Proof that $\lim \sup (a_n · b_n) = \lim \sup a_n · \lim \sup b_n $

Question

With $a_n, b_n \geq 0$ or non-negative for all $n$ in $\mathbb N$.

Proof that for $n \to \infty $

$\lim \sup (a_n · b_n) = \lim \sup a_n · \lim \sup b_n $
, if $a_n$ and $b_n $converge.

My start would be:

Since $a_n$ converges and $\lim a_n = a$, so is $\lim \sup a_n = a$
and since $b_n$ converges and $\lim b_n = b$, so is $\lim \sup b_n = b$.

So that, $\lim \sup (a_n · b_n) = a · b$.

Answer 1

I't not correct ! take $a_n=\boldsymbol 1_{2\mathbb N}(n)$ and $b_n=\boldsymbol 1_{2\mathbb N+1}(n)$. Then $$\limsup (a_nb_n)=0$$ whereas $$\limsup(a_n)\limsup(b_n)=1.$$