# Points around which one expands and the radiuses of convergence

I'm trying to make sense of the following passage:

Let $f(x)=\frac{1}{x+1}$ and $R_0$ the radius of convergence of the Taylor series of $f$ around $x_0=0$, analogously: $R_1$ — around $x_1=\frac{1}{2}$. According to the theorem: $R_1 \geq R_0-\frac{1}{2}$.

My expansions:

\begin{align} T_{x_0=0}(x) &= \sum_{n=0}^{\infty}(-1)^n x^n \ \ \ \ \ \mbox{ (so $R_0=1$)} \\ T_{x_1=\frac{1}{2}}(x) &=\sum_{n=0}^{\infty}(-1)^n \Big(\frac{2}{3}\Big)^{n+1} (x-\frac{1}{2})^n \ \ \ \ \ \mbox{ (so $R_1=\frac{3}{2}$)} \end{align}

Generally, for the series of $f$ around $x_2: R_2=|x_2+1|$

What might be the general statement of the theorem they are referring to?

Why did we get "strictly greater" here $(\frac{3}{2}>1-\frac{1}{2})$? When would there be an equality $(R_1 = R_0-\frac{1}{2})$? Does it just depend on whether $x_1<-1$ or $x_1>-1$?

Any help is much appreciated.

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The general theorem I have in mind is the following: The radius of convergence of a power series around a point $p$ is equal to the the radius of the largest disk centered at $p$ on which the function has an analytic extension.
Let $R_0$ be the radius of convergence at $x_0$, and let $x_1$ be another point. Then the ball of radius $(R_0 - |x_0 - x_1|)$ centered at $x_1$ is contained inside the ball of radius $R_0$ centered at $x_0$ (by the triangle inequality). (Of course this radius may not even be positive, but then the statement is vacuous.)
Since the function extends to be analytic on $B(x_0, R_0)$, it also extends to be analytic on the subset $B(x_1, R_0 - |x_0 - x_1|)$. Therefore, if $R_1$ is the radius of convergence at $x_1$, then $$R_1 \geq R_0 - |x_0 - x_1|$$ which is exactly what you have quoted.
This inequality can be strict, as your calculation shows, because it certainly might be the case that the function extends to be analytic on a larger ball around $x_1$. (Draw a picture!)