}{3-u_{n-1}} = u_n$$

The inequality is of course based on the assumption. For a full proof you should add the base case as well.