Radius of Divergence: Find R & Interval of Convergence

[Abner]

Homework Statement

I have this problem to consider the power series,
[itex]\sum_{n=1}^{\infty}\frac{(-4)^{n}}{\sqrt{n}}(x+4)^{n}[/itex]
So, i need to find the [itex]R[/itex] and interval of convergence.

Homework Equations

The Attempt at a Solution

This is what i did:
[itex]\lim_{n\rightarrow \infty} {\frac{(-4)^{n+1}(x+4)^{n+1}}{\sqrt{n+1}}}\frac{\sqrt{n}}{(-4)^{n}(x+4)^{n}}[/itex]

and this is what i get after i finished calculating for [itex] R [/itex] [itex]= 4|x+4|[/itex] [itex]\rightarrow R = 1/4[/itex] and the interverval for convergence [itex]= (-17/4, -15/4)[/itex]

When i submitted this answer into webwork, the system said it was wrong. So, can somebody please guide me to the correct path of calculating this question please.

 

[STEMucator]

Your answer looks correct to me.

 

[Abner]

Yes, the answer is correct but i just noticed that i enter the interval in the wrong notations. It supposed to be [itex](-17/4,-15/4][/itex]. I just don't understand why one interval is open, and the other one is closed.

 

[Panphobia]

When you use the Ratio test for interval of convergence, you have to check the end points, this is because the test is inconclusive when the limit is 1.

 

[mfb]

You have to check the borders separately. If you do that, you'll see one gives a convergent series, the other one does not.

Edit: Didn't see Panphobia's post before.

 

[Abner]

So, when one point is convergent we use the closed interval, and open if it diverges?

 

[Panphobia]

Yes, because it is included in the interval.

 

[Abner]

ok that makes sense. Thanks for the replies and the help.

 

[Mark44]

Yes, the answer is correct but i just noticed that i enter the interval in the wrong notations. It supposed to be [itex](-17/4,-15/4][/itex]. I just don't understand why one interval is open, and the other one is closed.

Technical point. (-17/4, -15/4] is one interval. The two numbers are endpoints of this interval, not intervals themselves.