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I am interested in evaluating the following infinite sum \begin{equation} \sum_{n=0}^{\infty} \frac{\alpha^{n}}{n!}n^{\beta} \end{equation}

where both $\alpha$ and $\beta$ are real numbers. However, in addition, $\alpha$ is always positive.

Clearly the sum converges for any value of $\alpha$ and $\beta$ since the factorial kills both exponential and power terms for sufficiently large $m$'s. Does the sum have a closed form?

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    $\begingroup$ $$ \sum_{n=0}^\infty \frac{e^{\gamma n}}{n!}n^\beta = \frac{\partial^\beta}{\partial \gamma^\beta} \sum_{n=0}^\infty \frac{e^{\gamma n}}{n!} = \frac{\partial^\beta}{\partial \gamma^\beta} \exp (e^\gamma) $$ $\endgroup$
    – uranix
    Commented Sep 11, 2015 at 19:50
  • $\begingroup$ @uranix Nice one, but is there any method you could do to compute the $\beta$-th derivative ? $\endgroup$
    – Oussama Boussif
    Commented Sep 11, 2015 at 19:54
  • $\begingroup$ Especially $\beta$ is not an integer! It seems to me that using fractional derivatives wouldn't be that easy! Smart trick though @uranix, thanks $\endgroup$
    – Ben
    Commented Sep 11, 2015 at 19:55
  • $\begingroup$ @OussamaBoussif For integer $\beta$ the answer is expressed using Touchard_polynomials maybe they could be analytically continued for complex $\beta$ $\endgroup$
    – uranix
    Commented Sep 11, 2015 at 19:58
  • $\begingroup$ @uranix Thanks for sharing that with us ! $\endgroup$
    – Oussama Boussif
    Commented Sep 11, 2015 at 20:03

2 Answers 2

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Assuming $\alpha=e^{\gamma}$,

$$\sum_{n\geq 0}\frac{\alpha^n}{n!}n^\beta = \sum_{n\geq 0}\frac{e^{\gamma n}}{n!}n^{\beta}=\left.\frac{\partial^\beta}{\partial u^{\beta}}\,\exp(\exp( u))\right|_{u=\gamma}.$$ If $\beta\not\in\mathbb{N}$, see fractional calculus.

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Indeed, it is likely that anything approaching a closed form can be only achieved for $b=0,1,2,3,\dots$

Here's some additional information not provided in the answer (though touched upon in uranix's comment).

Writing the series as:

$$f(a,b)=a \sum_{n=0}^{\infty} \frac{(n+1)^b}{(n+1)!} a^n$$

We consider the ratio of successive terms and find by definition that for $b=1,2,3,4,5,\dots$ the series equals to a generalized hypergeometric function:

$$f(a,b)=a \cdot~{_{b-1} F_{b-1} } (\mathbf{2};\mathbf{1};a)$$

Where $\mathbf{2}=2,2,\ldots,2$ repeated $b-1$ times.

Fortunately, these functions can be expressed as polynomials (which can also be directly obtained by taking derivatives as shown by uranix and Jack D'Aurizio):

$$f(a,1)=a~e^a$$

$$f(a,2)=a~e^a~(a+1)$$

$$f(a,3)=a~e^a~(a^2+3a+1)$$

$$f(a,4)=a~e^a~(a^3+6a^2+7a+1)$$

$$f(a,5)=a~e^a~(a^4+10a^3+25a^2+15a+1)$$

And so on. The coefficients are Stirling numbers of 2nd kind, and can be found in OEIS: http://oeis.org/A008278.

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