Ratio simplification using ratio series.

[rcmango]

Homework Statement

here is the problem: http://img155.imageshack.us/img155/5175/15399391yy7.png

how do i simplify the fraction?

Homework Equations

The Attempt at a Solution

i got the correct answer but I'm not sure how to simplify the fraction correctly, especially with the n+1 with a exponent 4.

please help.

 

[Defennder]

Have you learned L'Hospital's rule yet? If you have then all you need to do is to evaluate \lim_{n\to \infty} \frac{n^4+16}{(n+1)^4 +16} separately, then put that result back into the expression then you'll get the answer.

 

[rcmango]

Okay, you after applying l'hospital's rule, i get 1. also, i believe i could change this (n+1)^4 to this n^4 + 1^4 i believe?

thanks.

 

[Defennder]

I don't think that's allowed. What rule are you following there?

 

[VietDao29]

Okay, you after applying l'hospital's rule, i get 1. also, i believe i could change this (n+1)^4 to this n^4 + 1^4 i believe?

thanks.

No, it's not correct at all.

Here's a simple counter example. If n = -1, then your LHS will be 0, whereas your RHS is 2, they are not equal.

To expand the terms that have the form: (a + b)ⁿ (where n is a natural number), one should use Binomial Theorem.

Or, you can just divide both numerator, and denominator by n⁴ (the greatest power), like this:

\lim_{n \rightarrow \infty} \left| (2x - 1) \frac{n ^ 4 + 16}{(n + 1) ^ 4 + 16} \right|

= |2x - 1| \lim_{n \rightarrow \infty} \left| \frac{\frac{n ^ 4 + 16}{n ^ 4}}{\frac{(n + 1) ^ 4 + 16}{n ^ 4}} \right| (since 2x - 1 is a constant, independent of n, we can "pull" it out)

= |2x - 1| \lim_{n \rightarrow \infty} \left| \frac{1 + \frac{16}{n ^ 4}}{\frac{(n + 1) ^ 4}{n ^ 4} + \frac{16}{n ^ 4}} \right|

= |2x - 1| \lim_{n \rightarrow \infty} \left| \frac{1 + \frac{16}{n ^ 4}}{\left( \frac{n + 1}{n} \right) ^ 4 + \frac{16}{n ^ 4}} \right|

= |2x - 1| \lim_{n \rightarrow \infty} \left| \frac{1 + \frac{16}{n ^ 4}}{\left( \frac{n + 1}{n} \right) ^ 4 + \frac{16}{n ^ 4}} \right| = ...

Can you go from here? :)

 

[rcmango]

Viet Dao, thankyou for that thorough explanation! That was exactly my question, and it helped me a great deal!

 

[rcmango]

just skipping to the end, i get the interval of convergence to be [0, 1] where i plug these into the original equation and they both converge, correct? just need a confirmation.

also the radius of convergence came to 1/2

thanks.

 

[VietDao29]

just skipping to the end, i get the interval of convergence to be [0, 1] where i plug these into the original equation and they both converge, correct? just need a confirmation.

also the radius of convergence came to 1/2

thanks.

What is the series you are working on? You haven't shown us any series at all.