Weird Sum Identity I came across a weird sum identity that I would like to prove:
$$
\sum_{i=1}^{k-2}\frac{(-1)^i}{(i-1)!(k-2-i)!(n-k+2+ij)}=\frac{-\Gamma\left(1+\frac{n-k+2}{j}\right)}{j\Gamma\left(k-1+\frac{n-k+2}{j}\right)}.
$$
I would like to know how one can prove this. It seems like direct computations, induction, combinatorial interpretation and other simple proof techniques won't work.
However, software like Wolfram Alpha can evaluate it right away so there must be some easy way to go at it.
Because of the look of the summands, I even tried using the Wilf-Zeilberger algorithm, but it fails in this case.
So the last thing I tried is showing that, if $S_m$ represents the partial sums up to m, then $S_m-S_{m-1}$ is equal to the summand. Then the sum we consider telescopes and we get $S_{k-2}$, which is just what we want. The problem with this approach is that $S_m$ is really nasty if $m\neq k-2$ (it includes hypergeometric series that vanish at $m=k-2$, which can be seen by plugging $S_m$ in Wolfram Alpha).
Any help proving this result will be appreciated
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\bbox[#ffd,10px]{\sum_{i = 1}^{k - 2}{\pars{-1}^{i} \over
\pars{i - 1}!\pars{k - 2 - i}!\pars{n - k + 2 + ij}} =
-\,{\Gamma\pars{1 + \bracks{n - k + 2}/j} \over
j\,\Gamma\pars{k - 1 + \bracks{n - k + 2}/j}}}:\ {\Large ?}}$.

With $\ds{a \equiv \pars{n - k + 2}/j}$:
\begin{align}
&\bbox[#ffd,10px]{\left.\sum_{i = 1}^{k - 2}{\pars{-1}^{i} \over
\pars{i - 1}!\pars{k - 2 - i}!\pars{n - k + 2 + ij}}
\,\right\vert_{\ k\ \in\ \mathbb{N}_{\large\ \geq\ 3}}}
\\[5mm] = &\
{1 \over j\pars{k - 2}!}\sum_{i = 0}^{k - 2}{\pars{k - 2}! \over
i!\pars{k - 2 - i}!}\,\pars{-1}^{i}\,{i \over i + a}
\\[5mm] = &\
{1 \over j\pars{k - 2}!}\sum_{i = 0}^{k - 2}{k - 2 \choose i}
\pars{-1}^{i}\,\pars{1 - {a \over i + a}}
\\[5mm] = &\
{1 \over j\pars{k - 2}!}\bracks{\delta_{k,2} -
a\sum_{i = 0}^{k - 2}{k - 2 \choose i}\pars{-1}^{i}
\int_{0}^{1}t^{i + a - 1}\,\dd t}
\\[5mm] = &\
{\delta_{k,2} \over j} -
{a \over j\pars{k - 2}!}\int_{0}^{1}t^{a - 1}
\sum_{i = 0}^{k - 2}{k - 2 \choose i}\pars{-t}^{i}\,\dd t
\\[5mm] = &\
{\delta_{k,2} \over j} -
{a \over j\pars{k - 2}!}\int_{0}^{1}t^{a - 1}
\pars{1 - t}^{k - 2}\,\dd t
\\[5mm] = &\
{\delta_{k,2} \over j} -
{a \over j\pars{k - 2}!}\,{\Gamma\pars{a}\Gamma\pars{k - 1}
\over \Gamma\pars{a + k - 1}}
\\[5mm] = &\,\,\,
\bbox[10px,#ffd,border:1px groove navy]{
-\,{1 \over j}\,{\Gamma\pars{1 + a} \over
\Gamma\pars{k - 1 + a}}}\,,
\qquad a \equiv {n - k + 2 \over j}
\\ &
\end{align}
because $\ds{k = 3,4,5,\ldots}$
