Evaluating $\sum_{m=j_2-j}^{j_1} \frac{(j_1+m)!(j_2+j-m)!}{(j_1-m)!(j_2-j+m)!}$ using combinatorial identities I'm trying to prove the identity
$$
\sum_{m=j_2-j}^{j_1} \frac{(j_1+m)!(j_2+j-m)!}{(j_1-m)!(j_2-j+m)!}=
\frac{(j+j_1-j_2)!(j-j_1+j_2)!(j+j_1+j_2+1)!}{(j_1+j_2-j)!(2j+1)!} \tag{1}
$$
where $2j_1,2j_2\in \mathbb{Z}^+$ and $\vert 2j_1-2j_2\vert\le 2j\le 2(j_1+j_2)$.  The summation over $m$ is in steps of $1$ from $j-j_2$ to $j_1$. This summation occurs as part of a normalization condition on $su(2)$ Clebsch-Gordan coefficients.
Apparently to prove this one must use the binomial identities 
\begin{align}
\sum_{x} {a\choose x}{b\choose c-x}&={a+b\choose c} \tag{2}\\
{u \choose v}&= (-1)^v {v-u-1\choose v}\, .  \tag{3}
\end{align}
I found in Eq.(6.1) of this volume that (2) is known as the integral Vandermonde Convolution.  (3) can be found as Eq.(2.1) of this paper.
My difficulty is in seeing the relevance of (2) since the left hand side of
(1) contains the summation index $m$ in the numerator and the denominator while the dummy index in (2) appears only in the denominator.
I have to deal with several variants of this summation formula so any help in bridging the gap between the left and right hand side of (1) would be appreciated.  
 A: Here is a streamlined variant.

The following is valid for integers $j_1,j_2,j$ with $0\leq j_2-j\leq j_1$.
  \begin{align*}
\frac{1}{(j+j_1-j_2)!(j-j_1+j_2)!}\sum_{m=j_2-j}^{j_1}\frac{(j_1+m)!(j_2+j-m)!}{(j_1-m)!(j_2-j+m)!}=
\frac{(j+j_1+j_2+1)!}{(j_1+j_2-j)!(2j+1)!}
\end{align*}

We obtain

\begin{align*}
&\color{blue}{\frac{1}{(j+j_1-j_2)!(j-j_1+j_2)!}\sum_{m=j_2-j}^{j_1}\frac{(j_1+m)!(j_2+j-m)!}{(j_1-m)!(j_2-j+m)!}}\\
&\qquad=\sum_{m=j_2-j}^{j_1}\binom{j_1+m}{j_2-j+m}\binom{j_2+j-m}{j_1-m}\tag{1}\\
&\qquad=\sum_{m=j_2-j}^{j_1}\binom{-j_1+j_2-j-1}{j_2-j+m}(-1)^{j_2-j+m}\binom{-j_2-j+j_1-1}{j_1-m}(-1)^{j_1-m}\tag{2}\\
&\qquad=(-1)^{j_1+j_2-j}\sum_{m=0}^{j_1+j_2-j}\binom{-j_1+j_2-j-1}{m}\binom{-j_2-j+j_1-1}{j_1+j_2-j-m}\tag{3}\\
&\qquad=(-1)^{j_1+j_2-j}\binom{-2j-2}{j_1+j_2-j}\tag{4}\\
&\qquad\color{blue}{=\binom{j_1+j_2+j+1}{j_1+j_2-j}}\tag{5}
\end{align*}
  and the claim follows.

Comment:


*

*In (1) we collect the factorials and write them as binomial coefficients.

*In (2) we apply twice the binomial identity
\begin{align*}
\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q
\end{align*}

*In (3) we shift the index to start from $m=0$.

*In (4) we apply Vandermonde's identity.

*In (5) we apply again the identity as in (2).
A: Please note that if this answer is not as rigorous as you may like then I urge you to fill in the details.
I'm thinking you can rearrange $(1)$ into the form
$$\sum_{m=j_2-j}^{j_1} \frac{(j_1+m)!}{(j_2-j+m)!(j+j_1-j_2)!}\frac{(j_2+j-m)!}{(j_1-m)!(j-j_1+j_2)!}\stackrel{?}{=}
\frac{(j+j_1+j_2+1)!}{(j_1+j_2-j)!(2j+1)!}$$
or
$$\sum_{m=j_2-j}^{j_1} \binom{j_1+m}{j_2-j+m}\binom{j_2+j-m}{j_1-m}\stackrel{?}{=}
\frac{(j+j_1+j_2+1)!}{(j_1+j_2-j)!(2j+1)!}\tag{*}$$
Using $(3)$ on the left hand side summation
$$\sum_{m=j_2-j}^{j_1} (-1)^{j_2-j+m}\binom{j_2-j_1-j-1}{j_2-j+m}(-1)^{j_1-m}\binom{j_1-j_2-j-1}{j_1-m}$$
gives
$$(-1)^{j_1+j_2-j}\sum_{m=j_2-j}^{j_1} \binom{j_2-j_1-j-1}{j_2-j+m}\binom{j_1-j_2-j-1}{j_1-m}$$
then call $x=j_2-j+m$ and $c=j_1+j_2-j$ so we have from $(2)$
$$(-1)^{j_1+j_2-j}\sum_{x=2(j_2-j)}^{j_2+j_1-j} \binom{j_2-j_1-j-1}{x}\binom{j_1-j_2-j-1}{c-x}=\binom{-2j-2}{j_1+j_2-j}\tag{**}$$
Now use $(3)$ on the right hand side of $(\text{**})$
$$\begin{align}(-1)^{j_1+j_2-j}\sum_{x=2(j_2-j)}^{j_2+j_1-j} \binom{j_2-j_1+j+1}{x}\binom{j_2-j_1+j-1}{c-x}=&(-1)^{j_1+j_2-j}\binom{j+j_1+j_2+1}{j_1+j_2-j}\\[1ex]\implies\sum_{x=2(j_2-j)}^{j_2+j_1-j} \binom{j_2-j_1+j+1}{x}\binom{j_2-j_1+j-1}{c-x}&=\frac{(j+j_1+j_2+1)!}{(j_1+j_2-j)!(2j+1)!}\end{align}$$
which is the right hand side of $(\text{*})$.
Notice I am playing fast and loose with limits on the Vandermonde summation. I'll leave that to you to make sure they obey $(2)$.
