connection between $\limsup[a_n, b_n]$ and $[\limsup a_n, \limsup b_n]$

Question

If $\{a_n\}$ and $\{b_n\}$ are sequences of real numbers, what is the connection between $\limsup[a_n, b_n]$ and $[\limsup a_n, \limsup b_n]$?

Is $\limsup[a_n, b_n] = [\limsup a_n, \limsup b_n]$?

I would appreciate a proof or an example (if it exists) that it is not always true.

Thank you.

Answer 1

In general, $\emptyset\subseteq\limsup\limits_{n\to\infty}[a_n,b_n]\subseteq\left[\liminf\limits_{n\to\infty} a_n,\limsup\limits_{n\to\infty} b_n\right]$.

For $a_n=2^{-n-1}$ and $b_n=2^{-n}$, both sequences converge and the $\limsup$ of the intervals is empty (though this does not happen when $\lim_na_n<\lim_nb_n$). For $a_n=\cos (\pi n)$ and $b_n=1+\cos(\pi n)$, neither of the sequences converges and the other equality holds.

The sequences $a_n=\cos(\pi n)+2^{-n-1}$, $b_n=\cos(\pi n)+2^{-n}$ are an instance when $\emptyset =\limsup\limits_{n\to\infty} [a_n,b_n]\subsetneqq \left(\liminf\limits_{n\to\infty}a_n, \limsup\limits_{n\to\infty} b_n\right)$.