Proof that $\limsup (a_nb_n)=\limsup a_n \lim b_n$

Question

Let $a_n$ be a bounded sequence and let $b_n$ be a convergent sequence. Prove that $\limsup⁡〖(a_n b_n )=\lim sup⁡〖a_n 〗 \lim⁡b_n 〗$.

Here is what I got

Proof

Let $a_n$ be a bounded sequence and let $ b_n$ be a convergent sequence. Since $a_n$ be a bounded sequence, the largest accumulation point exist, so there exists some a such that

$a=\limsup⁡〖a_n 〗$

Since $b_n$ be a convergent sequence, there exist a unique $b$ such that

$\lim⁡〖b_n 〗=b$.

Observe that

$\limsup⁡〖a_n lim⁡〖b_n 〗=b* \limsup⁡〖a_n.〗 〗= \limsup⁡〖〖(ba〗_n)$

How can I show that $\limsup⁡〖〖(ba〗_n)=\limsup⁡〖(a_n b_n)〗.〗$

Answer 1

Let's first show that $\limsup a_nb_n\leq \limsup a_n\lim b_n$:

We take $M>0$ such that $|a_n|\leq M$, and since $b_n\to b$ we can take for all $\epsilon>0$ an index $N$ such that for all $m>N$ we have $|b_m-b|<\frac{\epsilon}{2M}$ and $a_m\leq \limsup a_n + \frac{\epsilon}{2b}$. Now for such $m$: $$a_mb_m\leq a_mb + \frac{\epsilon}{2}\leq b\limsup a_n + \epsilon\implies\limsup a_nb_n\leq b\limsup a_n + \epsilon.$$ Note that the first inequality holds whether $a_n$ is positive or negative (at least it does for sufficiently small $\epsilon$). Finally, this holds for all $\epsilon>0$, and we're done.

Now let's show that $b\limsup a_n\leq\limsup a_nb_n$:

Take a subsequence $(a_{m_k})_{k\geq 1}$ such that $a_{m_k}\to\limsup a_n$. Then it's (presumably) known that $a_{m_k}b_{m_k}\to b\limsup a_n$. This shows that $b\limsup a_n$ is an accumulation point of $(a_nb_n)$, and as such smaller than the maximal accumulation point, i.e. $\limsup a_nb_n$.