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Showing that lim_sup a_n b_n = lim a_n lim_sup b_n$

OhMyMarkov

Member
Mar 5, 2012
83
Hello everyone! :)

I'm trying to prove the following:
$a_n$ and $b_n$ are real sequences. If $a_n \rightarrow M \in R$, and $\lim \sup a_n b_n = L$ then $\lim \sup a_n b_n = \lim a_n \cdot \lim \sup b_n$.

This is what I got so far:
(1) Let $c_n = a_n b_n$, and let $C_n$ = $\sup \{c_k \; | \; k\geq n \}$. We know that $\lim C_n = L$.
Hence, for a given $\epsilon > 0, \; \exists N \in \mathbb{N} \; s.t. \; \forall n \geq N$, we have that
$|C_n - L| < \epsilon$ and thus $a_n b_n - L < \epsilon$

(2)$a_n \rightarrow M$. Let $\epsilon ' >0, \; \exists N_1 \in \mathbb{N} \; s.t. \; \forall n \geq N_1$, we have that
$M-\epsilon ' < a_n < M+\epsilon '$

(3) Combining (1) and (2), we get
$a_n b_n - L < (M+\epsilon ')b_n - L$

Now I'm stuck here, I want to show the RHS of the equation above is less the $\epsilon$, but I don't know how!

Any directions/suggestions are appreciated!
 
Last edited by a moderator:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Re: Showing that $\lim \sup a_n b_n = \lim a_n \lim \sup b_n$

Hello everyone! :)

I'm trying to prove the following:
$a_n$ and $b_n$ are real sequences. If $a_n \rightarrow M \in R$, and $\lim \sup a_n b_n = L$ then $\lim \sup a_n b_n = \lim a_n \cdot \lim \sup b_n$.
This is not true without some further restriction. For example, suppose that $a_n=-1$ for all $n$, and that $b_n = 2+(-1)^n.$ Then $\lim a_n = -1$, $\limsup b_n = 3$ and $\limsup a_nb_n = \limsup\bigl(-2-(-1)^n\bigr) = -1$. Then $$ -1 = \limsup a_n b_n \ne \lim a_n \cdot \limsup b_n = -3.$$ You can avoid this by requiring that $M>0.$ In that case, I would set about the proof like this:

Given $\varepsilon>0$, $a_nb_n<L + \varepsilon$ and $a_n>M - \varepsilon$, for all sufficiently large $n$. Thus $b_n < \dfrac{L + \varepsilon}{M - \varepsilon}$, for all sufficiently large $n$. Use that to show that $\limsup b_n \leqslant L/M.$ Next, there are infinitely many values of $n$ such that $a_nb_n>L - \varepsilon$ and $a_n<M + \varepsilon$. Thus there are infinitely many values of $n$ such that $b_n > \dfrac{L - \varepsilon}{M + \varepsilon}$. Use that to show that $\limsup b_n \geqslant L/M.$