From Rudin's Principles of Mathematical Analysis.
Problem 3.4:
Find the upper and lower limits of the sequence $\left\{s_n\right\}$ defined by $$ s_1=0 ; \quad s_{2 m}=\frac{s_{2 m-1}}{2} ; \quad s_{2 m+1}=\frac{1}{2}+s_{2 m} $$
Provided Solution:
We can obtain that $s_{2 m+1}=1-\left(\frac{1}{2}\right)^m$ and $s_{2 m+2}=\frac{1}{2}\left(1-\left(\frac{1}{2}\right)^m\right)$, for $m \geq 0$.
Thus $\lim \sup _{n \rightarrow \infty} s_n=1$ and $\liminf _{n \rightarrow \infty} s_n=\frac{1}{2}$.
Question:
I'm struggling to see how we can conclude from $s_{2 m+1}=1-\left(\frac{1}{2}\right)^m$ and $s_{2 m+2}=\frac{1}{2}\left(1-\left(\frac{1}{2}\right)^m\right)$ that $\lim \sup _{n \rightarrow \infty} s_n=1$ and $\liminf _{n \rightarrow \infty} s_n=\frac{1}{2}$.
As far as I understand, an upper limit of a sequence (as defined by Rudin) is the supremum of the set containing the limits of all (convergent) subsequences. Now, are we assuming that the only convergent subsequences of $\{s_n\}$ are the odd and even numbers respectively? If so, how can one see this?
I appreciate any insights or clarifications!
For direct answer to your question let's consider $\mathbb{N}=\mathbb{N}_1\cup \mathbb{N}_2$ with $\mathbb{N}_1\cap \mathbb{N}_2=\emptyset$ and both infinite. If $f:\mathbb{N}_1\to \mathbb{R}$ and $f:\mathbb{N}_2\to \mathbb{R}$ are converged sequences, with limits $a_1$ and $a_2$ respectively, then any subsequence of $f:\mathbb{N}\to \mathbb{R}$ have only limit points $a_1$ and $a_2$.
This property can be extended to finite number of subsequences.
Addition.
To proof let's consider any subsequence $f_0:\mathbb{N}_0\to \mathbb{R}$ of sequence $f:\mathbb{N}\to \mathbb{R}$ i.e. some $\mathbb{N}_0\subset\mathbb{N}$, where $\mathbb{N}_0$ is infinite. Suppose $f_0$ have limit point $a_0$ different from $a_1$ and $a_2$. Then there exists a subsequence of $f_0$, which converges to $a_0$. But infinite number of members of this subsequence should be in $\mathbb{N}_1$ or $\mathbb{N}_2$, which is impossible, as they have their own different limit.