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$\{a_{n}\}$,$\{b_{n}\}$ are two bounded sequences.How can we prove that $\limsup(a_{n}b_{n})$ = $\lim(a_{n})\limsup(b_{n})$. Is it equal to $\lim(a_{n})\lim(b_{n})$?

If the sequences are not convergent but bounded (for example $a_{n} =(-1)^{n}$ and $b_{n}=(-1)^{n-1}$) How can this result hold.

I have seen the proof for $\limsup(a_{n}b_{n}) \leq \limsup(a_{n}) \limsup(b_{n})$. Can we say that $\limsup(a_{n})\leq \lim(a_{n})$ ?

Martin Sleziak
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Madhu
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  • If ${a_n}$ is only assumed to be bounded, how do you define $\lim(a_n)$?

    Either way, this should answer your question: http://math.stackexchange.com/questions/113121/lim-sup-inequality-limsup-a-n-b-n-leq-limsup-a-n-limsup-b-n

     – Ian May 01 '14 at 02:41
  • Do you know when the limit exists? – Mhenni Benghorbal May 01 '14 at 02:47

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In general $\limsup a_nb_n=\limsup a_n\limsup b_n$ does not hold. Just consider the sequences $a_n=(1,0,1,0,\dots)$ and $b_n=(0,1,0,1,\dots)$. In this case $\limsup a_n=\limsup b_n=1$ and $\limsup a_nb_n=0$.

However, if you know that a sequence $(x_n)$ is convergent, then $\limsup x_n=\lim x_n$. So if both $(a_n)$ and $(b_n$) are convergent, then $$\limsup a_n\limsup b_n = \lim a_n \lim b_n = \lim (a_nb_n) = \limsup (a_nb_n).$$

If you only know that $b_n$ is convergent and $a_n,b_n\ge 0$, then you have $$\limsup (a_nb_n)=\limsup a_n\limsup b_n = \limsup a_n\lim b_n.$$

There are already several posts concerning this equality:

It is also mentioned in the post you linked to: lim sup inequality $\limsup ( a_n b_n ) \leq \limsup a_n \limsup b_n $

You can also obtain this from a more general result that $$\limsup a_n \liminf b_n \le \limsup (a_nb_n) \le \limsup a_n \limsup b_n$$ which holds if you assume that both $a_n,b_n\ge 0$. (Together with using that $\lim b_n=\liminf b_n=\limsup b_n$, if the sequence $(b_n)$ is convergent.)

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Martin Sleziak
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  • If in my question it is given that limit$a_{n} \geq 0 $ can we say that $limitsup (a_{n}b_{n}) = lim(a_{n})lim(b_{n}) $ – Madhu May 03 '14 at 06:11
  • You can say that $\limsup(a_nb_n)=\lim a_n \limsup b_n$. (This is shown in the first question I linked.) You cannot use $\lim b_n$ unless you know, that the limit exists. – Martin Sleziak May 03 '14 at 06:41