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I am having doubt regarding the power series of the function $g(x) = \frac{1}{1+x^2}$. Since the function $f(x) = {1+x^2}$ is analytic everywhere and by using the theorem that if $f(x)$ is analytic everywhere and has no real zeroes then the reciprocal of the function $f(x)$ will also be everywhere analytic.
Therefore, I can conclude that the radius of convergence of the power series of $g(x) = \frac{1}{1+x^2}$ should be infinite around every point $x_{0}$ $\in$ $\mathbb{R}$ but then if I write the power series of $\frac{1}{1+x^2}=1-x^2+x^4-x^6+....$ and by using the root test the radius of convergence is coming out to be $1$.
I am confused why these two concepts are contradicting because the power series should have the radius of convergence of $\infty$, but the radius of convergence from the root test is coming to be $1$. I cannot figure out where am I doing wrong? Thanks.

Mathpdegeek497
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    An analytic function can be locally represented by a power series everywhere, and there exist analytic continuations of the Taylor series / geometric series that give you a power series for every unit disk centred around any $x_0$ – FShrike Aug 01 '21 at 13:41
  • So just to confirm, telling that the Taylor series of an everywhere analytic function has a radius of convergence infinity is wrong? – Mathpdegeek497 Aug 01 '21 at 13:57
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    Yes - not every function is a polynomial, so it follows that being able to locally represent a function by a(n) (infinite) polynomial is a local thing for an analytic function; some analytic functions' power series converge everywhere, some don't. See Laurent series and other things like that - Taylor series aren't the only series we can use – FShrike Aug 01 '21 at 14:03
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    The answers here helped me greatly on a very similar question: https://math.stackexchange.com/q/2539520/464495 – Randall Aug 01 '21 at 14:05
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    Your function is not analytic everywhere since it has singularities at $\pm i$ - in general when we talk about analytic functions, we have to look at them in the complex plane for most general properties like radius of convergence of the Taylor series etc – Conrad Aug 01 '21 at 14:12
  • @Conrad Oh I used to thought $1/(1+x^2)$ is analytic everywhere from the theorem that its reciprocal is everywhere analytic and also never zero on the "real" line. Can you tell why even we see the complex singularities when talking about the "real" analyticity of the function? – Mathpdegeek497 Aug 01 '21 at 14:23
  • If a taylor series converge at a point $x \ne 0$ it then converges absolutely on the disc $|z| <x$ so even if $x$ is real and the coefficients are real, complex numbers get involved; also most powerful properties of analytic functions come from the complex realm where Cauchy theorem holds ( any real analytic function extends by the above to a complex domain that encloses its real domain but that domain can be irregular in general) – Conrad Aug 01 '21 at 14:28

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