is (infinitely often p) ∨ (infinitely often ¬p) valid?
The LTL formula for the given problem could be: $G(F(p)) \lor G(F(\lnot p))$. Consider a $\omega$-word $w = w_1w_2w_3 \ldots$ over $\{\{\}, \{p\}\}$: 

Let $w_i = w_iw_{i+1}\ldots$
\begin{align*}
&amp; \; \; \; \; \; \;  \; \; \; w \not\models G(F(p))\\ 
&amp;\implies \exists i . w_i \not\models F(p)\\
&amp;\implies \exists i. \forall j \geq i . w_j \models \lnot p\\ 
&amp;\implies \exists i. \forall j \geq i . w_j \models F(\lnot p)\\ 
&amp;\implies \exists i. w_i \models G(F(\lnot p))\\ 
&amp;\implies w \models G(F(p)) \lor G(F(\lnot p))
\end{align*}

Therefore, $G(F(p)) \lor G(F(\lnot p))$ is valid. 

Note that this doesn't mean that $G(F(p)) \land G(F(\lnot p))$ is not satisfiable. For example, consider the word $w=\{p\}\{\}\{p\}\{\}\{p\}\{\}\ldots$