TITLE: Taylor series convergence for $e^{-1/x^2}$
QUESTION [5 upvotes]: Consider the Taylor series for $e^{-1/x^2}$ around $0$:
$$e^{-1/x^2}=1-\dfrac{1}{x^2}+\dfrac{1}{2!x^4}-\dfrac{1}{3!x^6}+\ldots$$
For which $x$ does the series on the right converge to $e^{-1/x^2}$?

REPLY [4 votes]: For which $y$ does $e^y = 1+y+\frac{y^2}{2!} + ...$ converge? Set $-\frac{1}{x^2} = y$ and you can find the corresponding values of $x$ which allows your expansion to converge.