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let be a positive number 'k' positive integer then let be the power series

$$ f(x)= \sum_{n=0}^{\infty} \frac{(-x)^{n}}{(n!)^k} $$

  • if $k=0 $ we have $ (1+x)^{-1} $

  • if $k=1$ we have $ \exp(-x) $

  • if $k=2$ we have the Bessel function $ J_{0} (2\sqrt{x}) $

but what happens for $ k \ge 3 $

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    For $k\to\infty$, $f(x)\to1-x$ :) – 2017-01-08
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    i suppose you assume $ 1^{\infty} =1 $ but this doesn't hold always – 2017-01-08
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    hem, $\lim_{k\to\infty}1^k=1$. Always. – 2017-01-08
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    The function seems to be alternating (forever ?), with the lobes wider and wider, higher and higher as $k$ grows. This is how it tends to $1-x$. – 2017-01-08

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