# Determine the convergence of a sequence given by $a_n= \frac{a_{n-1} + a_{n-2}}2$

let $a_0$ and $a_1$ be any two real numbers, and define

$a_n= \dfrac{a_{n-1} + a_{n-2}}{2}$

Determine the convergence of a sequence.

Alright, what I have so far. I have two cases $a_0$ > $a_1$ or $a_0$ < $a_1$

for $a_0 > a_1$

The sequence converges to $\frac{1}{3}$($a_0 - a_1$) + $a_0$

for $a_0 < a_1$

The sequence converges to $\frac{2}{3}$($a_1 - a_0$) + $a_0$

• Are you just supposed to determine whether it converges, or what the limit is? Your question seems to be about the former. – robjohn Mar 4 '14 at 23:57
• I'm supposed to determine the limit, sorry for the he ambiguity. – Benji_Bombadill Mar 4 '14 at 23:59
• Why is the answer for $a_0\gt a_1$ different than the answer for $a_0\lt a_1$? – robjohn Mar 5 '14 at 0:31
• I posted an overkill solution involving generating functions. Let me know if I need to clarify stuff. ~R – Rustyn Mar 5 '14 at 1:06
• Additional input: you have the answer, but not the proof that it works. In a math course, that may mean you get zero credit... – GEdgar Aug 10 '17 at 13:16

Hint 1: $$a_n-a_{n-1}=-\tfrac12(a_{n-1}-a_{n-2})$$ Show by induction that $$a_n-a_{n-1}=\left(-\tfrac12\right)^{n-1}(a_1-a_0)$$ Hint 2: $$a_n=a_0+\sum_{k=1}^n\left(-\tfrac12\right)^{k-1}(a_1-a_0)$$
$a_n = \dfrac{a_{n-1}+a_{n-2}}{2}$. One overkill way of finding the limit, is getting the $a_n$ in terms of $n$ via a generating function.
Write $F(x) = \sum_{i=0}^{\infty}a_i x^i$. Thus,
$$\frac{F(x)-a_0}{x}=\sum_{i=0}^{\infty}a_{i+1}x^{i}$$ $$\frac{F(x)-a_0-a_1}{x^2} = \sum_{i=0}^{\infty}a_{i+2}x^{i}$$ So, $$\frac{F(x)-a_0-a_1}{x^2} = \frac{1}{2}\cdot \frac{F(x)-a_0}{x} + \frac{1}{2}F(x)$$ Solving for $F(x)$, one obtains: $$\frac{(x-2)a_0-2a_1}{x^2+x-2}=F(x)$$ Using partial fractions, one then obtains: $$F(x) = \frac{-(a_0+2a_1)}{3(x-1)}+\frac{2(2a_0+a_1)}{3(x+2)}$$ Now write each term as a power series, obtaining: $$F(x) =\frac{(a_0+2a_1)}{3}\sum_{k=0}^{\infty} x^k +\frac{(2a_0+a_1)}{3}\sum_{k=0}^{\infty} \left(-\frac{1}{2}\right)^k(x)^k$$
Now, $a_{n+1}$ is the $(n+1)^{st}$ coefficient and therefore the coefficient of $x^n$ given above. Explicitly, $$a_{n+1} = \frac{(a_0+2a_1)}{3}+\frac{(2a_0+a_1)}{3}\left(\frac{-1}{2}\right)^n$$ e.g. if $a_0=0, a_1=1$, we have: $$a_2=\frac{1}{2},a_3=\frac{\frac{1}{2}+1}{2}=\frac{3}{4},a_4=\frac{\frac{3}{4}+\frac{1}{2}}{2}=\frac{5}{8}$$ Now we expect that $$a_4 = \frac{(a_0+2a_1)}{3}+\frac{(2a_0+a_1)}{3}\left(\frac{-1}{2}\right)^3 = \frac{5}{8}$$ And $$\frac{(a_0+2a_1)}{3}+\frac{(2a_0+a_1)}{3}\left(\frac{-1}{2}\right)^3= \\\frac{2}{3}+\frac{1}{3}\left(\frac{-1}{2}\right)^3=\frac{5}{8}$$
Now, it is easy to see that $$\lim_{n\to \infty}a_{n+1} = \lim_{n\to \infty} \frac{(a_0+2a_1)}{3}+\frac{(2a_0+a_1)}{3}\left(\frac{-1}{2}\right)^n = \frac{(a_0+2a_1)}{3}$$