Equality involving combinatorics Does anyone know of any relationship involving sums and factorials that can help me prove this identity?
I want to prove that for $j, k \in \mathbb{N}$
$$\sum_{i=1}^{2j+1}\frac{(-1)^{i+1}}{(-i+2j+1)!(i+2k)!}= \frac{1}{(2k)!(2j)!(2k+2j+1)}.$$
My attempt was to separate the positive and negative terms and try to write as binomials, as follows
\begin{equation}
\begin{split}
\sum_{i=2}^{2j+1}\frac{(-1)^{i+1}}{(-i+2j+1)!(i+2k)!}&= \sum_{i=1}^{j}\frac{1}{(2k+2i+1)!(2j-2i)! }- \sum_{i=1}^{j}\frac{1}{(2k+2i)!(2j+1-2i)! }\\
&= \frac{1}{(2k+2j+1)!}\left[ \sum_{i=1}^{j}\binom{2k+2j+1}{2k+2i+1} - \binom{2k+2j+1}{2k+2i} \right].
\end{split}
\end{equation}
The first term I left to be added at the end. This way, the amount of positive and negative factors is the same and i can join in the same sum. From here I have no idea how to continue.
 A: Let the quantity you want to evaluate be $X$. Multiplying, we have
$$(2k+2j+1)!X=\sum_{i=1}^{2j+1}\frac{(-1)^{i+1}(2k+2j+1)!}{(-i+2j+1)!(i+2k)!}=\sum_{i=1}^{2j+1}(-1)^{i+1}\binom{2k+2j+1}{2j+1-i}.$$
We want to show that this is $\binom{2k+2j}{2j}$.
Let
$$S_{n,m}=\sum_{i=0}^m(-1)^\ell\binom{n}{\ell},$$
so that
$$(2k+2j+1)!X=\sum_{i=0}^{2j}(-1)^\ell\binom{2k+2j+1}{\ell}=S_{2k+2j+1,2j};$$
we will show that
$$S_{n,m}=(-1)^m\binom{n-1}m.$$
We can show this by induction on $m$. For $m=0$, the sum consists only of the summand $1$, so it's true there. Now, assume it's true for $m$; then
$$S_{n,m+1}=S_{n,m}+(-1)^{m+1}\binom n{m+1}=(-1)^{m+1}\left(\binom n{m+1}-\binom{n-1}m\right)=(-1)^{m+1}\binom{n-1}{m+1},$$
as desired.
A: We seek to verify that
$$\sum_{p=1}^{2j+1} (-1)^{p+1} {2j+2k+1\choose 2j+1-p}
= {2j+2k\choose 2j}.$$
As there are no odd multiples of $j$ and $k$ we may alternatively prove
the more general
$$\sum_{p=1}^{j+1} (-1)^{p+1} {j+k+1\choose j+1-p}
= {j+k\choose j}.$$
which includes OP's question as a special case with $j$ and $k$ being
even. The LHS is
$$[z^{j+1}] (1+z)^{j+k+1} \sum_{p=1}^{j+1} (-1)^{p+1} z^p.$$
Now here the coefficient etractor enforces the upper range of the sum
and we may write
$$[z^{j+1}] (1+z)^{j+k+1} \sum_{p\ge 1} (-1)^{p+1} z^p
= [z^{j+1}] (1+z)^{j+k+1} z \sum_{p\ge 0} (-1)^p z^p
\\ = [z^j] (1+z)^{j+k+1} \frac{1}{1+z}
= [z^j] (1+z)^{j+k} = {j+k\choose j}$$
as claimed.
