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What do we know about the series

$$\sum_{n=0}^{\infty} \left(\frac{1}{n!} \right)^m$$

where $m$ is a positive constant integer?

We know when $m = 1$, we get the famous $e$. But I became curious if the series in general for other exponents has been studied before, and whether the sums would converge to a well-known number. Also, does it have any applications in math other than the well-studied case of $m=1$.

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  • $\begingroup$ If you type the sums into Wolfram Alpha for certain (maybe all) values of $m$, you get answers in terms of hypergeometric functions... which you will then probably need to Google... :) $\endgroup$
    – David
    Jan 25, 2023 at 5:23
  • $\begingroup$ I didn't see this kind of sum. But it looks very interesting. I've tried Wolfarm alpha (wolframalpha.com/…) and it seems like there is some appearance of Bessel and Hypergeometric functions. $\endgroup$ Jan 25, 2023 at 5:24
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    $\begingroup$ For $m = 2$, the series sum to $I_0(2)$ where $I_\alpha(x)$ is modified bessel function of first kind. $\endgroup$ Jan 25, 2023 at 5:24
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    $\begingroup$ For $m=2$, it is $f(1)$ where $\frac d{dx}(x\frac{df}{dx})=f(x)$ and $f(0)=\frac{df}{dx}(0)=1$. The differential equation generalises easily, but not its solution. $\endgroup$
    – Empy2
    Jan 25, 2023 at 5:32
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    $\begingroup$ As others pointed out, it is hopeless to expect that this sum has an elementary closed form for $m\neq1$. $\endgroup$ Jan 25, 2023 at 9:05

1 Answer 1

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There is something interesting in $$f_m=\sum_{n=0}^{\infty} \left(\frac{1}{n!} \right)^m$$ For any real $m$, $(m f_m)$ is almost a straight line with a slope very close to $2$ which is the asymptotic value of $f_m$. $$f_{100}=2+7.89\times 10^{-31}\qquad\qquad f_{1000}=2+9.33\times 10^{-302}$$

What can be observed is that

$$\color{blue}{\log\left(f_m-2\right)\sim -m \log(2)-\frac{3}{40 m}+\frac{\log (2)}{m^2}-\frac{1}{4 m^3}+O\left(\frac{1}{m^4}\right)}$$ which seems to be decent $$\left( \begin{array}{ccc} m & \text{approximation} & \log\left(f_m-2\right) \\ 1 & -0.32500 & -0.33089 \\ 2 & -1.28176 & -1.27445 \\ 3 & -2.03668 & -2.04251 \\ 4 & -2.75192 & -2.76027 \\ 5 & -3.45501 & -3.46163 \\ 6 & -4.15329 & -4.15751 \\ 7 & -4.84933 & -4.85157 \\ 8 & -5.54421 & -5.54503 \\ 9 & -6.23844 & -6.23827 \\ 10 & -6.93229 & -6.93145 \\ 11 & -7.62590 & -7.62461 \\ 12 & -8.31935 & -8.31776 \\ 13 & -9.01269 & -9.01091 \\ 14 & -9.70597 & -9.70406 \\ 15 & -10.3992 & -10.3972 \\ \end{array} \right)$$

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    $\begingroup$ Since $f_m=2+2^{-m}+6^{-m}+24^{-m}+\cdots$ I expected the first correction term to be $O(3^{-m})$ rather than $O(1/m)$ $\endgroup$
    – Empy2
    Jan 25, 2023 at 11:11
  • $\begingroup$ @Empy2. Me too, be sure ! Cheers :-) $\endgroup$ Jan 25, 2023 at 13:39

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