For real sequences $\{a_n\}$ and $\{b_n\}$ with $a_n\to a>0$ and $\limsup_{n\to \infty}b_n=-\infty$, prove $\limsup_{n\to \infty}a_nb_n=-\infty$ 
Suppose that $\{a_n\}$ and $\{b_n\}$ are real sequences such that $a_n\to a$, where $a>0$ and $\limsup \limits_{n\to \infty}
 b_n=-\infty$. Prove that $\limsup \limits_{n\to \infty}a_n
 b_n=-\infty$.

Probably this question has been asked before, but most proofs which I have seen assumed that $\limsup \limits_{n\to \infty}b_n$ is finite. The case when it is finite I can handle quite easily and I've already proved it. I'd like to see the proof for the case when limit superior is $-\infty$.
 A: It suffices to prove that for any $\xi\in\mathbb{R}$,
$$a_nb_n<\xi$$
for all sufficiently large $n$. (If $\lim\sup a_nb_n=x>-\infty$, then, for any y<x, one can find infinitely many $a_nb_n>y$.)
Since $a_n\rightarrow a$,
$$\frac{1}{2}a<a_n<\frac{3}{2}a$$
for all $n$ from some fixed index $N$ onward. Since $\lim\sup b_n=-\infty$, there exists $N_1>N$ such that
$$b_n<\min(0,\frac{2\xi}{a})$$
for all $n>N_1$. Hence for $n>N_1$,
$$a_nb_n<\xi$$
as was to be proved.
A: When $\limsup b_n=-\infty$, what can be said about $b_n$? 
$\limsup b_n$, by definition is the supremum of set of all limit points of the sequence $(b_n)$ (i.e., supremum of set of all subsequential limits of the sequence $(b_n)$). 
Let $(b_{n_k})$ be any subsequence of $(b_n)$. One of the following must be true: 
$1): b_{n_k}\to c \in\mathbb R\cup \{\infty\}$ 
Clearly this is not possible because if so then by definition of $\limsup$, $\limsup b_n\geq a$ which is a contradiction. 
$2):b_{n_k}$ oscillates (finitely or infinitely)
In either case, we can find subsequence of $(b_{n_k})$ converging to $b\in \mathbb R\cup \{\infty\}$ which is a contradiction as argued in $1)$. 
$3):b_{n_k}\to -\infty$ 
So $(b_{n_k})$ has only one choice left viz. $b_{n_k}\to -\infty$. We have shown that every subsequence of $(b_n)$ diverges to $-\infty$. 
Now we know the behaviour of all subsequences of $(b_n)$. 
Let $(a_{n_m}b_{n_m})$  be any subsequence of the sequence $(a_nb_n)$. 
For large enough $m$, we do have $\frac a2\lt a_{n_m}\lt\frac{3a}2\implies a_{n_m}b_{n_m}\to -\infty$ (you may fill in the details here). 
We have $\limsup (a_nb_n)=\sup\{a_{n_m}b_{n_m}:\text{$a_{n_m}b_{n_m}$ is a subsequence of $(a_nb_n)$} \}=-\infty$ . Proved.
