Simplifying sum with rising and falling factorials Let $(x)^{(n)}=x(x+1)\cdots(x+n-1)$ be the rising factorial and $(x)_{(n)}=x(x-1)\cdots(x-n+1)$ be the falling factorial.
I am sure that the sum
$$\sum_{n, p, q \geqslant 0}\sum_{ l =0}^{n}\frac{\gamma^{n + p + q}}{\left( p + q + n + 1\right)}     \frac{(-1)^l\left(q + l \right)!\left(p + n - l\right)!}{p! q!} \frac{  \left(N\right)^{(p)} \left(N\right)_{(q)}}{\left(M + 1\right)^{(p + n - l)}\left( M - 1 \right)_{(q + l)}}$$
is equal to the simpler sum
$$\sum_{m \geqslant 0}\sum_{k = 0}^m \frac{\gamma^m}{m + 1} \frac{\left(N + 1\right)^{(m - k)}\left( N  - 1\right)_{(k)}}{\left(M + 1\right)^{(m - k)}\left(M - 1\right)_{(k)}},$$
order by order in $\gamma$, and I would like some help proving this.
The obvious idea would be to write $n+p+q=m$ and replace the sum over $n$, or $p$, or $q$ with a sum over $m$. However, the summand is so convoluted that I am not going anywhere.
 A: In the first sum, the coefficient of $\frac{\gamma^m}{m+1}$ is
$$
\sum_{n+p+q=m}\sum_{ l =0}^{n} \frac{(-1)^l\left(q + l \right)!\left(p + n - l\right)!}{p! q!} \frac{  \left(N\right)^{(p)} \left(N\right)_{(q)}}{\left(M + 1\right)^{(p + n - l)}\left( M - 1 \right)_{(q + l)}}.
$$
We set $k=q+l$; as $l$ goes from $0$ to $n$, $k$ goes from $q$ to $q+n = m-p$:
$$
\sum_{n+p+q=m}\sum_{k=q}^{m-p} \frac{(-1)^{k-q} k!\left(m-k\right)!}{p! q!} \frac{  \left(N\right)^{(p)} \left(N\right)_{(q)}}{\left(M + 1\right)^{(m-k)}\left( M - 1 \right)_{(k)}}.
$$
Now we rearrange this sum to depend on $k$ first. For fixed $m$ and $k$, our constraints are that $q \le k$, $p \le m-k$, and $n = m-p-q$. So we have
$$
   \sum_{k=0}^m \sum_{q=0}^k \sum_{p=0}^{m-k} \frac{(-1)^{k-q} k!\left(m-k\right)!}{p! q!} \frac{  \left(N\right)^{(p)} \left(N\right)_{(q)}}{\left(M + 1\right)^{(m-k)}\left( M - 1 \right)_{(k)}}.
$$
Looking only at the $k^{\text{th}}$ term of this sum, we can factor it as
$$
   \frac{(-1)^k k! (m-k)!}{\left(M + 1\right)^{(m - k)}\left(M - 1\right)_{(k)}} \left(\sum_{q=0}^k (-1)^q \frac{(N)_{(q)}}{q!}\right) \left(\sum_{p=0}^{m-k} \frac{(N)^{(p)}}{p!}\right).
$$
The sum over $p$ is known to simplify to $\frac{(N+1)^{(m-k)}}{(m-k)!}$. The sum over $q$ is nearly as well-behaved: we can rewrite $(-1)^q \frac{(N)_{(q)}}{q!}$ as $\frac{(-N)^{(q)}}{q!}$ and then simplify the sum to $\frac{(-N+1)^{(k)}}{k!}$ or $(-1)^k \frac{(N-1)_{(k)}}{k!}$.
These results cancel with the factors that only depend on $k$, so the $k^{\text{th}}$ term of our sum simplifies to
$$
   \frac{\left(N + 1\right)^{(m - k)}\left( N  - 1\right)_{(k)}}{\left(M + 1\right)^{(m - k)}\left(M - 1\right)_{(k)}}.
$$
Remembering that we are summing this as $k$ goes from $0$ to $m$, we get exactly the coefficient of $\frac{\gamma^m}{m+1}$ in the sum we wanted to get.
