What is this series relating to the residues of the Gamma function? I was messing around with the Gamma function and it's poles when I happened upon the following series:
$$f(x)=\sum_{n=0}^\infty\frac{(-1)^n}{n!(x+n)}$$
Does anyone know what this is?  This graph (red line) appears to be very close to the Gamma function (black line):
full graph here

Is there any way to turn this exactly into the Gamma function or derive the closed form of that series?  I thought about adding $e^x$ to get the positive side more fixed and similar, but I'm also interested in an actual way to get this working.
 A: It is $f(s) = \gamma(s,1)$ the lower incomplete gamma function. You have separated $\Gamma(s) = \int_0^\infty x^{s-1} e^{-x}dx$ into $$\Gamma(s) = \int_0^1 x^{s-1} e^{-x}dx+\int_1^\infty x^{s-1} e^{-x}dx$$ where $\int_1^\infty x^{s-1} e^{-x}dx$ is entire and $\int_0^1 x^{s-1} e^{-x}dx=\sum_{n=0}^\infty \frac{(-1)^n}{n!}\int_0^1 x^{s+n-1}dx$ is a locally uniformly convergent series of poles 
(using say the Riemann-Lebesgue lemma and $\Gamma(s+1) = s \Gamma(s)$ we can see that $\gamma(s,1)\to 0$  when $s$ moves away from the negative real axis)
A: Let's compute 
$$f(x+1)=\sum_{n=0}^\infty\frac{(-1)^n}{n!(x+1+n)}=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{(n-1)!(x+n)}$$
and observe that
$$f(x+1)=\sum_{n=1}^\infty\frac{(-1)^{n-1}[(x+n)-x]}{n!(x+n)}=1-\frac 1e+x\sum_{n=1}^\infty\frac{(-1)^n}{n!(x+n)}=1-\frac 1e+x\left(f(x)-\frac 1x\right)$$
We get :
$$\boxed{\forall x>0,\,f(x+1)=x\,f(x)-\frac 1e}$$
This is not the same functional equation than the one verified by $\Gamma$, but looks like ...
A: This series differs from $\Gamma(x)$ by an analytic function. For $\Re(x)>0$,
$$ \Gamma(x) = \int_0^{\infty} t^{x-1} e^{-t} \, dt \\
= \int_0^1 t^{x-1} e^{-t} \, dt + \int_1^{\infty} t^{x-1} e^{-t} \, dt \\
= f(x) + \Gamma(1,x), $$
by expanding the exponential as a power series and integrating term-by-term. $\Gamma(1,x)$ is the upper incomplete Gamma-function, and is an analytic function of $x$, while your series is the lower incomplete Gamma-function $\gamma(1,x)$. Meromorphic continuation implies that the equality derived above holds whenever $x$ is not a nonpositive integer: one can show that $f(x)$ is locally uniformly convergent on domains avoiding the nonpositive integers, so is a meromorphic function, and the result follows.
Notably, it is easy to check using integration by parts that $\Gamma(1,x)$ is exponentially small for $x \ll 0$, hence why the dominant behaviour on the left is captured by $f$. On the other hand, $f$ is much smaller than the exponentially large $\Gamma(1,x)$ for $x \gg 0$, so $\Gamma(1,x)$ is dominant on the right.
For more information, you can also see this answer I wrote a while ago.
A: Similar to @reuns but "from left to right".
$$\begin{align}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!(n+x)}= & \int_{0}^{\infty}\sum_{n=0}^{\infty}\frac{(-1)^{n} e^{-t(n+x)}}{n!}\,dt  \\
= & \int_{0}^{\infty} e^{-tx}\sum_{n=0}^{\infty}\frac{(-e^{-t})^{n} }{n!}\,dt \\
=& \int_{0}^{\infty} e^{-tx-e^{-t}}\,dt \\
& \overset{t\to -\log(s)} = \int_{0}^{1} s^{x-1} e^{-s}\,ds\\
&= \gamma(x,1)
\end{align}$$
Here $\gamma(x,1)$ is the incomplete gamma function.
Generalization
We can find the more general sums
$$S_{k}(x) = \sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!(n+x)^{k}}, k=2, 3, ...$$
by calculating the  $k-1$-st derivative of $\gamma(x,1)$
