Let $a\leq b \in \mathbb{R}$. Show that the sequence $a_1 = a, a_2=b$ and $a_{n+2}=\frac{a_{n+1}+a_n}{2}$ for $n\geq 1$ is Cauchy and find it's limit.
I did for $n>m$: $$|a_n-a_m|=\frac{1}{2}|(a_{n-1}+a_{n-2})-(a_{m-1}+a_{m-2})|$$ $$\leq |a_n-a_{n-1}|+|a_{n-1}-a_{n-2}|+...+|a_{m+2}-a_{m+1}|+|a_{m+1}-a_{m}|$$
But I have no idea what to do next
Hint: The recurrence relation gives you immediately $|a_{n+2}-a_{n+1}| = \frac12|a_{n+1}-a_n|$.
This lets you express $|a_{n+1}-a_n|$ in terms of $b-a$. This in turn gives you an upper bound on $|a_n-a_m|$ for $m,n \ge N$.
You could proceed finding the general expression of $a_n$. To do so, consider the characteristic polynomial of the equation:
$$2x^2-x-1 = 0$$
Whose roots are $1$ and $-1/2$. Therefore, the general solution is of the form
$$a_n = A_1 + A_2(-1/2)^n$$
Now, imposing initial conditions for $n = 1$ and $n = 2$, where we know that $a_1 = a$, and $a_2 = b$, we get:
$$a_n = \left(\frac{2}{3} b +\frac{1}{3} a\right) + \left(\frac{4}{3}(b-a)\right)\left(-\frac{1}{2}\right)^n$$
Now succession converges to $\frac{2}{3} b +\frac{1}{3} a$, therefore, it is Cauchy.
Hint: Try writing out the first few terms of $a_n$:
$\displaystyle a_1 = a, a_2 = b, a_3 = \frac{1}{2}(a+b), a_4 = \frac{1}{2^2}(a + 3b), a_5 = ...$
and deduce the closed form for arbitrary $a_n$
