Find radius of convergence and interval of convergence for the series

emk
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x^n/(2n-1) is the series. It starts at 1 and goes to infinity.

I did the ratio test on it and got abs.(x)

So the radius of convergence=1, and then I plugged -1 and 1 into the original series and got that they both converged. But the answer is [-1,1). Why aren't they both hard brackets?
 
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emk said:
x^n/(2n-1) is the series. It starts at 1 and goes to infinity.

I did the ratio test on it and got abs.(x)

So the radius of convergence=1, and then I plugged -1 and 1 into the original series and got that they both converged. But the answer is [-1,1). Why aren't they both hard brackets?

Pay special attention to x=1. When you plug it into your series you get ##\sum \frac{1}{2n-1}##.

You know that diverges...

Hint : Compare it to something you know diverges already.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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