Find the interval of convergence.

[Sabricd]

Homework Statement

\sum n^(1/2)*x^n where n=1 and goes to infinity.

Sorry, I'm new at this. I was kind of hoping to get help finding the interval of convergence.
After using the Ratio Test I got for an answer the absolute value of x. I know I have to prove whether or not the endpoints (-1,1) converge or not. Can I just treat it as an alternating series and apply its absolute value and test for divergence?

 

[micromass]

So basically, you'll need to check if the series

\sum{\sqrt{n}}~\text{and}~\sum{(-1)^n\sqrt{n}}

converge. Does the sequence \sqrt{n} converge to 0? What can you conclude from this?

 

[Sabricd]

Well if I take the limit of it and if the limit = 0 then it is convergent. If not, it is divergent, correct?

 

[Sabricd]

So I could basically take the absolute value and say that it diverges by the Test of Divergence...right?

 

[micromass]

You can only say, if it doesn't converge to 0, then our serie is divergent.
If it does converge to 0, then you have to apply another test.

 

[micromass]

So I could basically take the absolute value and say that it diverges by the Test of Divergence...right?

right!

 

[Sabricd]

Thank you!

 

[Sabricd]

How about \sum x^n/n^3 where n=1 and goes to infinity. I know the radius for convergence is 1. However, I have to prove whether the endpoints (-1, 1) converge. Hence, would the answer be no, it diverges since x does not have a limit?

 

[micromass]

You'll need to see whether the series

\sum{\frac{1}{n^3}}~\text{and}~\sum{\frac{(-1)^n}{n^3}}

These series will both converge (absolutely)...

 

[Sabricd]

Yes! p series! :)