# Find the interval of convergence.

## 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?

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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?

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

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

You can only say, if it doesnt converge to 0, then our serie is divergent.
If it does converge to 0, then you have to apply another test.

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

Thank you!

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?

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)...

Yes!! p series!!! :)