Uniformly convergence of holomorphic funcitons

Question

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

Answer 1

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*}

Answer 2

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 ...

Answer 3

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 $\text{Re}(z)>0$.