Recursive sequence convergence.: $s_{n+1}=\frac{1}{2} (s_n+s_{n-1})$ for $n\geq 2$, where $s_1>s_2>0$

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

$\lim_{n\to\infty} s_n=a$.

$(s_{2n})$ 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 $n\to\infty.$

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 $$ s_{2n+2}-s_{2n+1}={1\over4}(s_{2n}-s_{2n-1})<0. $$