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I was wondering if we have a formula for the following function:

$$ f(x) = \frac{1}{x^{0!}} + \frac{1}{x^{1!}} + \frac{1}{x^{2!}} + \frac{1}{x^{3!}} + ... = \sum_{n=0}^{\infty}x^{-n!} $$

(Like we have for the geometric series):

$$ \sum_{k=0}^{\infty}x^{-k} =\frac{1}{1-x} $$

Or even if we have nice values for similar functions (infinite sums that has factorials in the exponents) or if we can evaluete any value of f(x) like: $$f(e),f(2)...$$

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    $\begingroup$ No hope even using special functions (I am afraid). $\endgroup$ – Claude Leibovici Feb 18 '18 at 5:41
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    $\begingroup$ Your function is a good example of a function with a natural boundary on the unit circle in the complex plane. In that regard, it is completely unlike the geometric series. $\endgroup$ – GEdgar Feb 18 '18 at 11:03
  • $\begingroup$ regardless that there is no closed form solution, it might be interesting to consider the limiting behaviour as $x\rightarrow 0 ,1_-$ $\endgroup$ – tired Feb 18 '18 at 16:44

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