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

  • 5
    Looks good to me. What is your question?2012-09-17
  • 0
    Is this proof even reasonable? That is, I feel like I'm missing something.2012-09-17
  • 2
    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.2012-09-17
  • 0
    What do you mean exactly?2012-09-17
  • 0
    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$2012-09-17
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    The missing argument is that $(A_n)$ and $(B_n)$ are nonnegative (here) nonincreasing (always) sequences.2014-05-01
  • 0
    See also [lim sup inequality $\limsup ( a_n b_n ) \leq \limsup a_n \limsup b_n $](http://math.stackexchange.com/q/113121)2017-03-03

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