How to show this one is monotone?

Question

PROBLEM

Suppose $a_1>a_2>0$. For $n\geq2$, set $a_{n+1}=\frac{1}{2}(a_n+a_{n-1})$. Prove that

(a) $\{a_{2k+1}\}$ is monotone decreasing.

(b) $\{a_{2k}\}$ is monotone increasing.

(c) $\{a_n\}$ converges.

For (a), $a_{2k+1}=\frac{1}{2}(a_{2k}+a_{2k-1})$. And I tried to show this is decreasing by substracting $a_{2k-1}$ from $a_{2k+1}$, $a_{2k+1}-a_{2k-1}=\frac{1}{2}(a_{2k}+a_{2k-1})-\frac{1}{2}(a_{2k-2}-a_{2k-3})$. But I'm not sure if this is right method to compare $a_{2k+1}$ and $a_{2k-1}$..

Answer 1

(a) First, observe $$a_{2k}-a_{2k-1}=-\frac{1}{2}(a_{2k-1}-a_{2k-2})=\frac{1}{4}(a_{2k-2}-a_{2k-3}).$$ Therefore since $a_2-a_1<0$, for all $k$, $a_{2k}-a_{2k-1}<0$. Then, for all $k$ $$a_{2k+1}-a_{2k-1}=\frac{1}{2}(a_{2k}-a_{2k-1})<0,$$ i.e. $\{a_{2k+1}\}$ is monotone decreasing.

(b) Similar to (a).

(c) Note that $\{a_{2k+1}\}$ and $\{a_{2k}\}$ are bounded from below and above, respectively. This is because $$a_{2k}<a_{2k-1},$$ hence $$a_2<a_{2k-1},\quad\text{and }\quad a_{2k}<a_1,$$ for all $k$. Therefore, each $\{a_{2k+1}\}$ and $\{a_{2k}\}$ converges with $k\rightarrow \infty$, i.e. there exists $\alpha,\beta\in\mathbb{R}$, such that $$\lim_{k\rightarrow\infty}a_{2k+1}=\alpha,\text{ and }\lim_{k\rightarrow\infty}a_{2k}=\beta.$$Then, $$\alpha=\lim_{k\rightarrow\infty}a_{2k+1}=\lim_{k\rightarrow \infty}\frac{a_{2k}}{2}+\frac{a_{2k-1}}{2}=\frac{\beta}{2}+\frac{\alpha}{2}\Rightarrow \alpha=\beta,$$ which means $\{a_n\}$ converges.

Answer 2

You can use an induction argument, but you don't need to go that far back. Draw the zig-zag picture and prove some intuitive results.

Note that $a_{2n+3}-a_{2n+1} = ½ (a_{2n+2}-a_{2n+1}). $

From the picture, you see that, as you traverse the sequence (starting at $a_1$), you go down-up-down-up-.... In particular, when you go from a number with odd index (say, $a_{2n+1}$) to the next number ($a_{2n+2}$), the number decreases.

To prove this, just use another inductive argument. You can prove that $a_{2n+1}-a_{2n+2}=¼ (a_{2n-1}-a_{2n})$, and so on...

Part (b) is very similar.

Answer 3

As highlighted in the comments, the recursion can be solved by putting $x^n$ in place of $a_{n}$. This gives the equation $2x^2-x-1$. Since both roots of this equation (characteristic equation) are different, the general solution of $a_{n}$ is $p1^{n}+q(-0.5) ^{n}$. The condition $a_{1}>a_{2}$ gives $q<0$. From this the given results are obvious.