Express $\sum_{k=0}^{\infty}\frac{(-N)^k}{k!(x+k)}$ using $e^{-x}$

Question

You can see that this sum

$$\sum_{k=0}^{\infty}\frac{(-N)^k}{k!(x+k)}$$

looks shockingly like the Taylor expansion of $e^{-x}$ except for that $x+k$ term in the denominator. So, can anyone express this sum somehow using $e^{-x}$?

Answer 1

The series is equal to $N^{-x}\gamma(x,N)$, where $$ \gamma(x,N):=\int_0^Nt^{x-1}e^{-t}\,dt \tag{1} $$ is the lower incomplete gamma function.

Proof: \begin{align} N^{-x}\int_0^Nt^{x-1}e^{-t}\,dt &=N^{-x}\int_0^N\sum_{k=0}^{\infty}\frac{(-1)^kt^{x-1+k}}{k!}dt \\ &=N^{-x}\sum_{k=0}^{\infty}\frac{(-1)^kN^{x+k}}{k!(x+k)} \\ &=\sum_{k=0}^{\infty}\frac{(-N)^{k}}{k!(x+k)}. \tag{2} \end{align}

Answer 2

Another form for it is:$$\begin{split} \sum_{k=0}^{\infty}\frac{(-N)^k}{k!(x+k)} &= \sum_{k=0}^{\infty}\frac{(-N)^k}{k!}\int_0^1 u^{x+k-1}du\\ &= \int_0^1u^{x-1}\sum_{k=0}^{\infty}\frac{(-Nu)^k}{k!}du\\ &=\int_0^1u^{x-1}e^{-Nu}du \end{split} $$