Recursive sequence convergence.: $s_{n+1}=\frac{1}{2} (s_n+s_{n-1})$ for $n\geq 2$, where $s_1>s_2>0$ The problem is the following: suppose $s_1>s_2>0$ and let $s_{n+1}=\frac{1}{2} (s_n+s_{n-1})$ for $n\geq 2$. Show that ($s_n$) converges.
Now, here is what I figured out:


*

*$s_2<s_4$: Base Case for induction that $s_{2n}$ is an increasing sequence.

*Assume $s_{2n-2}<s_{2n}$.

*Induction step: $s_{2n}<s_{2n+2}$

*$s_1>s_3$: Base Case for induction that $s_{2n-1}$ is a decreasing sequence.

*Assume $s_{2n-1}<s_{2n-3}$.

*Induction step: $s_{2n+1}<s_{2n-1}$.


I have proved those two. However arguing in favor of convergence has me going around in circles. Since $s_1>s_2$ and (as discovered during the formulation of Base Cases) $s_3>s_4$, I figured it might be a good idea ot assume that if every odd member of the original sequence ($s_n$) is greater than the following even member, then the limit would be somewhere in between, the two (odd and even) sequences won't cross. Hence the upper and lower bounds would be $s_1$ and $s_2$ respectively. Here is how I approach this:


*

*Assume $s_{2n-1}>s_{2n}$

*Show that $s_{2n+1}>s_{2n+2}$


The proof as I mentioned has me running in circles. Any assistance?

From the definition of $s_n$ one has
$$
s_{n+1}-s_n=-{1\over2}(s_n-s_{n-1}).
$$
It follows that if $s_{2n}-s_{2n-1}<0$ then 
$$
s_{2n+2}-s_{2n+1}={1\over4}(s_{2n}-s_{2n-1})<0.
$$
