Continuity and differentiability of $f(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!(x+n)}$ Given the series: $$\sum_{n \ge 0} \frac{(-1)^n}{n!(x+n)}$$
Let $f_n(x)$ denote its general term.
Let $f(x)$ denote its sum (when exists).
The question asks to:
$i)$ Find the domain $\mathbb D$ on which $f$ exists.
$ii)$ Show that $\forall x \in \mathbb D$, $xf(x) - f(x+1)$ is independent of $x$.
$iii)$ Show that $f$ is continuous on $\mathbb D$, and that it is derivable on $\mathbb D$.
The first two parts are not important, I'll just write down the results:


*

*$\mathbb D = \mathbb R \setminus (\mathbb Z_- \cup \{0\})$


*$xf(x) - f(x+1) = e^{-1}$

For the third part, this is my attempt:
On $(0, \infty)$, we have:
$$||f_n||_{\infty} = \frac{1}{nn!}$$
And the series having this general term clearly converges, hence we have normal convergence and so uniform convergence. The $f_n$'s are continuous, therefore $f$ is continuous on $(0,\infty)$.
We can write:
$$\mathbb D = (0,\infty) \cup \left( \bigcup_{n=1}^{\infty} (-n,-n+1) \right)$$
We use the relation of part $ii)$ to prove the continuity of $f$ on each $(-n,-n+1)$, by induction on $n$:
For $n=1$: let $a \in (-1,0)$. Since $ii)$ is true for any $x \in \mathbb D$, we may just choose $x = a$, and the result follows due to the continuity of $f$ on $(0,\infty)$.
Etc.
Can you check my work? Is there a simpler/better approach?
Moreover, how would I prove that it is differentiable?
Thank you.
 A: Through the functional equation, you may also check that:
$$ \sum_{n\geq 0}\frac{(-1)^n}{n!(x+n)}=\Gamma(x)-\Gamma(x,1)\tag{1}$$
where:
$$ \Gamma(x,1) = \int_{1}^{+\infty} u^{x-1}e^{-u}\,du \tag{2}$$
is the incomplete $\Gamma$ function, regular (i.e. $C^{\infty}$) on the whole real line. Hence our function has the same regularity of the $\Gamma$ function, and it is a $C^\infty$ function over $\mathbb{R}\setminus(-\mathbb{N})$ as claimed.
A: If $x$ is a real number for which $x \not \in (\mathbb{Z} \cup \{0\})$, then there is a closed neighborhood of $x$, $D_x$ for which $D_x \cap (\mathbb{Z} \cup \{0\}) = \emptyset$.
Note that each of the functions: $f_M(y) = \sum_{n=0}^M \frac{(-1)^n}{n!(y+n)}$ are continuous, and that this sequence converges uniformly to $f$ on $D_x$. You can show this by bounding $(y+n)^{-1}$ for $y\in D_x$ and using $1/n!$ to your advantage. Thus $f$ is continuous on $D_x$.
Moreover, $f'_M(y) = \sum_{n=0}^M \frac{(-1)^{n+1}}{n!(y+n)^2}$ also converges to a function uniformly, and therefore $f'(y)$ exists and is equal to $$f'(y) = \lim_{M\to\infty} f'_M(y)$$ for all $y \in D_x$. This uses Theorem $7.17$ in Rudin's Principles of Mathematical Analysis.
