i need to proove that lim inf(An*Bn)>=liminfAn * liminfBn
i got the opposite resolt
where is my mistake:
http://img187.imageshack.us/img187/6705/44112683ho1.gif
i need to proove that lim inf(An*Bn)>=liminfAn * liminfBn
i got the opposite resolt
where is my mistake:
http://img187.imageshack.us/img187/6705/44112683ho1.gif
Here's an example where the inequality is strict. You can test your argument against this example, to pinpoint where you went wrong.
Let $\displaystyle a_n = 2+(-1)^n$, $\displaystyle b_n = 2-(-1)^n$. Thus a_n is alternately 1 and 3, b_n is alternately 3 and 1. The lim inf for both sequences is 1. But when you multiply the sequences together you get a sequence where each term is 3. So the lim inf of the product is 3.
So $\displaystyle \liminf_{n\to\infty}a_nb_n = 3 > 1 = \liminf_{n\to\infty}a_n\liminf_{n\to\infty}b_n$.
i cant see the mistake
i am done this proof strictly on this definition
http://img515.imageshack.us/img515/5666/47016823jz1.gif
why its not true??
in each of the following sequence our temporary inf
gets bigger or stays the same
so the function of the inf's goes up
and their limit is the least upper bound(b)
and the least upper bound is bigger or equal than every member
in the sequence.
can you tell me whats wrong with what i said?
ok the limit is bigger or equal then any temporary inf of this sequence.
so what could i say the relations of the lim inf with the members of the sequence.
here i was given a solution
http://www.mathhelpforum.com/math-he...-question.html
and i was told that liminf is smaller than any temporary inf.
how he came to this conclusion??