<s>[INST] Prob. 4, Chap. 3 in Baby Rudin: How to show that these are the limit superior and the limit inferior? Here's Prob. 4, Chap. 3 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Find the upper and lower limit of the sequence $\left\{ s_n \right\}$ defined by $$s_1 = 0; \ s_{2m} = \frac{s_{2m-1}}{2}; \ s_{2m+1} = \frac{1}{2}+ s_{2m}.$$

My effort: 

We note that $s_1 = 0$, $s_2 = 0$, $s_3 = \frac{1}{2}$, $s_4 = \frac{1}{4}$, $s_5 = \frac{1}{2} + \frac{1}{4}$, $s_6 = \frac{1}{4} + \frac{1}{8}$, $s_7 = \frac{1}{2} + \frac{1}{4} + \frac{1}{8}$, $s_8 =  \frac{1}{4} + \frac{1}{8} + \frac{1}{16}$, $s_9 = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16}$, $s_{10} = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32}$, $s_{11} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32}$, $s_{12} = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \frac{1}{64}$, and continuing in this way we obtain 
  $$s_{2m} = \frac{1}{2}- \frac{1}{2^m} \ \mbox{ and } \ s_{2m+1} = 1- \frac{1}{2^m}  $$ for $m = 0, 1, 2, 3, \ldots$. So $$\lim_{m \to \infty} s_{2m} = \frac{1}{2} \ \mbox{ and } \ \lim_{m \to \infty} s_{2m+1} = 1.$$

Am I right?
Using the above, how to rigorously prove that $$\lim\inf_{n \to \infty} s_n = \frac{1}{2} \ \mbox{ and } \ \lim\sup_{n\to\infty} s_n = 1?$$
 [/INST] Hint. Show inductively that
$$
s_{2n}=\frac{1}{2}-\frac{1}{2^n}\quad\text{and}\quad s_{2n+1}=1-\frac{1}{2^n}.
$$
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