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So I was messing around and realized that the geometric series and exponential series are really similar when written as a power series.

$$\sum_{n=0}^\infty \frac{x^n}{(n!)^0}=\sum_{n=0}^\infty {x^n}=\frac{1}{1-x}$$ $$\sum_{n=0}^\infty \frac{x^n}{(n!)^1}=\sum_{n=0}^\infty \frac{x^n}{n!}=e^x$$

By adding a simple exponential in to the denominator of the power series, you can switch between the two depending on if it is a $1$ or a $0$. That leads to the question of, what would it be if it was a different value.

$$\sum_{n=0}^\infty \frac{x^n}{(n!)^a}= ???$$

What happens when $a$ is $.5$, or something else like $2$.

user1447952
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    At least when $a=2$ you get the Bessel function $I_0$. – A rural reader Mar 07 '24 at 20:51
  • Related: https://math.stackexchange.com/q/2902724/442 and https://math.stackexchange.com/q/2866642/442 – GEdgar Mar 07 '24 at 21:03
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    And also since the geometric series has radius of convergence $R = 1$ and the exponential series has $R = \infty$, what happens in between? At what threshold value of $a$ does $R$ become infinite? – Sammy Black Mar 07 '24 at 22:47
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    @SammyBlack: With Stirling’s approximation of the factorial, the general term becomes

    $$ (2\pi n)^{-\frac a2}\left(x\left(\frac{\mathrm e}n\right)^a\right)^n;. $$

    For any $a\gt0$ the value in parentheses goes to $0$, so the terms eventually decay more than exponentially. So the radius of convergence is infinite for all positive $a$.

     – joriki Mar 08 '24 at 01:29

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