Prove that $\limsup\left(b_{n}a_{n}\right)=\limsup\left(a_{n}\right) $ when $\lim_{n\rightarrow\infty}\left(b_{n}\right)=1$

Question

I'm having trouble with this homework question:

"Let there be a sequence $\left(b_{n}\right)_{n=1}^{\infty} $ such that $\lim_{n\rightarrow\infty}\left(b_{n}\right)=1$. Let there also be some bounded sequence $\left(a_{n}\right)_{n=1}^{\infty}$

Prove that the following statement is correct: $\limsup\left(b_{n}a_{n}\right)=\limsup\left(a_{n}\right) $"

Alright so I sat a good two hours on this question with no major breakthroughs as of yet. I have some semi-proofs all of which use some arbitrary assumptions which I can't make.

Any hints?

Thank you!

Answer 1

Assume $c>\limsup a_n$. Then there is $c'<c$ such that almost all $a_n<c'$ (e.g. $c'=\frac{c+\limsup a_n}2$). Also, almost all $b_n$ are positive. For what values of $b_n>0$ can it occur that $a_n<c'$ but $a_nb_n\ge c$?

If $c>0$, this needs $a_n>0$. Hence we have $c>c'>a_n>0$ and we obtain $b_n\ge \frac c{a_n} >\frac{c}{c'}$. As $\frac c{c'}>1$, this holds only for finitely many $n$.

And if $c\le 0$, we have $a_n<c'<c\le 0$ and $b_n\le \frac c{a_n}\le \frac c{c'}$. As now $\frac c{c'}<1$, this holds only for finitely many $n$ - again.

We conclude that $a_nb_n<c$ for almost all $n$ and hence $\limsup b_na_n\le c$. As a consequence $\limsup b_na_n\le \limsup a_n$. Since $b_n\to 1$ also implies $\frac1{b_n}\to 1$, we obtain $\limsup a_n\le \limsup b_na_n$ by the same argument.

Answer 2

First note that as there is some $K > 0$ such that $-K < a_i < K$ for all $i \in \mathbb{N}$ you know that $\lim \sup_{n \rightarrow \infty}a_n=R \in \mathbb{R}$. In particular you can restrict yourself to sequences $q_n$ which converge against $R$ from below to show the statement.

Now we want to show that $\lim_{n \rightarrow \infty}b_n q_n=\lim_{n\rightarrow \infty}q_n$. But $\lim_{n\rightarrow \infty}q_n=1\cdot \lim_{n\rightarrow \infty}q_n = \lim_{n\rightarrow \infty}b_n \cdot\lim_{n\rightarrow \infty}q_n$.

Now use the theorem that if you have two convergent sequences you can merge the limits.