# Thread: Prove result for limits?

1. ## Prove result for limits?

If a sequence of numbers $\{x_n\}$ is determined by the equality $x_n = \frac{x_{n - 1} + x_{n - 2}}{2}$ and the values $x_0$ and $x_1$, prove that:

$\lim_{n\rightarrow \infty} x_n = \frac{x_0 + 2x_1}{3}$

2. Since $x_{n}$ is the mean between $x_{n-2}$ and $x_{n-1}$ is...

$x_{n} = x_{0} + (x_{1}-x_{0}) (1-\frac{1}{2} + \frac{1}{4} - ... + \frac{(-1)^{n}}{2^{n}} )$ (1)

... so that is...

$\lim_{n \rightarrow \infty} x_{n} = x_{0} + \frac {x_{1}-x_{0}}{1+\frac{1}{2}} = x_{0} -\frac{2}{3}\cdot x_{0} + \frac{2}{3}\cdot x_{1} = \frac{1}{3}\cdot x_{0} + \frac{2}{3}\cdot x_{1}$ (2)

Kind regards

$\chi$ $\sigma$

3. Could you explain the second line please? I couldn't understand it.

4. ## Rewrite

Rewrite $x_n=x_0+(x_1-x_0)a_n$, for $a_0=0$ and $a_1=1$. By substituting into the recursive formula, you get $a_n=\frac{a_{n-1}+a_{n-2}}2$. Since we are taking iterative midpoints, the gap between successive values of $a_n$ are decreasing by half, so $$|a_n-a_{n-1}|=\left(\frac12\right)^{n-1}$$ . It should also be apparent that $a_{n}>a_{n-1}$ when n is odd, and $a_{n} when n is even, again, by virtue of iterating the midpoint formula, making the value of the gap alternate between positive and negative. Therefore, $a_n=\sum_{i=1}^{n-1}\left(-\frac12\right)^i$ for $n\geq1$, the limit of which as n gets large is $\frac23$.

5. There is no need for such heavy explanation.
$2x_{2}=x_{1}+x_{0}$
$2x_{3}=x_{2}+x_{1}$
$2x_{4}=x_{3}+x_{2}$
$2x_{5}=x_{4}+x_{3}$
$2x_{6}=x_{5}+x_{4}$
$2x_{7}=x_{6}+x_{5}$
$2x_{8}=x_{7}+x_{6}$
$2x_{9}=x_{8}+x_{7}$
-----------------------------------
$2x_{n-2}=x_{n-3}+x_{n-4}$
$2x_{n-1}=x_{n-2}+x_{n-3}$
$2x_{n}=x_{n-1}+x_{n-2}$

$x_{n-1}+2x_{n}=2x_{1}+x_{0}$
$\lim_{n\to \infty}(x_{n-1}+2x_{n})$= $2x_{1}+x_{0}$
If $\lim_{n\to \infty}x_{n}=x,then \lim_{n\to\infty}x_{n-1}=x$
$3x=2x_{1}+x_{0}$
$