How do I calculate a sum containing binomial coefficients? Let $a \in (0,1)$, $l\le k-3-p$, $p\ge0$. The question is to show the following identity.
\begin{equation}
\sum\limits_{l_1=l+2}^{k-1-p} C^{k-l_1-1}_{p} \frac{l_1^p}{a^{l_1}}  
= \frac{\left(\sum\limits_{s=p+1}^{2p+1} a^{s-k} \mathcal{A}_s\right) +
  \left(\sum\limits_{s=-1}^{2p-1} a^{s-l} \mathcal{B}_s \right)}{(-1 + a)^{2 p+1}}
\end{equation}
where $\mathcal{A}_s$ and $\mathcal{B}_s$ depend on $k$ and $l$ only.
Find explicit expressions for those coefficients.
 A: Let us substitute for $\alpha := 1/a$. Using the usual identities:
\begin{equation}
\frac{1}{p!}\left. \frac{d^p}{d x^p} x^n \right|_{x=1} =C^n_p
\end{equation}
and
\begin{equation}
\left. \frac{d^p}{d x^p} \exp(x n) \right|_{x=0} =n^p
\end{equation}
we can sum up the series over $l_1$ and the following identity holds:
\begin{equation}
(1) \quad \mathcal{S} := \sum\limits_{l_1=l+2}^{k-1-p} C^{k-l_1-p}_{p} l_1^p \alpha^{l_1} = \left. \frac{1}{p!} \frac{d^p}{d x^p} \frac{d^p}{d y^p} \left(\frac{\alpha^{l+2} e^{(l+2) y} x^{k-l-2} - \alpha^{k-p} e^{(k-p) y} x^p}{x - \alpha e^y}\right) \right|_{x=1,y=0}
\end{equation}
Now, the only thing we need to do is to compute the derivatives. Here we  compute the derivatives sequentially at each time applying the chain rule of differentiation. The calculations are pretty straightforward and yet tedious. The final result is the following:
\begin{equation}
(2) \quad  \mathcal{S} := \sum\limits_{q=0}^p (-1)^q \sum\limits_{q_1=0}^p C^p_{q_1} \left[\alpha^{l+2} C^{k-l-2}_{p-q} (l+2)^{p-q_1} - \alpha^{k-p} C^p_q (k-p)^{p-q_1}\right] \sum\limits_{q_2=q+1}^{q+1+q_1} \frac{\mathcal{A}^{(q)}_{q_1,q_2}}{(1-\alpha)^{q_2}}
\end{equation}
where the coefficients $\left\{\mathcal{A}_{n,p}\right\}_{n=0,p=q+1}^{\infty,q+n+1}$ are defined recursively as follows:
\begin{eqnarray}
\mathcal{A}_{0,q+1} = 1 \\
\mathcal{A}_{n+1,p} = (-p) \left. \mathcal{A}_{n,p} \right|_{p\le q+n+1} + (p-1) \left. \mathcal{A}_{n,p-1} \right|_{p \ge q+2}
\end{eqnarray}
Now, all what remains is to reduce equation (2) to the common denominator. We can see that the biggest exponent in the denominator is equal to $2p + 1$. This coincides with the conjecture stated in the formulation of the question.
