borel cantelli application

[claire...@gmail.com]

so given lim n-> infinity, X_n / Y_n = 1 ,then sum(X_n) < infinty
i.f.f. sum(Y_n) < infinity. To go about showing this, it must be the
case that the Probability ( | (X_n/Y_n) -1 | > 1 infinitely often )
must be 0. By Borel Cantelli, this must mean that sum of the
probablities for the individual terms must be less than infinity. How
do I proceed from here, I can set up some sort of bound for the
summation which must be less than infinity but how does the violation
of " i.f.f. sum(Y_n) < infinity" disrupt the limit.

[Robert Israel]

claire...@gmail.com writes:

For this to make sense, X_n >= 0 and Y_n >= 0. I assume your
lim_{n->infty} X_n/Y_n = 1 is meant to be true almost surely. Then
it is indeed true that Prob(|X_n/Y_n - 1| > 1 infinitely often) = 0,
but it's better to use, say, Prob(|X_n/Y_n - 1| > 1/2 infinitely often).
In particular, with probability 1, 2/3 X_n < Y_n < 2 X_n for all
sufficiently large n, and when this is true sum_n X_n < infty iff
sum_n Y_n < infinity. There's no need to use Borel-Cantelli.
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

[claire...@gmail.com]

On Oct 5, 11:04 pm, Robert Israel
<isr...@math.MyUniversitysInitials.ca> wrote:

Thanks for your help Dr. Israel. I was trying to show that the Prob(|
X_n/Y_n - 1| > 1/2 infinitely often) is true for any 1/n but used the
same subscript 'n' and ran into some trouble. After realizing it
should just be a fixed constant k, everything worked out fine. I
guess I just need to look more carefully at when b-c is really
necessary, especially after doing a problem on finding bounds for the
cdf using b-c lemma. Thanks again!