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?

Consider the characteristic equation of your recurrence:
$X^2-\frac1{2}X-\frac1{2}=0$
Which has solutions $1$ and $-\frac1{2}$.
Therefore the general expression for $s_n$ is:
$s_n=a\cdot (1)^n+b\cdot(-\frac1{2})^n=a+b\cdot(-\frac1{2})^n$
You can determine the value of the constants $a$ and $b$ using
$s_1$ and $s_2$.
Therefore
$\lim_{n\to\infty} s_n=a$.
