Find $\limsup_{n\to\infty}{s_n}$ and $\liminf_{n\to\infty}{s_n}$

Question

Let $s_n$ be defined by $s_1=0$, $s_{2n}=\frac{1}{2}s_{2n-1}$, $s_{2n+1}=\frac{1}{2}+s_{2n}$, $n \in \mathbb{N}$. Find $\limsup_{n\to\infty}{s_n}$ and $\liminf_{n\to\infty}{s_n}$

I've found that $s_{2n-1}=s_{2(n-1)+1}=\frac{1}{2}+s_{2(n-1)}=\frac{1}{2}+s_{2n-2}$. You can then show that $s_{2n}=\frac{1}{2}(\frac{1}{2}+s_{2n-2})$ ; however, I'm not really sure if this helps or where to go from here.

Answer 1

Let $t_k = s_{2k+1}$. Then $t_0 = s_1 = \frac12$, and

$$t_k = \frac12 + s_{2k} = \frac12 + \frac12s_{2k-1} = \frac12 + \frac12 t_{k-1}$$ It is easy to get $t_k$ in closed form: Let $t_k = 1-u_k$, then $$ 1-u_k = \frac12 +\frac12 -\frac12 u_{k-1} \implies u_k = \frac12 u_{k-1} $$ with $u_0 = \frac12$, thus $u_k = 2^{-(k+1)}$ and $t_k =1- 2^{-(k+1)}$.

And let $r_k = s_{2k}$. Then $$r_k = \frac12 t_{k-1} = \frac12 -\frac12 2^{-k} = \frac12 - 2^{-(k+1)}$$

Now, since $s_n$ is, for any $n$, either $t_n$ or $r_n$, and $s_n$ includes all the $t_n$ and $r_n$, $$\lim \sup (s_n) = \max \{ \lim \sup (t_n) , \lim \sup (r_n) \} = \max\{ 1,\frac12 \} = 1$$ and $$\lim \inf (s_n) = \min \{ \lim \inf (t_n) , \lim \inf (r_n) \} = \min\{ 1,\frac12 \} = \frac12$$