No matter the function, it seems the interval of convergence of the Maclaurin series is always centered at $0$. I have found that the interval can be open on one side and closed on the other, but that seems to be the only possible difference between the two sides. It is not clear to me why this should be the case, especially since the addition of each term often seems to effect each side of the function differently. (e.g. for $\frac1{1-x}$, the left side oscillates while the right does not, but both sides converge within the same radius.)

How can this be proven?

Edit: More generally, why is the interval of convergence of a Taylor series centered about the point the Taylor series is centered about? It is not obvious to me why the two sides of the interval necessarily have the same length.

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    $\begingroup$ Well, the Maclaurin series itself is always center on $0$. $\endgroup$ – David G. Stork Aug 22 at 0:30
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    $\begingroup$ If it's centred anywhere else, it's a Taylor series. $\endgroup$ – Angina Seng Aug 22 at 1:50
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    $\begingroup$ @AnginaSeng I conjecture that the intended question was not why the series is centered at $0$, i.e., uses powers of $x$ rather than of some $x-a$, but rather why the interval of convergence is, except for endpoints, symmetric about that center. $\endgroup$ – Andreas Blass Aug 22 at 2:20
  • $\begingroup$ @AndreasBlass Yes, that is what I am asking. Thanks for finding a more clear way to say it. $\endgroup$ – Polygon Aug 22 at 3:01

The root test implies that $\sum_{j=0}^{\infty}a_j \cdot (x-x_0)^j$ converges when $r < 1$ and diverges when $r>1,$ where $r=\lim (|a_j|^{\frac{1}{j}} \cdot |x-x_0|)$. So, we can be sure that when $|x-x_0|<\lim |a_j|^{\frac{-1}{j}}$ the series converges and when $|x-x_0|>\lim |a_j|^{\frac{-1}{j}}$ it diverges. It is not hard to convince yourself that the region $|x-x_0|<\lim |a_j|^{\frac{-1}{j}}$ is symmetrical in relation to $x_0$, which answers your question.

It is interesting to add, however, that the root test doesn't say anything about the case $r=1$. So we know nothing about the behavior at $|x-x_0| = \lim |a_j|^{\frac{-1}{j}}$ (i.e, at the points $x = x_0 + \lim |a_j|^{\frac{-1}{j}}$ and $x = x_0 - \lim |a_j|^{\frac{-1}{j}}$). There isn't always simetry at those points, it can sometimes converge at one and diverge at the other.

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    $\begingroup$ Quibbles: perhaps it should be "it is easy to convince yourself that" (and no double "that"). And either "can say nothing" or "can't say anything"... Such small things can sometimes confuse questioners who are confused in the first place, hence asking the questions. $\endgroup$ – paul garrett Aug 22 at 20:30
  • $\begingroup$ @paulgarrett, is it better now? $\endgroup$ – Anderson Brasil Aug 22 at 21:18
  • $\begingroup$ Yes, I think better. :) $\endgroup$ – paul garrett Aug 22 at 21:42

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