If $a_n \to a < 0$, then $\lim \inf a_nb_n = a \lim \sup b_n$ I got this proof that I can't show for some hours now. Does anyone have a hint?
If $a_n, b_n$ are bounded sequences and $\lim_{n\to\infty} a_n = a < 0$ then
$$\liminf_{n\to\infty} a_nb_n = a \cdot \limsup_{n\to\infty} b_n$$
Thanks!
 A: I would try something like this:
Let $\varepsilon > 0$. Then there exists $n_0 \in \Bbb{N}$ s.t.
$$
n \geq n_0 \implies a - \varepsilon < a_n < a + \varepsilon
$$
Now if $n \geq k_0$
\begin{align*}
(a - \varepsilon) \sup_{k \geq n} b_k & \overset{(*)}{=} \inf_{k \geq n} ( (a - \varepsilon)b_k) \\
&\leq \inf_{k \geq n} (a_k b_k) \\
& \leq \inf_{k \geq n} ((a+\varepsilon)b_k) \\
&\overset{(*)}{=} (a+\varepsilon) \sup_{k \geq n} b_k
\end{align*}
Then by taking the limit $n \to \infty$ we'd get
\begin{align*}
(a-\varepsilon) \limsup_{n \to \infty} b_n &\leq \liminf_{n \to \infty} (a_n b_n) \leq (a + \varepsilon) \limsup_{n \to \infty} b_n \\
\iff -\varepsilon \limsup_{n \to \infty} b_n &\leq \liminf (a_nb_n) - a \limsup_{n \to \infty} b_n \leq \varepsilon \limsup_{n \to \infty} b_n \\
\iff \left| \liminf (a_nb_n) - a \limsup_{n \to \infty} b_n \right| &\leq \varepsilon |\limsup_{n \to \infty} b_n|
\end{align*}
The sequence $(b_n)$ is bounded, so $|\limsup b_n| < \infty$, and the result follows.
The $(*)$ equalities feel right, but need some more justification.
