How to solve the recursion: $f(n)=3-\frac{1}{f(n-1)}$? I have some problems with the recursion: $f(n) = 3 - \frac{1}{f(n-1)} $, and the initial is $f(1)= \frac{5}{2}$.
Can anyone give me some hints of this?Thanks!!!

The given equation is 
$$f(n)f(n-1)-3f(n-1)+1=0 \implies f(n-1)(f(n)-3)+1=0.$$
Let $$f(n)-3=\frac{g(n-1)}{g(n)},$$ then
we get $$g(n)+3g(n-1)+g(n-2)=0.$$
Now we take $g(n)= x^n$, then
$$x^2+3x+1=0 \implies x= a,b =\frac{-3 \pm \sqrt{5}}{2}$$
so $g(n)=p a^n+q b^n$.  So $$f(n)=\frac{p a^{n-1}+ q b^{n-1}}{pa^n+qb^n}+3= \frac{1}{b}\frac{rc^{n-1}+1}{r c^n+1}+3, c=\frac{a}{b}=\frac{7-3\sqrt{5}}{2}, r=\frac{p}{q}$$
Now, if $f(1)=\frac{5}{2}$, then can find the value of $r$.
