I don't know how to solve the following question, your help is very much appreciated:
Let $(a_n)$ be a sequence where $1\le (a_n) \le 2$ for all $n$.
Prove or disprove: if $(a_n)$ is divergent, $$\overline{ \lim_{n \to \infty}} a_n * \overline{ \lim_{n \to \infty}}\left(\frac{1}{a_n}\right)>1$$
Where $$\overline{ \lim_{n \to \infty}}$$
Means $\limsup$.
As $(a_n)$ is divergent, $2 \ge \alpha := \lim \sup a_n > \lim \inf a_n =: \beta \ge 1$ $\implies \exists \ \gamma\in (1, \alpha)$, such that $\beta < \gamma < \alpha$.
On the other hand:
$$\lim \sup \frac1{a_n} = \frac1{\lim \inf a_n} > \frac1{\gamma}$$
Therefore:
$$\lim \sup a_n \times \lim \sup \frac1{a_n} > \gamma \times \frac1{\gamma} = 1$$