Infinite series Is it possible to compute explicitly or in an approximate way the infinite series:
$$ S(x) = \sum_{k=0}^{+\infty}{\frac{x^k}{k!} (a+k)^{-(n+1)}}$$
where $n \in \mathbb{N}$.
 A: hint for $a^{-(n+1)}(1+(\frac{k}{a}))^{-(n+1)}$ condition is $|k/a|<1$ so we can approximately take first $4,5$ terms according to significant figures and in $x^k$ we can treat $x$ as constant and it can be easily observed that after some terms $x^k/k!$ becomes small. But if we take k sufficiently large then it isnt possible. 
A: I found a closed formulation using Touchard polynomials: $$
\sum_{k=0}^{\infty}\frac{x^k}{k!}(a+k)^{-(n+1)}=a^{-(n+1)}xe^xT_{-n}(x)
$$
too bad there is no definition I found for Touchard polynomials with negative index.
Anyway you can maybe find useful the relation I found using negative binomial expansion: $$
\sum_{k=0}^{\infty}\frac{x^k}{k!}(a+k)^{-(n+1)}=\sum_{k=0}^{\infty}\frac{1}{a^{n+1}}\frac{x^k}{k!}\left (1+\frac{k}{a}\right )^{-(n+1)}=\sum_{k=0}^{\infty}\frac{1}{a^{n+1}}\frac{x^k}{k!}\sum_{j=0}^{\infty}\binom{-(n+1)}{j}\left (\frac{k}{a}\right )^{j}=\sum_{j=0}^{\infty}\binom{-(n+1)}{j}\frac{1}{a^{n+j+1}}\sum_{k=0}^{\infty}\frac{x^k}{k!}{k}^{j}=\sum_{j=0}^{\infty}\binom{-(n+1)}{j}\frac{1}{a^{n+j+1}}T_j(x)e^x
$$
If you have a way to evaluate rapidly the Touchard polynomials the factor inside the sum should decrease as $(x/a)^j$.
