If $\limsup_{n\to\infty}a_n<\infty$, then $\limsup_{n\to\infty}(-a_n)>-\infty$?

Question

Assume $(a_n)$ is a positive sequence. I stuck to find if the following always true?

If $\limsup_{n\to\infty}a_n<\infty$, then $\limsup_{n\to\infty}(-a_n)>-\infty$.

Answer 1

Yes, it is true, because \begin{align} \limsup(-a_n) = -\liminf (a_n) \geq -\limsup(a_n)>-\infty. \end{align} The first equality should be clear from definitions, while the other steps follow from $\liminf(a_n)\leq \limsup(a_n)<\infty$. The proof even shows that this is true for all sequences $\{a_n\}_{n=1}^{\infty}$ in $[-\infty,\infty]$; positivity is not required (a little reflection should convince you that your "if then" is even an "if and only if").

Answer 2

From scratch:

Applying the definitions:

$\tag1 \limsup_{n\to\infty}(-a_n)=\underset{k\to \infty}\lim g_k\quad \text{where}\quad g_k=\sup_{n\ge k}(-a_n).$

It suffices then to show that

$\tag2 -\sup (-S)=\inf S\ \text{for any set}\ S$

The definition of $\sup$ implies that if $\alpha=\sup (-S)<\infty$, then

$\tag3 \alpha\ge -s\quad \text{for each}\ s\in S\quad \text{and if}\quad \epsilon>0\ \text{there is a}\ -s\in -S\ \text{such that}\ \alpha-(-s)<\epsilon.$

Or equivalently,

$\tag4 -\alpha\le s\quad \text{for all}\quad s\in S\quad \text{and} -\alpha-s>\epsilon$

The cases $\alpha=\pm \infty$ are even easier.