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