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clooneyisagenius

Member

Registered: 2007-03-25

Posts: 56

Help reducing complicated fractions (with limits? maybe?)

I have that

and

for the two following:

Now I can plug these into the equations and I get:
(for

)

(for

)

Okay, so now my real task is to take the limit of these r(n) and p(n) as n goes to infinity. Should i simplify these, or is there an easier way to do this? THANKS FOR READING AND ANY HELP!

clooneyisagenius

Member

Registered: 2007-03-25

Posts: 56

Re: Help reducing complicated fractions (with limits? maybe?)

Okay, new question (similar i suppose)... If i calculated the limits of the r(n) and p(n) from above using Mathematica and I get an interval, what exactly does that mean?

For

r(n) gives no response
p(n) gives interval from 1/2 to 2.

For

r(n) gives interval from 0 to infinity.
p(n) gives interval from 1/3 to 1/2.

Last edited by clooneyisagenius (2008-03-18 17:10:52)

TheDude

Member

Registered: 2007-10-23

Posts: 361

Re: Help reducing complicated fractions (with limits? maybe?)

Let's start with the sequence a(n).  Notice that for any n, a(n) will be either (4/2)^(-n) = 2^(-n) or (6/2)^(-n) = 3^(-n).  This means that for any n,

Now, note that starting at n = 2, 3^n > 2^(n + 1).  This means that in cases where the numerator of r(n) is 3^(-(n + 1)) r(n) is less than 1, whereas when the numerator is 2^(-(n + 1)) r(n) is greater than 1.  This all means that r(n) has no limit for the sequence a(n), so Mathematica was right.

As for p(n), I think you switched your a(n) and your b(n).  The limit of p(n) using the series a(n) once again doesn't exist.  It will continually flip between 1/2 and 1/3.  This must have been what Mathematica was trying to tell you with the interval.

Use a similar trick when working on b(n).  For any n, b(n) = 2^n or b(n) = 2^(-n).  You can see from this that the function r(n) will alternate between approaching 0 or approaching infinity, depending on which version of b(n) is in the numerator, and you can also see that p(n) will alernate between 1/2 and 2.

These proofs are far from rigorous so you still have quite a bit of proving to do, but this should get you started.

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