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Let $B(x_{0},R) $ designate the open ball of center $x_{0}$ and radius $R>0$ in $\mathbb{R}^{k}$ ($k\geq1$). Let $f$ and $g$ be two real analytic functions on a neighborhood of the closure of $B(x_{0},R) $. Suppose we have $$0\leq f(x)\leq g(x)$$ on the whole of $B(x_{0},R) $. Now, if the radius of convergence of the Taylor's series of $g$ is $R$, can we conclude that the radius of convergence for the Taylor's series of $f$ is also $R$?

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Let $n=1$, $x_0=0$, $R=1$ and $$ f(x)=\frac{1}{1+4\,x^2},\quad g(x)=\frac{1}{1+x^2}. $$ Both $f$ and $g$ are real analytic on $\Bbb R$ and $f(x)\le g(x)$ for all $x$. The radius of convergence of $g$ around $0$ is $1$, but the radius of convergence of $f$ around $0$ is $1/2$.

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  • $\begingroup$ Thank you Julian Aguirre. $\endgroup$ – M. Rahmat Oct 20 '18 at 4:57

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