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 $$\lim\inf_{n \to \infty} s_n = \frac{1}{2} \ \mbox{ and } \ \lim\sup_{n\to\infty} s_n = 1?$$
Suppose there exists a sub sequence of $\{s_n\}$ converges to a point $s$. That is $\{s_{n_k} \}\to s$. Define $S_1$ as the set of terms with odd indices. That is, $n_k$ is odd and similarly $S_2$ for even terms. Note that either $S_1$ or $S_2$ has infinite elements, but not both. Hence, $s=1$ or $s=1/2$. Thus for any limit $s$ of sub sequence we have $1/2\leq s\leq 1$ and the result follows.