Question:
If $a_n$ upholds $a_n \geq 0$ for all $n \in \mathbb {N}$ with $\lim_{n \to \infty}a_n = a$ and $b_n$ is a bounded sequence:
Prove $\limsup_{n \to \infty}(a_nb_n)$ = a $\cdot \limsup_{n \to \infty}b_n$.
So there is one point in the proof where I am stuck.
This is what I have so far:
$b_n$ is bounded $\implies$ exists some $b_{n_k}$ s.t $\lim_{k \to \infty}b_{n_k} = \limsup_{n \to \infty}bn =$ sup $b_n$.
Notice, $a_{n_k}b_{n_k}$ is a subsequence of $a_nb_n$, then from this knowledge and arithmetics of limits we have:
$\limsup_{n \to \infty}(a_nb_n) \geq \limsup_{k \to \infty}(a_{n_k}b_{n_k}) = a \cdot $ sup$b_n = a \cdot \limsup_{n \to \infty}b_n$.
$\implies \limsup_{n \to \infty}(a_nb_n) \geq a \cdot \limsup_{n \to \infty}b_n$ and this is the first inequality.
The second one is where I am struggling.
Since $b_n$ is bounded and $a_n$ converges to $a$ we have $\exists \limsup_{n \to \infty}(a_nb_n)$.
$\implies \exists a_{n_p}b_{n_p}$ subsequence of $a_nb_n$ s.t $\lim_{p \to \infty}(a_{n_p}b_{n_p}) = \limsup_{n \to \infty}(a_nb_n)$.
Also, $\exists i \in \mathbb {N}$ s.t $\forall x \gt i$ we have $a_n \lt a+\epsilon$ for some $\epsilon \gt 0$. We know $b_n \lt$ sup$b_n$.
$\implies a_{n_p}b_{n_p} \lt (a+\epsilon) \cdot$ sup$b_n$.
Here lies the part I'm insecure about:
Taking limit on both sides we will get:
$\limsup_{n \to \infty}(a_nb_n) = \lim_{p \to \infty}(a_{n_p}b_{n_p}) \leq a \cdot$ sup$b_n = a \cdot \limsup_{n \to \infty}b_n$.
$\implies \limsup_{n \to \infty}(a_nb_n) \leq a \cdot \limsup_{n \to \infty}b_n$ which is the second inequality.
$\implies \limsup_{n \to \infty}(a_nb_n) = a \cdot \limsup_{n \to \infty}b_n$ as needed.