Sum of terms in closed form I've reduced a problem that I'm doing to showing that for all $n \geq r$ natural numbers, $$\sum_{i=o}^{n-r}\frac{(-1)^i}{(r+i+1) \ i! \ (n-r-i)!} = \frac{r!}{(n+1)!}$$
Or equivalently writing $n = N +r$ that either $$\sum_{i=o}^{N}\frac{(-1)^i}{(r+i+1) \ i! \ (N-i)!}  = \frac{r!}{(N+r+1)!}$$ or $$\frac{1}{N!}\sum_{i=o}^{N}\frac{(-1)^i}{(r+i+1)} {N \choose i}  = \frac{r!}{(N+r+1)!}$$
I thought about trying induction, with the $n=r$ ie $N=0$ case being very easy, but it seems like the inductive step wouldn't work with the $(r+i+1)$ term not fitting well with the binomial coefficient or factorials...
Any help is appreciated!
 A: Let's define $f$ as
$$f(x) := x^r\sum_{i=0}^N (-x)^i\binom{N}{i} = x^r(1-x)^N.$$
Moreover, we define $$F(x):= \int_0^x f(y)dy = x^{r+1}\sum_{i=0}^N \frac{(-x)^i}{r+i+1}\binom{N}{i} = \int_0^xy^r(1-y)^Ndy.$$
Using $x=1$, we just need to prove that $$\int_0^1y^r(1-y)^Ndy = \frac{N!r!}{(N+r+1)!}$$
We'll do this inducting on $N$ (for all $r$ at the same time). If $N=0$, this is trivial.
Now, $$\int_0^1y^r(1-y)^{N+1}dy = \Big[\frac{y^{r+1}}{r+1}\cdot(1-y)^{N+1}\Big]_0^1+\frac{N+1}{r+1}\cdot\int_0^1y^{r+1}(1-y)^Ndy = \frac{N+1}{r+1}\cdot \frac{N!(r+1)!}{(N+r+2)!} = \frac{(N+1)!r!}{(N+r+2)!}.$$
So, induction is completed.
A: We prove that
$$S_N = \sum_{q=0}^N \frac{(-1)^q}{q+r+1} {N\choose q}
= \frac{1}{r+1} {N+r+1\choose N}^{-1}$$
by introducing
$$f(z) = (-1)^N N! \frac{1}{z+r+1} \prod_{p=0}^N \frac{1}{z-p}$$
which has the property that for $0\le q\le N$
$$\mathrm{Res}_{z=q} f(z)
= (-1)^N N! \frac{1}{q+r+1}
\prod_{p=0}^{q-1} \frac{1}{q-p}
\prod_{p=q+1}^N \frac{1}{q-p}
\\ = (-1)^N N! \frac{1}{q+r+1}
\frac{1}{q!} \frac{(-1)^{N-q}}{(N-q)!}
= \frac{(-1)^q}{q+r+1} {N\choose q}.$$
Now residues sum to zero and the residue of $f(z)$ at infinity is zero
by inspection and we have
$$S_N = \sum_{q=0}^N \mathrm{Res}_{z=q} f(z)
= - \mathrm{Res}_{z=-r-1} f(z)
= - (-1)^N N! \prod_{p=0}^N \frac{1}{-r-1-p}
\\ = N! \prod_{p=0}^N \frac{1}{r+1+p}
= N! \frac{r!}{(N+r+1)!}
= \frac{1}{r+1} {N+r+1\choose N}^{-1}.$$
This is the claim. Here we have supposed that $r$ is a non-negative
integer. The closed form is valid for $r$ a complex number that is not a
negative integer  $-N-1\le r\le -1$, it does not simplify to factorials
in the general  case, however.
