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:


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*$s_2<s_4$: 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?

$(s_{2n})$ is increasing and $(s_{2n-1})$ is decreasing and we have $s_2\leqslant s_{2n}<s_{2n+1}<s_{2n-1}\leqslant s_1$ so $(s_{2n})$ converges as well as $(s_{2n-1}),$ say to $a$ and $b,$ respectively. Now since $s_{2n}=\frac{1}{2}(s_{2n-1}+s_{2n-2})$ then $a=\frac{1}{2}(b+a)$ so $a=b.$ Thus, $s_n\to a=b$ as $n\to\infty.$
