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 $$\limsup(a_n)\limsup(b_n)=1.$$

  • $\begingroup$ Thanks for the quick reply! Sorry, I don't really understand the series of a_n and b_n above. Could you elaborate a bit further? $\endgroup$ – KoolAidx Jun 11 '20 at 10:06
  • $\begingroup$ $a_n$ 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. $\endgroup$ – Gary Jun 11 '20 at 10:28
  • $\begingroup$ i might be wrong, and now this boils down to a basic question about convergence, which i beforehand might not have understood it fully. Wouldn't it make a_n and b_n to a divergent series since they both have 2 accumulation points, which nulls the requirement where a_n and b_n have to converge? Or could and should i work with subseries instead? Thanks. @Gary $\endgroup$ – KoolAidx Jun 11 '20 at 10:47
  • $\begingroup$ Do you know the definition of $\lim \sup$? $\endgroup$ – Gary Jun 11 '20 at 10:52
  • 1
    $\begingroup$ The your question is equivalent to prove that $\lim (a_nb_n)=\lim(a_n)\lim(b_n)$.@KevinAdi $\endgroup$ – kola Jun 11 '20 at 11:05

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