# lower bound for the radius of convergence of Taylor series [duplicate]

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Radius of convergence of power series

In 4 years of studying physics I came across a lot of Taylor series. All of them converged in a disc with a radius equal to the distance to the nearest singular point. I know, that this distance is an upper bound for the radius of convergence, but is it also a lower bound? Is there some work on that for the general case? If not: what are the limitations for this to work or is there another lower bound for the radius in the general case?

A friend pointed out, that the Taylor expansion of $f(x) = e^{-\frac{1}{x^2}}$ does not converge anywhere but at $x=0$. This is not a contradiction though, because the exponential function is singular at $\infty$ and thus $f(x)$ is singular at $x=0$, right?

(A search here and via google only resulted in concrete examples, I am concerned with a general observation though)

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## marked as duplicate by example, LVK, Aang, Hagen von Eitzen, BelgiSep 30 '12 at 0:00

This question was marked as an exact duplicate of an existing question.

See this thread: math.stackexchange.com/q/46829/5363 – t.b. Apr 4 '12 at 9:38
@t.b. Thanks. Will search more thoroughly next time -.- – example Apr 4 '12 at 9:52

Your function (if defined on $\mathbf R$ and with $f(0)=0$) is smooth ($\mathcal C^\infty$) at $x=0$, and in fact everywhere. And its (zero) Taylor expansion does converge, only not to $f$.
However, every (formal) power series is the Taylor series of some smooth function $\mathbf R\to\mathbf R$ (in fact of many of them), so there is nothing one can deduce about the radius of convergence, or about any other property of the series for that matter.