# Proof of $\lim\sup(a_nb_n)\leq \lim\sup(a_n)\limsup(b_n)$

Let $a_n>0$ and $b_n\geq 0$, then $\lim\sup(a_nb_n)\leq \lim\sup(a_n)\limsup(b_n)$

My attempt at a proof is as follows. Let $A_n=\sup\{a_n, a_{n+1},...\}$, $B_n=\sup\{b_n, b_{n+1},...\}$, and $C_n=\sup\{a_nb_n, a_{n+1}b_{n+1},...\}$.

Note: $a_mb_m \leq A_nB_n$ for all $m \geq n$.

Thus $\limsup(a_nb_n)=\lim C_n \leq \lim (A_nB_n) = (\lim A_n)(\lim B_n) = (\limsup a_n)(\limsup b_n).$

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Looks good to me. What is your question? – Alex Becker Sep 17 '12 at 3:50
Is this proof even reasonable? That is, I feel like I'm missing something. – emka Sep 17 '12 at 3:52
If you want to be super picky, to say $\lim(A_nB_n)=(\lim A_n)(\lim B_n)$, you need to first show the two individual limits exist. Of course this was probably done in defining what a limsup is in the first place. – Alex R. Sep 17 '12 at 3:57
What do you mean exactly? – emka Sep 17 '12 at 4:07
the limit of products factors into the product of limits iff both individual limits $\lim A_n$ and $\lim B_n$ exist by themselves. As a contradiction, consider: $1=\lim_{n\rightarrow\infty} \frac{n}{n}\neq (\lim n)(\lim \frac{1}{n})=undetermined$ – Alex R. Sep 17 '12 at 4:16

1. You need to assume the sequences are bounded. Otherwise the statement may be even nonsensical: how to interpret $1\le 0\cdot \infty$, for example?
2. $(A_n)$ and $(B_n)$ are nonnegative nonincreasing sequences; these properties imply the existence of the limits $\lim A_n$ and $\lim B_n$, which allows to conclude about the existence and the value of $\lim (A_nB_n)$.
You could have assumed $a_n\ge 0$ instead of $a_n>0$.