Upper limits for $s_n$ $\leq$ $t_n$

Question

Let $(s_n)$ and $(t_n)$ be two sequences of real numbers. Suppose there exists $N_0$ such that for all $n>N_0$, $s_n \leq t_n$.

Two Questions:

1. Suppose that $\lim s_n$ and $\lim t_n$ both exist. Show that lim $ s_n \leq $lim $ t_n$.

Is this a reasonable answer:

Proof: By assumption since there exists $N_0$ such that for all $n>N_0$ then $s_n \leq t_n$ then
sup{$t_n: n>N_o$} $\geq$ sup{$s_n: n>N_o$} and thus lim sup{$t_n: n>N_o$} $\geq$ lim sup{$s_n: n>N_o$} and since both of these limits exist and are equal to lim $t_n$ and lim $s_n$ respectively then the desired result follows.

2. Second question where I am stumped. show that lim inf $s_n$ $\leq$ lim inf $t_n$ and that lim sup $s_n$ $\leq$ lim sup $t_n$

Answer 1

For $1$:

No. Note that it suffices to prove: if $(a_n)$ is a convergent sequence and if there is $N_0$ such that for $n > N_0$, $a_n \ge 0$, then $\lim a_n \ge 0$.

Let $a$ be the limit. If $a < 0$, then $\epsilon = - a > 0$. We know that, by the definition of limit, there exists $N$ such that for $n > N$, $a_n < \epsilon + a = 0$. Let $N_1 = \max(N, N_0) + 1$, then $a_{N_1} \ge 0$ and $<0$. Contradiction. Therefore, $a \ge 0$.

For $2$:

Here is a hint. We have for a sequence $(a_n)$,

• $\liminf a_n =\lim_{n \to \infty} \inf_{k \ge n} a_k$

• $\limsup a_n = \lim_{n \to \infty} \sup_{k \ge n} a_k$

Try to apply the result of part $1$ to some special sequences.