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.
 A: 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.
A: As others have said, you don't need a formula for $u_n$ to show that it is strictly decreasing and converges to $\dfrac{3-\sqrt{5}}{2}$. 
However, You can get an exact formula for $u_n$. Lets assume $u_n = \dfrac{a_n}{b_n}$ for integers $a_n, b_n$. 
Then, $\dfrac{a_{n+1}}{b_{n+1}} = u_{n+1} = \dfrac{1}{3-u_n} = \dfrac{1}{3-\dfrac{a_n}{b_n}} = \dfrac{b_n}{3b_n-a_n}$. 
So, we can let $a_{n+1} = b_n$, and $b_{n+1} = 3b_n - a_n$. 
Thus, $a_{n+1} = b_n = 3b_{n-1} - a_{n-1} = 3a_n - a_{n-1}$
The solution to this recurrence is $a_n = C_1\left(\dfrac{3+\sqrt{5}}{2}\right)^n + C_2\left(\dfrac{3-\sqrt{5}}{2}\right)^n$. 
Since $u_1 = 2$ and $u_2 = 1$, we have $a_1 = 2$ and $a_2 = 1$. Solve for $C_1$ and $C_2$ to get: $a_n = \left(1-\dfrac{2}{\sqrt{5}}\right)\left(\dfrac{3+\sqrt{5}}{2}\right)^{n-1} + \left(1+\dfrac{2}{\sqrt{5}}\right)\left(\dfrac{3-\sqrt{5}}{2}\right)^{n-1}$. 
Then, $u_n = \dfrac{a_n}{b_n} = \dfrac{a_n}{a_{n+1}} = $ (really ugly expression). 
Needless to say, the problem doesn't intend for you to find an explicit formula for $u_n$, although a formula does exist. 
A: After seeing that $u_{1} > u_{2}$, the result follows since $u_{n} > u_{n + 1}$ implies
$$\frac{1}{3 - u_{n}} > \frac{1}{3 - u_{n + 1}}.$$
Once you know that there is a limit, say $L$, you have
$$\lim_{n\rightarrow\infty} \frac{1}{3-u_{n}} = L.$$
But as $n \rightarrow \infty$, $u_{n} \rightarrow L$, so also
$$\lim_{n \rightarrow\infty} \frac{1}{3-u_{n}} = \frac{1}{3 - L}.$$
Now you can just solve for $L$.
A: If there is a limit, it will be defined by $$L=\frac{1}{3-L}$$ which reduces to $L^2-3L+1=0$. You need to solve this quadratic and discard any root greater than $2$ since this is the starting value and that you proved that the terms are decreasing.
