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Say we are given the simple power series $$\sum_{i=0}^{\infty}(-1)^k\frac{(x-4)^k}{2^k}$$

The interval of convergence can easily be shown to be $x\in(2,6)$ using the Root Test, and, since absolute convergence implies conditional convergence, all is well. But what about points that only converge conditionally? We must take the absolute value of our expression before starting the Root Test, as it only works for positive monotonically decreasing sequences. So then, how are those points "accounted for"? The points that make our series converge because it has $(-1)^k$ and diverge when it doesn't? Wouldn't they be beyond our interval, as those (the points in the interval $(2,6)$) are the only points that make the un-alternating series converge? I hope I explained that in enough detail.

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I don't quite follow what you're asking. Power series always converge on some open interval (or at a single point) and diverge outside of that interval except possibly at the endpoints. So once you've established that the radius of convergence is $r=2$ so the series converges on $(2,6)$, it must diverge on $(-\infty,2)\cup(6,\infty)$. The points in question are $x=2$ and $x=6$. Plugging in those values for $x$ yield obviously divergent series. –  Bill Cook Nov 1 '11 at 1:44
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The root test shows that the radius of convergence is 2. It tells you the interval of convergence is AT LEAST $(2, 6)$. On that interval, you have absolute convergence. You then must check the endpoints separately.

Since you are looking at this series when $x=2$, plug in $x=2$. You have $$\sum_{i=0}^\infty (-1)^k \frac{(-2)^k}{2^k} = \sum_{i=0}^\infty 1.$$ and this diverges because the $n$th term does not go to 0. So, we don't have convergence at all when $x=2$. Now, test when $x=6$ by plugging in $x=6$. We have $$\sum_{i=0}^\infty (-1)^k \frac{(2)^k}{2^k} = \sum_{i=0}^\infty (-1)^k.$$ Again, we have divergence because the $n$th term does not go to $0$. Therefore, we can now conclude our interval of convergence is exactly $(2, 6)$.

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@Grapthth: Yes, but what I meant is, how do we know that the domain of the function $$f(x)=\sum_{k=0}^{\inf}(-1)^k\frac{(x-4)^k}{2^k}$$ is the same as the domain of the function $$g(x)=\sum_{k=0}^{\inf}\frac{(x-4)^k}{2^k}$$? –  Hautdesert Nov 1 '11 at 2:46
    
@Hautdesert: The domains of both $f$ and $g$ are guaranteed to include $(2,6)$ and to exclude $(-\infty,2)\cup(6,\infty)$; the only places where they can possibly differ are $2$ and $6$. In this case both the positive and the alternating series diverge at both $2$ and $6$, so both series have domain $(2,6)$. –  Brian M. Scott Nov 1 '11 at 4:08
    
@Brian M. Scott: What guarantees this? –  Hautdesert Nov 1 '11 at 4:34
    
@Hautdesert: The root test, in this example; in others, the ratio test. Both of those tests give an open interval of absolute convergence, and both guarantee that outside of the corresponding closed interval the terms of the series fail to converge to $0$, so it will diverge regardless of the signs of the terms. The only points about which these tests give no information are the endpoints of the interval; that’s why they have to be tested individually. And it might happen that the positive series converges at one of them while the alternating series does not. –  Brian M. Scott Nov 1 '11 at 4:46
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