Limit of Sequence $x_n = (x_{n-1} +x_{n-2})/2$

Question

I was unable to solve this particular question in my real analysis assignment and I am looking for help here.

Question: Let $x_1 =0$ and $x_2 =1$ and for n>2 define $x_n = (x_{n-1} +x_{n-2})/2$. Then which of following are true?

A ${x_n}$ is monotone

B ${x_n}$ is Cauchy

C Lim $n \to \infty $ $x_n=2/3$.

I was unable to prove $x_n$ is Cauchy and monotone. All I could prove was that $|x_{n+1} -x_n|$=$ 1/(2)^{n-2}$ and I put n tends to infinity in recurrence relation to get limit but I am only getting that x=2x/2 .

I am unable to prove/ doisprove any of the options.

Can you please help?

Answer 1

$$ x_{n+1}=\frac{1}{2}(x_{n}+x_{n-1}) \quad\Longrightarrow\quad x_{n+1}-x_n=-\frac{1}{2}(x_{n}-x_{n-1})=\cdots =\frac{(-1)^{n-1}}{2^{n-1}}(x_2-x_1)=\frac{(-1)^{n-1}}{2^{n-1}}. $$ and $$ x_{n+1}=\frac{1}{2}(x_{n}+x_{n-1}) \quad\Longrightarrow\quad x_{n+1}+\frac{x_n}{2}=x_n+\frac{x_{n-1}}{2}=\cdots=x_2+\frac{x_1}{2}=1 $$ Subtracting the above we obtain $$ x_n=\frac{2}{3}+\frac{(-1)^{n-1}}{3\cdot 2^{n-1}}\to \frac{2}{3} $$

Answer 2

You can write an explicit formula for $x_n$: write characteristic equation$$ 2x^2-x-1=0$$ It solutions are $1$ and $-1/2$ so it is of form $$x_n = a+b({-1\over 2})^n$$ where you can find $a$ and $b$ from initial conditions. Clearly it is convergent with $a$ as a limit and it is not monotone.

Answer 3

Since $x_1=0$, $x_2=1$, and $x_3=\frac12$, it is clear that $(x_n)_{n\in\Bbb N}$ is not monotonic.

For each $n\in\Bbb N$, you have$$\begin{bmatrix}x_{n+2}\\x_{n+1}\end{bmatrix}=\begin{bmatrix}\frac12&\frac12\\1&0\end{bmatrix}\begin{bmatrix}x_{n+1}\\x_n\end{bmatrix}$$ and therefore$$\begin{bmatrix}x_{n+1}\\x_n\end{bmatrix}=\begin{bmatrix}\frac12&\frac12\\1&0\end{bmatrix}^n\begin{bmatrix}1\\0\end{bmatrix}.$$But then, since$$\lim_{n\to\infty}\begin{bmatrix}\frac12&\frac12\\1&0\end{bmatrix}^n=\begin{bmatrix}\frac23&\frac13\\\frac23&\frac13\end{bmatrix},$$and since$$\begin{bmatrix}\frac23&\frac13\\\frac23&\frac13\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}\frac23\\\frac23\end{bmatrix},$$the limit of your sequence is indeed $\frac23$. And, since the sequence converges, it is a Cauchy sequence.