I've been working with proofs involving $\limsup$ and $\liminf$, and I'm a bit confused regarding their general methodology. More specifically, I'm unsure about whether my approach to the following problem makes sense.
Problem. Let $(s_n)$ and $(t_n$) be sequences and suppose that there exists $N_0$ such that $s_n \leq t_n$ for all $n > N_0$. Show that $\liminf s_n \leq \liminf t_n$ and that $\limsup s_n \leq \limsup t_n$.
The way I approached it was as follows:
Let $N > N_0$. Then $\limsup_{N \rightarrow \infty} \{ s_n : n > N \} \leq t_n$ as $s_n \leq t_n$, and $\limsup s_n $ is the largest possible limit of a subsequence of $s_n$. As $t_n : n > N$ is (by definition) less than $\limsup_{N \rightarrow \infty} \{ t_n : n > N \}$, the proof is complete.
I'm pretty sure this is incorrect, however, and I'm generally unclear about the method behind such a proof. Any help is appreciated!
Let $x_n = t_n - s_n$. Your problem is equivalent to the following: if $x_n \ge 0 \ \forall n \ge N$, then $\liminf x_n \ge 0$.
Remember that, for sequences of real numbers, $\liminf$ is the $\inf$ of the set of limit points of $\{x_n\} _{n \ge N}$. I find working with this description of $\liminf$ very visual, therefore very productive - give it a try! Therefore, let us focus on this set of limit points, first.
Let $x$ be a limit point of the set $\{x_n\} _{n \ge N}$. This means that there exist a subsequence of $(x_n) _{n \ge N}$ which tends to $x$. Since all the $x_n$ are positive, so will be the terms of the subsequence, hence so will be its limit: $x \ge 0$. We have discovered that the set of limit points is made up of only positive numbers.
Next, you have to take the $\inf$ of this set. But the $\inf$ of a set of positive numbers is itself positive, which is exactly what you needed to show.
A similar argument is for $\limsup$.
Notice that the essential ingredients that we have used here are that the limit of a sequence of positive numbers is itself positive, and that the $\inf$ of a set of positive numbers is itself positive.