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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 $\frac{2}{3}$($a_1 - a_0$) + $a_0$

Any additional input?

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  • $\begingroup$ Are you just supposed to determine whether it converges, or what the limit is? Your question seems to be about the former. $\endgroup$ – robjohn Mar 4 '14 at 23:57
  • $\begingroup$ I'm supposed to determine the limit, sorry for the he ambiguity. $\endgroup$ – Benji_Bombadill Mar 4 '14 at 23:59
  • $\begingroup$ Why is the answer for $a_0\gt a_1$ different than the answer for $a_0\lt a_1$? $\endgroup$ – robjohn Mar 5 '14 at 0:31
  • $\begingroup$ I posted an overkill solution involving generating functions. Let me know if I need to clarify stuff. ~R $\endgroup$ – Rustyn Mar 5 '14 at 1:06
  • $\begingroup$ Additional input: you have the answer, but not the proof that it works. In a math course, that may mean you get zero credit... $\endgroup$ – GEdgar Aug 10 '17 at 13:16
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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: $$ a_n=a_0+\sum_{k=1}^n\left(-\tfrac12\right)^{k-1}(a_1-a_0) $$

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$a_n = \dfrac{a_{n-1}+a_{n-2}}{2}$. 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 $$ \lim_{n\to \infty}a_{n+1} = \lim_{n\to \infty} \frac{(a_0+2a_1)}{3}+\frac{(2a_0+a_1)}{3}\left(\frac{-1}{2}\right)^n = \frac{(a_0+2a_1)}{3} $$

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