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Is there a periodic function $f:\mathbb{R}\to\mathbb{R}$ with period $2\pi$ such that $f\in C^\infty$(i.e. $f$ is infinitely differentiable) and $\sum\limits_{i=0}^\infty\frac{f^{(i)}(0)}{i!}$ does not converge?

My idea is that since there is an infinitely differentiable function $g$ such that \begin{align*} g(x)=\begin{cases} 1, & x\in[a,b],\\ 0, & x\in(-\infty,a']\cup[b',\infty), \\ (between\ 0\ and\ 1), & otherwise, \\ \end{cases} \end{align*} Where $a'b$, I can find some $a$,$a'$,$b$,$b'$ and multiply g with $\frac{1}{1-x}$ and extend the product to $\mathbb{R}$. Is such construction legal?

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    It is unclear. $\sum\limits_{k=0}^\infty\frac{f^{(k)}(0)}{k!}$ diverges means that $f$ can be analytic but its Taylor series at $x=0$ has a radius of convergence $< 1$. So that $f(x) = \frac{1}{x+i/2}$ works, and $F(x) =\sum_{n-\infty}^\infty f(x-n) e^{-(x-n)^2}$ is periodic. See also [a smooth function which is nowhere real analytic](https://en.wikipedia.org/wiki/Non-analytic_smooth_function#A_smooth_function_which_is_nowhere_real_analytic) – 2017-02-26
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    What does $f(x)=\frac{1}{x+\frac{i}{2}}$ mean? :O – 2017-02-26
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    A function with a pole at $-i/2$ (where $i$ is the imaginary unit) – 2017-02-26
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    Yes, such function works (and there is a typo around the $\sum$ sign I guess). The only problem is that I wish to find a real-valued function. Is that possible? – 2017-02-27
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    $f(x) = \frac{1}{x^2+1/4}=\frac{1}{(x+i/2)(x-i/2)}, F(x) = \sum_{n=-\infty}^\infty f(x-n) e^{-(x-n)^2}$ – 2017-02-27

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