How show that this sum $\sum^{n}_{i=0} \frac{(-1)^i}{n+k+i} \cdot \frac{(m+n+i)!}{i!(n-i)!(m+i)!}=0$ show that :
$$\sum^{n}_{i=0} \frac{(-1)^i}{n+k+i} \cdot \frac{(m+n+i)!}{i!(n-i)!(m+i)!}=0$$
This problem is from :  http://www.artofproblemsolving.com/Forum/viewtopic.php?p=239364&sid=2fbf367cb9fab8240df03e632a41085a#p239364
Evaluate a finite sum with four factorials
I think this problem have integral methods,But I can't ,Thank you
 A: We have:
$\displaystyle S_{n,m,k}=\sum_{i=0}^n\dfrac{(-1)^i}{n+k+i}\cdot\dfrac{(m+n+i)!}{i!(n-i)!(m+i)!}=\sum_{i=0}^n\dfrac{(-1)^i(m+n+i)...(n+k+i+1). (n+k+i-1)!}{i!(n-i)!(m+i)...(k+i+1).(k+i)!}$
$\displaystyle\quad =\sum_{i=0}^n\dfrac{(m+n+i)^{\underline{m-k}}}{(m+i)^{\underline{m-k}}}\cdot \dfrac{(-1)^i{n\choose i}{n+k+i-1\choose n-1}}{n}$
*Note: $(x)^{\underline{n}}=x(x-1)...(x-n+1)$
Now, We caculate differences
$\Delta\left[\dfrac{(m+n+i)^{\underline{m-k}}}{(m+i)^{\underline{m-k}}}\right]=\dfrac{(n+k+i+2)...(m+n+i+1)}{(k+i+2)...(m+i+1)}-\dfrac{(n+k+i+1)...(m+n+i)}{(k+i+1)...(m+i)}$
$\quad =n(k-m)\cdot\dfrac{(m+n+i)^{\underline{m-k-1}}}{(m+i+1)^{\underline{m-k+1}}}$
and
$\Delta\left[(-1)^{i-1}.{n-1\choose i-1}{n+k+i-1\choose n}\right]=\dfrac{n+k}{n}(-1)^i.{n\choose i} {n+k+i-1\choose n-1}$
Therefor by Summation by parts, we have:
$\displaystyle S_{n,m,k}=\dfrac{1}{n+k}\sum_{i=0}^n \dfrac{(m+n+i)^{\underline{m-k}}}{(m+i)^{\underline{m-k}}}\Delta\left[(-1)^{i-1}.{n-1\choose i-1}{n+k+i-1\choose n}\right]$
$\displaystyle = \dfrac{1}{n+k}\left.\left[\dfrac{(m+n+i)^{\underline{m-k}}}{(m+i)^{\underline{m-k}}}\cdot (-1)^{i-1}.{n-1\choose i-1}{n+k+i-1\choose n}\right]\right|_{i=0}^{n+1}$
$\quad +\dfrac{n(m-k)}{n+k}\sum_{i=0}^n (-1)^{i}.{n-1\choose i}{n+k+i\choose n} \cdot\dfrac{(m+n+i)^{\underline{m-k-1}}}{(m+i+1)^{\underline{m-k+1}}}$
$\displaystyle =\dfrac{(m-k)}{n+k}\sum_{i=0}^{n-1} \dfrac{(-1)^i(m+n+i)!}{(n+k+i+1) i!(n-1-i)!(m+i+1)!}$
$\Rightarrow S_{n,m,k}=\dfrac{m-k}{n+k}S_{n-1,m+1,k+2}=\dfrac{m-k}{n+k}\cdot\dfrac{m-k-1}{n+k+1}S_{n-2,m+2,k+4}=...=$
$=\dfrac{(m-k)(m-k-1)...(m-k-n+1)}{(n+k)(n+k+1)...(2n+k-1)}S_{0,m+n,k+2n}$
inside 
$S_{0,m+n,k+2n}=\dfrac{(-1)^0 (m+n+0+0)!}{(0+k+2n+0) 0!(0-0)!(m+n+0)!}=\dfrac{1}{2n+k}$
Result: $\boxed{\displaystyle\sum_{i=0}^n\dfrac{(-1)^i}{n+k+i}\cdot\dfrac{(m+n+i)!}{i!(n-i)!(m+i)!}=\dfrac{(m-k)^{\underline{n}}}{(n+k)_{n+1}}}$
*Note: Pochhammer symbol: $(x)_n=x(x+1)...(x+n-1)$
A: I will assume that $n+k \geq 1$. (That is, $n$ and $k$ are not both zero simultaneously.) Let $p_{n}(x)$ be the polynomial defined by

$$ p(x) = \displaystyle \sum_{i=0}^{n} \frac{(-1)^{i}}{n+k+i} \frac{(m+n+i)!}{i!(n-i)!(m+i)!} x^{m+i} . $$

