$\limsup{s_n}=\liminf{s_n}=\infty$,then $s_n$ diverges to infinity

Question

The proof given in Goldberg goes like this,

Since $\lim\inf{s_n}=\infty\;,\;$ given any $M>0$, then g.l.b.$\{s_n,s_{n+1},s_{n+2},\ldots \}>M\;(n\geqslant N).$

So $M$ is a lower bound for { $\{s_n,s_{n+1},\ldots\}$,but not the greatest (how can we claim this ? $M$ might not be a lower bound)

The proof ends with $s_n>M (n\geqslant N)$

Answer 1

If $\{s_n\}:\mathbb N\to\mathbb R$ is divergent,prove that:

$\lim\sup s_n=\infty=\lim\inf s_n$.

Now $s_n>M>0,\; n\geqslant N_1$

Since all the terms are $s_n>M$ so $M$ is the lower bound then g.l.b.$\{s_{N},s_{N+1},\ldots\}\geqslant M$

Since $M$ can be chosen arbitrarily large we can conclude that $\lim\inf{s_n}=\infty$. As $\lim\inf s_n\leqslant\lim\sup s_n$ . We can conclude that $\lim\sup s_n=\infty$.

Answer 2

Since $\liminf s_n\leq \limsup s_n$, it is sufficient to have that $\liminf s_n=\infty$.

By definition, $\lim s_n=\infty$ $\iff$ $\forall M>0$ $\exists N$: $n>N\implies |s_n|>M$.

Let $M>0$. By definition, $\liminf s_n=\sup_n \inf_{k\geq n} s_k$.

Since $\sup_n (\inf_{k\geq n} s_k)=\infty$, there exists $N$ such that $\inf_{k\geq N} s_k>M$... Can you continue?

Answer 3

Option:

$a_n= \inf \{ s_k : k\ge n \}$; an increasing sequence.

$\lim \inf s_n= \lim_{n \rightarrow \infty} a_n=\infty$.

For $M>0$ there is a $n_0$ s.t. $n\ge n_0$ implies

$a_n >M$, and since

$s_n \ge a_n >M$, we are done.