Let ${a_n}$ be a non-convergent (or divergent) sequence such that $1\le a_n \le 2$.
Prove or disprove:
$$\limsup a_n\cdot \limsup \frac 1 {a_n} \gt 1$$
My try:
$a_n$ must have at least two subsequential limits within the interval [1,2]. If that's the case then $\frac 1 {a_n}$ should have at least two subsequential limits within the interval $[\frac{1}{2}, 1]$. Thus, it follows that the inequality is true.
Is my approach correct?
Your approach is pretty good, but I would note that since the limit does not exist, $$ \limsup_{n\to\infty}a_n\gt\liminf_{n\to\infty}a_n $$ and that $$ \limsup_{n\to\infty}\frac1{a_n}=\frac1{\liminf\limits_{n\to\infty}a_n} $$ Therefore, $$ \limsup_{n\to\infty}a_n\cdot\limsup_{n\to\infty}\frac1{a_n} =\frac{\limsup\limits_{n\to\infty}a_n}{\liminf\limits_{n\to\infty}a_n}\gt1 $$