If $f_n = \frac12(f_{n-1} +f_{n-2})$ for every $n>2$, then show that the sequence $(f_n)$ converges If $ <f_n> $is a sequence of positive numbers such that $f_n = \frac{1}{2}(f_{n-1} +f_{n-2}), \forall \ n>2$, then show that $<f_n>$ converges. 
Any Hints on how to approach this problem? 
 A: It suffices to show that the sequence is Cauchy.
Let $a:=f_1$ and $b:=f_2$. Then show by induction that
$$|f_n-f_{n-1}|=\frac{1}{2^n}|a-b|.$$
Use this to prove that the sequence is Cauchy; i.e., that for every $\varepsilon>0$ there is $N\in\mathbb{N}$ such that
$$|f_{n+k}-f_n|<\varepsilon$$
whenever $n\geq N$ and $k\in\mathbb{N}$.
A: астон's answer is incorrect. This sequence is not eventually monotone. You need to use something else here.
Real numbers are complete. That means that every Cauchy sequence converges. You just need to proof that the sequence is indeed Cauchy. For this note that:
$$|f_{n+1}-f_n|=\frac{|f_n-f_{n-1}|}{2}$$
Which means that every next two points in the series are at half of the distance of the previous pair.
Using this you should be able to show that $\forall \epsilon>0$ there is an $N$ such that:
$$|f_n-f_m|<\epsilon\quad \forall n,m>N$$
i.e. the sequence is Cauchy.
A: If $\alpha, \beta$ are roots of
$$2x^2 -x - 1 = 0$$
Then $f_n = A\alpha^n + B \beta^n$ for some constants $A, B$ which depend only on $f_0, f_1$.
In this case $\alpha = 1, \beta = -\frac{1}{2}$ so $f_n$ is convergent to $A$ which if computed turns out to be
$$ \frac{2f_1 + f_0}{3}$$
So it is interesting to see that the limit actually depends on the initial values!
