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I'm looking for a closed expression (if it exists) of the following sum: $$\sum_{m=0}^{\infty} \frac{m^n}{m!}c^m$$ where $n \geq 1$ is a positive integer, and $c$ is a fixed constant. The series seems to be convergent for all positive integers $n$, but I can't see the pattern; first values are: $$\sum_{m=0}^{\infty} \frac{m}{m!}c^m=ce^c$$ $$\sum_{m=0}^{\infty} \frac{m^2}{m!}c^m=c(c+1)e^c$$ $$\sum_{m=0}^{\infty} \frac{m^3}{m!}c^m=c(c^2+3c+1)e^c$$ But for an arbitrary $n$? I can't find any reference. Thanks!

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marked as duplicate by Martin R, Yanior Weg, mrtaurho, Paul Frost, Lord Shark the Unknown May 23 at 4:08

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This problem comes down to expressing $m^n$ as the sum of terms of form $\prod\limits_{k=0}^{r} (m-k)$.

Note that in the $n=2$ case, you had $m^2 = m + m(m-1)$, which yielded $c^2 + c$ as the coefficient of $e^c$. Similarly, $m^3 = m + 3m(m-1) + m(m-1)(m-2)$, which yielded $c^3 + 3c^2 + c$. In general, if you have $m^n = \sum \limits_{r=1}^{n} \left(a_r \prod\limits_{k=0}^{r-1} (m-k) \right)$, for integers $a_n$, you will obtain $\left( \sum \limits_{r=1}^{n} a_r c^{n-r+1} \right)e^c.$

You can determine $a_r$ by starting with $m^n$ and successively subtracting $\prod\limits_{k=0}^{r} (m-k)$, starting with $r=n$ first, and working down to $r=1$, ensuring to reduce the degree of the equation by $1$ each time.

There is not a simple closed form for these $a_r$ and therefore the polynomial multiplied by $e^c$. These polynomials are called Touchard polynomials and can be found here: https://en.wikipedia.org/wiki/Touchard_polynomials

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