# Power series

1. Dec 2, 2013

### Jbreezy

1. The problem statement, all variables and given/known data
Find the radius of convergence and the interval of convergence

2. Relevant equations

A_n = Ʃ sum n =1 to infinity [((-1)^n) x^(2n+1)]/(2n+1)!

3. The attempt at a solution
All I thought was to use the ratio test so I did A_(n+1) /A_n

= ((x^(2n+1))/(2n+1)!) ( (2n+1)!)/ [x^(2n+1)]
I got |x^2| limit n ---> |x^2 / 4n^2 +10n +6 | when I simplified and such. So the limit is equal to 0 so my interval of convergence is 0 ? and R = 0?

2. Dec 2, 2013

### Staff: Mentor

If the limit of A_(n+1)/A_n is zero, the radius of convergence is not 0. Check how the ratio test works.

3. Dec 2, 2013

### Yosty22

It is true that the limit is 0. However, this does not mean that the radius of convergence is 0. You did the more difficult part correctly, take a other look at the definition of ratio test and how to find radius of convergence from that.

4. Dec 2, 2013

### Jbreezy

Less then one. Converges right? Radius of convergence umm |x^2| < 1 so x < +-1?
No?

5. Dec 2, 2013

### Yosty22

Not quite. Are you familiar with a simple equation relating limit and radius of convergence?

6. Dec 2, 2013

### Jbreezy

No. It converges for all values of x so radius of convergence is infinity for all x? What is the equation?

7. Dec 2, 2013

### Yosty22

That's correct. It does indeed converge for all values of x. And the small equation I was referring to is R (radius of convergence) = 1/L (the limit). You assume that 1/infinity = 0 and 1/0 = infinity. So in this case, your answer for the limit is 0, so 1/0 = Infinity which implies an infinite radius of convergence.

8. Dec 2, 2013

### Jbreezy

Question when using the ratio test for power series and your limit is equal to 1 then you just...like this

|x| limit n --> (n^2 +1)/(n^2 +2) = |x| (1) then you say

You need |x| <1 and go from there? Even though your limit is one you can still use it although with the ratio test if you get a limit of 1 it is inclusive?

9. Dec 2, 2013

### Yosty22

Yes. What you can say is if the limit is 1, the radius is also one, so you can find the interval by adding and subtracting 1 from the center, even though the ratio test for convergence is I conclusive. That is because in this case, you aren't looking for the limit (which would be an inconclusive convergence test). Rather, you want 1/L, so it doesn't matter in this case that the ratio test for convergence is inconclusive.

10. Dec 2, 2013

### Jbreezy

11. Dec 2, 2013

### Yosty22

I am currently in the same class (i assume integral calculus) and that is what we were taught.

12. Dec 2, 2013

### Staff: Mentor

It is true, and you can prove it. For a series around the origin (everything else is just more to type, it does not change anything):

Assume the ratio of $\displaystyle \frac{|A_{n+1}|}{|A_n|}$ approaches a limit of L.

Consider an arbitrary value of x with |x|<1/L.
Then the ratio $\displaystyle \frac{|x^{n+1}A_{n+1}|}{|x^n A_n|} = |x| \frac{|A_{n+1}|}{|A_n|}$ approaches |x|*L < 1. Therefore, the series converges at that point via the regular ratio tests for series.

Consider an arbitrary value of x with |x|>1/L.
Then the ratio $\displaystyle \frac{|x^{n+1}A_{n+1}|}{|x^n A_n|} = |x| \frac{|A_{n+1}|}{|A_n|}$ approaches |x|*L > 1. Therefore, the series does not converge at that point via the regular ratio tests for series.

Therefore, the radius of convergence is R=1/L.

With the definitions 1/0=infinity, 1/infinity=0 this proof works for the two special cases as well.