Uniformly convergence of holomorphic funcitons I need to prove that the sequence
$$
f_n = \sum_{i=0}^n\prod_{j=0}^i \left(z+j\right)^{-1} = \frac{1}{z}+\frac{1}{z(z+1)}+\cdots + \frac{1}{z(z+1)(z+2)\cdots  (z+n)}$$
converge uniformly to a function in every compact subset of $\mathbb{C}\setminus\{0,-1,-2,-3,\dotsc\}$. The problem has other questions, but this is what stop me. Obviusly, $f_n$ is holomorphic in that domain. 
In fact, I don't know even whats the limit of the sequence or closed form of $f_n$. Can you give me a hint to continue? Please, don't spoil me final solution. 
I think I could take a compact $K$ and consider the series that define every term of $f_n$ and sum, but I don't get a general formula. 
Edit: I get
$$
f_n(z)= \sum_{i=0}^n \frac{1}{z}\frac{\Gamma(z+1)}{\Gamma(z+1+i)}= \Gamma(z)\sum_{i=0}^n\frac{1}{\Gamma(z+1+i)}
$$
This smell like M Weierstrass test
 A: For a fixed conpact set $K$, suppose $K \subseteq B(0, r)$ and $m \in \mathbb{N}_+$, $m > r + 1$. Then for $z \in K$,
\begin{align*}
\left| \sum_{k = m}^\infty \prod_{j = 0}^k (z + j)^{-1} \right| &\leqslant \sum_{k = m}^\infty \prod_{j = 0}^k |z + j|^{-1} = \prod_{j = 0}^{m - 1} |z + j|^{-1} \sum_{k = m}^\infty \prod_{j = m}^k |z + j|^{-1}\\
&\leqslant \prod_{j = 0}^{m - 1} |z + j|^{-1} \sum_{k = m}^\infty \prod_{j = m}^k (j - |z|)^{-1}\\
&\leqslant \prod_{j = 0}^{m - 1} |z + j|^{-1} \sum_{k = m}^\infty \frac{1}{(m - r)^{k - m}} < +\infty.
\end{align*}
A: An alternative approach, related to the fact that the incomplete $\Gamma$ function has a simple Laplace transform. One may notice that the function defined by such series of reciprocal Pochhammer symbols fulfills the functional identity $z f(z) = 1+ f(z+1)$ and $f(1)=e-1$. The same happens for
$$ g(z)=\int_{0}^{1} x^{z-1} e^{1-x}\,dx $$
over $\mathbb{R}^+$ (or $\text{Re}(z)>0$), by integration by parts. $g(z)$ is trivially holomorphic over $\text{Re}(z)>0$ and its only singularity over $\text{Re}(z)=0$ is the simple pole at the origin. Then the integral representation and the functional identity provide an analytic continuation to $\mathbb{C}\setminus\{0,-1,-2,-3,-4,\ldots\}$.
The situation is indeed very similar to the well-known facts about the (complete) $\Gamma$ function, $ z\Gamma(z)=\Gamma(z+1)$ and $\Gamma(z)=\int_{0}^{+\infty}x^{z-1}e^{-x}\,dx$ over $\text{Re}(z)>0$.
A: For the negative Falling Factorials
we have the partial fraction expansion:
$$
\eqalign{
  & \left( {x - 1} \right)^{\,\underline {\, - n\,} }  = {1 \over {x^{\,\overline {\,n\,} } }} =   \cr 
  &  = \left[ {n = 0} \right] + \left[ {1 \le n} \right]{1 \over {\left( {n - 1} \right)!}}\sum\limits_{\left( {0\, \le } \right)\,k\,\left( {\, \le \,n - 1} \right)} {\left( { - 1} \right)^{\,k} \left( \matrix{
  n - 1 \cr 
  k \cr}  \right)} {1 \over {\left( {x + k} \right)}} \cr} 
$$
where $[P]$ denotes the Iverson bracket
$$
\left[ P \right] = \left\{ {\begin{array}{*{20}c}
   1 & {P = TRUE}  \\
   0 & {P = FALSE}  \\
 \end{array} } \right.
$$
Therefore your function $f_n(x)$ will be the sum over $n$ of the above, 
and since you asked not to be provided with the solution ...
