Prove that the recurrence $f(n+2)=2/3f(n+1)+1/3f(n)$ for $f(0)=5$ and $f(1)=1$ converges 
Prove that the recurrence $f(n+1)=2/3f(n)+1/3f(n-1)$ for $f(0)=5$ and $f(1)=1$ converges
HINT: derive an expression for $f(n)-f(n+1)$

I dont know exactly what to do here. I must prove this only using the algebra of limits and the theorem that says that a monotonic and bounded sequence converges.
I tried induction over the subsequences $f(n+1)=7/9 f(n-1)+2/9 f(n-2)$ to prove monotonicity but I dont get something useful. And the expressions that I derived for $f(n)-f(n+1)$ dont give me any clue. Some hint or solution will be appreciated, thank you.
 A: Here is a (sketch of a) proof that the sequence converges, without using the hint, and without explicitly finding the limit.  It does use the algebra of limits and the theorem about monotonic, bounded sequences converging.
The recurrence $f(n+1)={2\over3}f(n)+{1\over3}f(n-1)$ is a weighted average, so each term in the sequence lies between the previous two terms.  Since $f(1)=1\lt5=f(0)$, it's easy to see that the sequence of odd terms, $f(1), f(3),f(5),\ldots$, is an increasing sequence bounded above (by $5$), while the sequence of even terms, $f(0),f(2),f(4),\ldots$, is a decreasing sequence bounded below (by $1$).  These two monotonic and bounded sequences therefore have limits: 
$$\lim_{n\to\infty}f(2n-1)=A\le5\qquad\text{and}\qquad\lim_{n\to\infty}f(2n)=B\ge1$$
It remains to show that $A=B$.  But this follows from the recurrence:
$$B=\lim_{n\to\infty}f(2n)=\lim_{n\to\infty}\left({2\over3}f(2n-1)+{1\over3}f(2n-2)\right)={2\over3}\lim_{n\to\infty}f(2n-1)+{1\over3}\lim_{n\to\infty}f(2(n-1))={2\over3}A+{1\over3}B$$
