Estimating the series: $\sum_{k=0}^{\infty} \frac{k^a b^k}{k!}$

Question

Any idea on how to estimate the following series:

$$\sum_{k=0}^{\infty} \frac{k^a b^k}{k!}$$

where $a$ and $b$ are constant values. Greatly appreciate any respond.

Answer 1

I think I've found a closed-form of this series, but I don't know how to prove it yet. I've asked for a proof in this question.

Let $a \in \mathbb{N}$ and $b \in \mathbb{R}$. Then

$$\sum_{k=0}^{\infty} \frac{k^a\,b^k}{k!} \stackrel{?}{=} e^b \sum_{j=0}^a S(a,a-j+1)\,b^{a-j+1},$$

where $e$ is Euler's number and $S(n,k)$ are the Stirling numbers of the second kind, defined as

$$S(n,k) = \frac{1}{k!}\sum_{i=0}^k (-1)^{i}{k \choose i} (k-i)^n,$$

with ${n \choose k}$ a binomial coefficient.

Beside I don't know how to prove it, I also don't know how to generalize it to $a \in \mathbb{R}$, because I don't know about this type of generalization of the Stirling numbers of the second kind.

For now the best you could do to take the floor of $a$, and then using the closed-form. If you want to esimante there are good lower and upper bounds for $S(n,k)$, and we have results about its asymptotic behavior.

Answer 2

Let's suppose $a > 0$ and $b > 0$. If $g(k) = k^a b^k/k!$, then $$ \dfrac{g(k+1)}{g(k)} = \left(1 + \dfrac{1}{k} \right)^a \dfrac{b}{k+1} \to 0 \ \text{as}\ k \to \infty$$ Take $0 < r < 1$ and $K$ such that $\dfrac{g(k+1)}{g(k)} \le r$ for $k \ge K$. Then $$ \sum_{k=0}^\infty \dfrac{k^a b^k}{k!} \le \sum_{k=0}^{K-1} \dfrac{k^a b^k}{k!} + \dfrac{K^a b^K}{K!} \sum_{j=0}^\infty r^j = \sum_{k=0}^{K-1} \dfrac{k^a b^k}{k!} + \dfrac{K^a b^K}{(1-r)\; K!}$$

Answer 3

If my calculation is correct, we have

$$\sum_{k=0}^{\infty} \frac{k^a}{k!} x^{k} = e^{x} \left\{ x^{a} + \binom{a}{2}x^{a-1} + \mathcal{O}(x^{a-2}) \right\}$$

as $x \to \infty$. I have only an iPad currently in my hand, which is apparetly inadequate for $\TeX$ing. So I will elaborate my answer later.