# proof that $\lim \sup (a_n · b_n) = \lim \sup a_n · \lim \sup b_n$

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, $$\lim \sup (a_n · b_n) = a · b$$.

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 $$\limsup(a_n)\limsup(b_n)=1.$$
• $a_n$ is the indicator function of the positive even integers: it is $1$ if $n$ is even, $0$ if $n$ is odd. Similarly, $b_n$ is the indicator function of the positive odd integers: it is $1$ if $n$ is odd, $0$ if $n$ is even. – Gary Jun 11 '20 at 10:28
• Do you know the definition of $\lim \sup$? – Gary Jun 11 '20 at 10:52
• The your question is equivalent to prove that $\lim (a_nb_n)=\lim(a_n)\lim(b_n)$.@KevinAdi – kola Jun 11 '20 at 11:05