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.
Best Answer
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 $(2, 6)$.