Find radius of convergence and interval of convergence for the series

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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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  • #2
STEMucator
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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?
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.
 

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