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I wanna show that for $f:(0,\infty)\rightarrow\mathbb R, x\mapsto\exp(-\frac1{x^2})$ the sum of the derivates of $f$, so $\sum\limits_{n=0}^\infty f^{(n)}(x)$, converges to $0$, so $$\lim\limits_{x\rightarrow0}\sum\limits_{n=0}^\infty f^{(n)}(x)=0$$

It's intuitively clear but I have some issues on writting it down. Can anybody help? Thanks!

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    The individual terms go to $0$ as $x \to 0$, but I don't think it's at all clear that the infinite series converges for any $x > 0$.2012-06-20
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    Formally, $\sum_{n=0}^\infty f^{(n)}(x) = \int_0^\infty e^{-t} f(x+t)\ dt$, and as $x \to 0+$ that should go to $\int_0^\infty e^{-t} e^{-1/t^2}\ dt$ which is not $0$.2012-06-20
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    Note that, by Taylor's Theorem with Lagrange remainder, for any $\epsilon > 0$ and any positive integer $n$ there is $t_n$ with $0 < t_n < \epsilon$ and $f^{(n)}(t_n) = n! \epsilon^{-n} f(\epsilon)$. As $n \to \infty$, this goes to $+\infty$.2012-06-20

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