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Here is the problem statement:

Let $a$, $b$ $\in$ $\Bbb{R}$. A sequence $(a_{n})_{n \in \Bbb{N}}$ is defined recursively by $$a_{0}:=a, \qquad a_{1}:=b, \qquad a_{n} = \frac{1}{2}(a_{n-1}+a_{n-2}) \quad \text{for} \quad n \geq 2.$$ Prove that $\lim_{n\to\infty}a_{n}$ exists and compute its value.

Now I have proved that the limit exists. My idea was to use the Monotone Convergence Theorem as follows:

Proof: Without loss of generality, assume $a<b$. By inspection of the first two terms of $a_{n}$, we see that $$a_{3} = \frac{1}{2}(a_{2}+a_{1})=\frac{1}{2}\left[\frac{1}{2}(a_{1}+a_{0}) + a_{1}\right] = \frac{1}{2}(a_{1}+a_{0})+\frac{1}{4}(a_{1}-a_{0})=a_{2}+\frac{1}{4}(a_{1}-a_{0})>a_{2} \quad (\because a_{1}>a_{0}).$$ Therefore we need to show that the sequence is increasing, so our inductive hypothesis is $a_{n}<a_{n+1}$. $$a_{n}<a_{n+1}\implies a_{n+1}=\frac{1}{2}(a_{n}+a_{n-1})<\frac{1}{2}(a_{n+1}+a_{n})=a_{n+2}.$$ Thus $a_{n}$ is increasing for $n\geq 2$. Now we need to show that it is bounded. By inspection, $a_{3}=\frac{1}{2}(a_{2}+a_{1})<\frac{1}{2}(a_{1}+a_{1}) = a_{1}$, and $a_{4}<a_{1}$ and so on. So again assume by induction $a_{n}<a_{1}$ and we have $$a_{n+1}=\frac{1}{2}(a_{n}+a_{n-1})<\frac{1}{2}(a_{1}+a_{1}) = a_{1}.$$ Thus $a_{n}$ is bounded above. Hence by the Monotone Convergence Theorem, sequence is convergent and $\lim a_{n}$ exists.$\quad\square$

However, I'm stuck in finding the limit. Any hints would be appreciated.

tmaj
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  • A general method for solving second order recurrence equations is by solving their associated characteristic equation. – Jean Marie Jan 22 '21 at 18:40
  • Another technique is that of generating functions. But there is a geometric way of interpreting your recurrence that should lead you directly to the limit. – While I Am Jan 22 '21 at 18:46
  • @JeanMarie I'm not conversant with solving recurrence equations, but thank you for the hint. – tmaj Jan 22 '21 at 18:50
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    The solution is almost never monotone. Instead, you'll get a pingpong so for example if $a_0 < a_1$, then $a_0 < a_2 < a_4 < \cdots < a_5 < a_3 < a_1$. So, to apply the monotone convergence theorem, you would need to consider the subsequences $a_{2n}$ and $a_{2n+1}$, and then produce some argument why the limits of the two subsequences must be equal. – Daniel Schepler Jan 22 '21 at 19:04
  • @DanielSchepler ah, I had not considered that. In that case would the Bolzano-Weierstrass theorem be helpful? – tmaj Jan 22 '21 at 19:13

3 Answers3

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The characteristic equation is $$\lambda^2=\frac{1}{2}(\lambda+1)$$ Its solutions are $$\lambda=1;\;\lambda=-1/2$$ The general solution of the recurrence is $$a_n=m+n\left(-\frac{1}{2}\right)^n$$ for $n=0$ we get $a_0=m+n=a$

for $n=1$ we have $a_1=m-\frac12n=b$

so we get $$m= \frac{1}{3} (a+2 b),n= \frac{2 (a-b)}{3}$$ and the recurrence can be written as $$a_n=\frac{1}{3} (a+2 b)+ \frac{2 (a-b)}{3}\left(-\frac{1}{2}\right)^n$$ as $n\to\infty$ we have $a_n\to \frac{1}{3} (a+2 b)$.

Hope this is useful

Raffaele
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$$ \begin{bmatrix} a_{n}\\ a_{n-1} \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & \frac{1}{2}\\ 0 & 1 \end{bmatrix} \begin{bmatrix} a_{n-1}\\ a_{n-2} \end{bmatrix} $$

$$ \begin{bmatrix} \frac{1}{2} & \frac{1}{2}\\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 1\\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0\\ 0 & \frac{1}{2} \end{bmatrix} \begin{bmatrix} 0 & 1\\ 1 & -1 \end{bmatrix} $$

$$ \begin{bmatrix} \frac{1}{2} & \frac{1}{2}\\ 0 & 1 \end{bmatrix}^{n} = \begin{bmatrix} 1 & 1\\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0\\ 0 & \frac{1}{2}^{n} \end{bmatrix} \begin{bmatrix} 0 & 1\\ 1 & -1 \end{bmatrix} $$

$$ \begin{bmatrix} a_{n}\\ a_{n-1} \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & \frac{1}{2}\\ 0 & 1 \end{bmatrix} ^ {n} \begin{bmatrix} a_{1}\\ a_{0} \end{bmatrix} = \begin{bmatrix} 1 & 1\\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0\\ 0 & \frac{1}{2}^{n} \end{bmatrix} \begin{bmatrix} 0 & 1\\ 1 & -1 \end{bmatrix} \begin{bmatrix} a_{1}\\ a_{0} \end{bmatrix} $$

John_Krampf
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Hint:

$$a_n=\frac12(a_{n-1}+a_{n-2})\implies a_n-a_{n-1}=-\frac{1}{2}(a_{n-1}-a_{n-2}).$$ Can you use this to find a closed form for $a_n-a_{n-1}$, and then for $a_n$?

jlammy
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