# limsup of the product of two sequences, of which one converges

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?

• Don't you mean $\limsup_{n\to\infty}$? – dannum Sep 23 '14 at 1:36
• It's not true unless $\lim a_n\geq 0$. Even then, if $\lim a_n=0$ you'll need $\limsup b_n<+\infty$ for the right hand side to be well-defined. – Thomas Andrews Sep 23 '14 at 1:46
• And everywhere you've written $x\to\infty$, it should be $n\to\infty$. – Thomas Andrews Sep 23 '14 at 1:46
• possible duplicate of lim sup inequality $\limsup ( a_n b_n ) \leq \limsup a_n \limsup b_n$ (the equality case is also treated there) – user147263 Sep 23 '14 at 1:50
• @ThomasAndrews which is also proved there – user147263 Sep 23 '14 at 1:52

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.