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?

  • $\begingroup$ Don't you mean $\limsup_{n\to\infty}$? $\endgroup$ – dannum Sep 23 '14 at 1:36
  • $\begingroup$ 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. $\endgroup$ – Thomas Andrews Sep 23 '14 at 1:46
  • $\begingroup$ And everywhere you've written $x\to\infty$, it should be $n\to\infty$. $\endgroup$ – Thomas Andrews Sep 23 '14 at 1:46
  • $\begingroup$ 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) $\endgroup$ – user147263 Sep 23 '14 at 1:50
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    $\begingroup$ @ThomasAndrews which is also proved there $\endgroup$ – user147263 Sep 23 '14 at 1:52

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.


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