Our goal is to evaluate $p(1)$. As a first step, we rewrite $p(x)$ as
\begin{align*}
p(x)
&= \sum_{i=0}^{n} \frac{(-1)^{i}}{n+k+i} \frac{1}{i!(n-i)!} \frac{d^{n}}{dx^{n}} x^{m+n+i} \\
&= \frac{d^{n}}{dx^{n}} \left( \frac{1}{n!} \sum_{i=0}^{n} \binom{n}{i} \frac{(-1)^{i}}{n+k+i} x^{m+n+i} \right) . \tag{1}
\end{align*}
Keep manipulating the expression inside the bracket of $\text{(1)}$, we get
\begin{align*}
\frac{1}{n!} \sum_{i=0}^{n} \binom{n}{i} \frac{(-1)^{i}}{n+k+i} x^{m+n+i} 
&= \frac{x^{m-k}}{n!} \sum_{i=0}^{n} \binom{n}{i} (-1)^{i} \frac{x^{n+k+i}}{n+k+i} \\
&= \frac{x^{m-k} }{n!} \sum_{i=0}^{n} \binom{n}{i} (-1)^{i} \int_{0}^{x} t^{n+k+i-1} \, dt  \\
&= \frac{x^{m-k}}{n!} \int_{0}^{x} t^{n+k-1} (1 - t)^{n} \, dt. \tag{2}
\end{align*}
Now we evaluate the integral $\text{(2)}$. Note that if $m \geq 1$, integration by parts shows that for any $m, n \geq 0$ the following identity holds:
\begin{align*}
\int_{0}^{x} t^{m} (1 - t)^{n} \, dt
&= \frac{m!n!}{(m+n+1)!} \left( 1 - \sum_{j=n+1}^{m+n+1} \binom{m+n+1}{j} x^{m+n+1-j} (1-x)^{j} \right).
\tag{3}
\end{align*}
Since $n + k \geq 1$, we can apply $\text{(3)}$ to $\text{(2)}$ and we get
\begin{align*}
\frac{x^{m-k}}{n!} \int_{0}^{x} t^{n+k-1} (1 - t)^{n} \, dt
&= \frac{(n+k-1)!}{(2n+k)!} \left( x^{m-k} - \sum_{j=n+1}^{2n+k} \binom{2n+k}{j} (1-x)^{j} x^{2n+m-j} \right)
\end{align*}
Then $\text{(1)}$ now reduces to
\begin{align*}
p(x)
&= \frac{(n+k-1)!}{(2n+k)!} \frac{d^{n}}{dx^{n}} \left( x^{m-k} - \sum_{j=n+1}^{2n+k} \binom{2n+k}{j} (1-x)^{j} x^{2n+m-j} \right) \\
&= \left( \frac{(n+k-1)!}{(2n+k)!} \prod_{j=0}^{n-1} (m-k-j) \right) \, x^{m-k-n} - q(x),
\end{align*}
where $q(x)$ is the polynomial given by
$$ q(x) = \frac{(n+k-1)!}{(2n+k)!} \sum_{j=n+1}^{2n+k} \binom{2n+k}{j} \frac{d^{n}}{dx^{n}} (1-x)^{j} x^{2n+m-j}. $$
But since $(1-x) \mid \frac{d^{i}}{dx^{i}} (1-x)^{j}$ whenever $i \leq n < j$, it follows that $(1-x) \mid q(x)$ and hence $q(1) = 0$. Therefore we obtain

$$ p(1) = \frac{(n+k-1)!}{(2n+k)!} \prod_{j=0}^{n-1} (m-k-j) \underset{m \geq k}{=} \frac{(n+k-1)!(m-k)!}{(2n+k)!(m-k-n)!}. $$

In particular, if $n > m-k$ then $p(1) = 0$. Clearly, if $ 1 \leq k \leq m \leq n$ then $n > m - k$ is automatically satisfied and the desired conclusion follows as corollay.
