Let $\{a_n\}_{n=1 }^{\infty}$ and $\{b_n\}_{n=1}^{\infty} $ be two sequences in $\mathbb{R}$, with the first sequence convergent . Prove that $$ \limsup\limits_{n\to \infty} a_n b_n =\lim\limits_{n\to\infty} a_n \limsup\limits_{n\to\infty} b_n$$
I tried following:
$ \limsup\limits_{n\to \infty} a_n b_n \leq \limsup\limits_{n\to\infty} a_n \limsup\limits_{n\to\infty} b_n$.
Since $\{a_n\}$ is convergent, $\limsup\limits_{n\to\infty} a_n =\lim\limits_{n\to\infty} a_n $ gives one inequality along with one of the property of the limsup of the product of two sequences i.e
$ \limsup\limits_{n\to \infty} a_n b_n \leq \lim\limits_{n\to\infty} a_n \limsup\limits_{n\to\infty} b_n$.
I want to prove
$ \limsup\limits_{n\to \infty} a_n b_n \geq \limsup\limits_{n\to\infty} a_n \limsup\limits_{n\to\infty} b_n$.
Can anyone help me on this?
Best Answer
It's not true.
Let $a_n=-1$ for all $n$, and let $b_n=(-1)^n$. Then $a_nb_n=(-1)^{n+1}$ so $$\limsup_{n\to\infty} a_nb_n = 1\\\lim_{n\to\infty} a_n=-1\\\limsup_{n\to\infty} b_n=1$$
It is true if $\lim_{n\to\infty} a_n > 0$.
If $\lim_{n\to\infty} a_n<0$ then the result is:
$$\limsup a_nb_n = \lim a_n \liminf b_n$$
If $\lim a_n = 0$, it gets more complicated.