If $a_n\to \infty$ and $b_n$ diverges then $a_nb_n$ diverges.
I got this statement and need to show whether it's true or false (by proving it or showing counterexample, respectively).
First of all I don't know for sure if $a_n\to\infty$ means the standard "$a_n\to +\infty$" or just "for any $c>0$ there exists $N$ such that $n\ge N \implies |a_n|>c$". I just try assuming both cases but I'm geting nowhere with any.
If $b_n\to \infty$ or $b_n\to +\infty$ or $b_n\to -\infty$ it's pretty direct in any of the cases. But I don't know how to deal with $b_n$ just assuming that it doesn't converge to any real $a$.
Best Answer
You don't have $\lim_{n\to\infty}b_n=0$. Therefore, there is some $\varepsilon>0$ such that $\lvert b_n\rvert\geqslant\varepsilon$ infinitely often. In other words, there is a subsequence $(b_{n_k})_{k\in\mathbb N}$ of $(b_n)_{n\in\mathbb N}$ such that$$(\forall k\in\mathbb N):\lvert b_{n_k}\rvert\geqslant\varepsilon.$$But then $(a_{n_k}b_{n_k})_{k\in\mathbb N}\rightarrow\infty$.