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I understand how the answer key took the root test to determine the interval of convergence. I'm just wondering, would it also be okay to state that it's a geometric series that converges for

abs ((x-2)/n) < 1, and then solve for abs (x). can you do this if the counter starts at 1? because when we test convergence, the counter doesn't matter, right? the counter only matters when we are asked to find the sum, right?

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Please state the question (briefly).

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do you need to use the root test or ration test to determine the interval of convergence? or can you just use the fact that it's a geometric series?

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When I said "state the question", I was referring to 12.8 #19. Here it is: Radius, interval of convergence of Sigma [(x-2)^{n}]/(n^{n}).

Now, to answer your query: your argument has a large gap or error. The geometric series is of the type Sigma (r^{n}), where r is a fixed number that does not depend on n. So, the series in #19 is NOT a geometric series, since we cannot take r =(x-2)/n because it does depend on n.

In general, please state briefly the problem (from the book or elsewhere) you are referring to.