In course of solving the problem How to show that $\lim \sup a_nb_n=ab$ I feel that I've probably made some mistake in my solution for I didn't use the fact that $a_n>0$ $\forall$ $n\geq1$.
The statement of the problem is: Let $a_n,b_n\in\mathbb R^+$ such that $\lim a_n=a>0$ and $\lim \sup b_n=b>0$. Show that $\lim \sup a_nb_n=ab.$
Please help me to find out the mistake I've made:
- $\exists$ a subsequence $\{a_{r_n}\}$ of $\{a_n\}$ such that $a_{r_n}\to a.$ Now $a_{r_n}\to a, b_{r_n}\to b\implies a_{r_n}b_{r_n}\to ab\implies ab$ is a subsequencial limit of {$a_nb_n$}. If possible let $\exists$ a subsequence $\{a_{p_n}b_{p_n}\}$ of $\{a_nb_n\}$ such that $a_{p_n}b_{p_n}\to m>ab.$ Since $b_n,b>0$ so $b_n^{-1}\to b^{-1}>0$ whence $a_{r_n}\to mb^{-1}>a,$ a contradiction to $\lim \sup a_n=a.$ Hence the result follows.
Thanks for voting up the question. But that didn't actually eliminate my confusion. I'm looking for some concrete comments and opinions.
You do need some extra assumption. Consider the following example (where $a$ is a finite real):
$a_n=a$ if $n$ is odd, $a_n=-n$ if $n$ is even,
$b_n=-1.$ (so in notation, $\lim b_n=b=-1$)
Then lim sup of $a_n$ is $a$, and lim $b_n$ is b, yet we have
$a_nb_n=-a$ if $n$ is odd, $a_nb_n=n$ if $n$ is even.
So in this example the limsup of $a_nb_n$ is $+\infty$, whereas $ab$ is not.
EDIT: I noted that in your proof you needed to use $b>0$ but that assumption should be stated in the problem.