{$a_n$} is a sequence that is recursively defined as the follows:
$$a_{n+2} =(\frac 12)(a_{n+1}+a_n)$$ for n = 1,2,3,... such that $a_1\gt a_2 \gt0$.
a) solving a) shows $$a_{n+2} -a_n = \frac {(-1)^n}{2^n}(a_1-a_2)$$
b) For any positive integers m and n, show that $$a_{2m}\lt a_{2n-1}$$
c) Show that the two sequences {${a_1, a_3, ...}$} and {${a_2, a_4, ...}$} converge to the same limit k, where $$k = \frac 13(a_1+2a_2)$$
Hi, we're supposed to answer this question only using the monotone convergence theorem (this chapter is introduced before learning derivatives and integration), but I can't seem to answer question b) and c).
I know that the odd subsequence is strictly decreasing and the even sequence is strictly increasing. And my intuition says that you prove that each sequence is bounded by a number z to answer b). However, how do you go and show that the limit is equal to k?
I understand this is not what you were asked to do, but maybe it can be useful for similar situations where the roadmap proposed in this question does not work. This is a linear recurrence and you can get an explicit formula for $a_n$ just by using the roots of the characteristic polynomial $p(\lambda)=\lambda^2-\frac 12 \lambda -\frac 12$. In fact, $$ a_n = c_1 \left(-\frac 12\right)^n + c_2 1^n $$
The constants $c_1, c_2$ are computed using the initial conditions, for instance the prescribed values $a_0, a_1$, yielding finally $$ a_n = \frac 23 a_1+\frac 13 a_0 + \frac 23 (a_0-a_1 )\left(-\frac 12\right)^n. $$
Now, you can directly compute the limit.
Part b) in your question does not make much sense... the condition $a_{2m}-a_{2n-1}$ does not necessarily hold for any positive $n,m$. In particular, depending on the choice of $a_1,a_2$, it does not hold when $n=m=1$. Did you accurately reproduce the question?