Let $l \ge 1$ and $m \ge 0$ and $k \ge l$ be integers. Then let ${\bf A}= \left( A_\eta \right)_{\eta=1}^l$ and ${\bf B}= \left( B_\eta \right)_{\eta=1}^l$ be parameters. We define the following multiple sum:
\begin{equation} {\mathfrak S}_{l,m}^{({\bf A},{\bf B})} (k):= \sum\limits_{1 \le j_1 < j_2 < \cdots < j_l \le k} \prod\limits_{\eta=0}^{l-1} (j_{\eta+1} + A_{\eta+1})^{(m+1)} (j_{\eta+1} + B_{\eta+1})^{(m)} \tag{1} \end{equation}
This sum appears in a natural way when finding the Frobenius power series solution to a certain class of ordinary differential equations with polynomial coefficients, see this post.
We define auxiliary coefficients as follows:
\begin{eqnarray} &&{\mathfrak C}_l= -\sum\limits_{\eta=1}^l \sum\limits_{\xi=0}^{(2(l-\eta)+1)m+l-\eta} \Lambda^{(l,\eta)}_\xi \cdot (A_\eta-l+\eta)^{(m+l-\eta+2+\xi)}\\ &&\Lambda^{(l+1,\eta)}_\xi = \sum\limits_{\xi_2=0 \vee \xi-( (2(l-\eta)+1)m+l-\eta )}^{(2m+1) \wedge \xi} \sum\limits_{\xi_1=0 \vee (\xi_2-m-1)}^m \\ && \left( \right. \\ && \left. \Lambda^{(l,\eta)}_{\xi-\xi_2} \cdot (B_{l+1} - A_{l+1}-m-1)^{(m-\xi_1)} (A_{l+1} - A_\eta-(m+\xi-\xi_2)-1)^{(m+1+\xi_1-\xi_2)} \cdot \binom{m+1+\xi_1}{\xi_2} \binom{m}{\xi_1} \cdot 1_{\eta \le l} + \right.\\ && \left. {\mathfrak C}_l (B_{l+1} - A_{l+1}-m-1)^{(m-\xi)} \cdot \binom{m}{\xi} \cdot 1_{\eta=l+1} \right. \\ && \left. \right) \cdot \frac{1}{m+l-\eta+3+\xi} \tag{2a} \end{eqnarray} for $l=1,2,\cdots$, $\eta=1,\cdots,l+1$ and $\xi=0,\cdots, (2(l+1-\eta)+1)m+l+1-\eta$ subject to $\Lambda^{(1,1)}_\xi = (B_1-A_1-m-1)^{(m-\xi)} \binom{m}{\xi}/(m+2+\xi) $.
We have shown that this multiple sum admits the following neat closed form expression below.
\begin{equation} {\mathfrak S}_{l,m}^{({\bf A},{\bf B})} (k) = \sum\limits_{\eta=1}^l \sum\limits_{\xi=0}^{(2(l-\eta)+1) m+l-\eta} \Lambda^{(l,\eta)}_\xi \cdot \left( (k+A_\eta-(l-\eta))^{(m+(l-\eta+2)+\xi)} - (0+A_\eta-(l-\eta))^{(m+(l-\eta+2)+\xi)} \right) \tag{2} \end{equation}
We have derived the result by decomposing the product of Pochhammer symbols into linear combinations of single Pochhammer symbols and then performing the sums bottom -up by using the Pascal triangle identity $\binom{n}{k} = \binom{n+1}{k+1} - \binom{n}{k+1}$.
Note:
We coefficient at the highest order term in $(2)$ are given below:
\begin{eqnarray} &&\Lambda^{(l,1)}_{\xi_m(l)} = \frac{1}{l! (2m+2)^l} \\ \end{eqnarray} where $\xi_m(l):= (2(l-1)+1)m+l-1$.
Proof by induction:
Fix $l$ and take the new factor in the term in the multiple sum and decompose that factor into a linear combination of upper Pochhammer symbols as follows:
\begin{equation} (j_{l+1} + A_{l+1})^{(m_1)} (j_{l+1} + B_{l+1})^{(m_2)} = \sum\limits_{\xi_1=0}^{m_2} (B_{l+1}-A_{l+1}-(m_1))^{(m_2-\xi_1)} \cdot \binom{m_2}{\xi_1} \cdot (j_{l+1} + A_{l+1})^{(m_1+\xi_1)} \tag{3a} \end{equation} for integer $m_1,m_2 \ge 0$.
Now we in $(1)$ we replace $k \rightarrow j_{l+1}-1$ and multiply both sides of it by $(3)$. Then we have:
\begin{eqnarray} &&\sum\limits_{1 \le j_1 < j_2 < \cdots < j_l < j_{l+1} } \prod\limits_{\eta=0}^l (j_{\eta+1} + A_{\eta+1})^{(m+1)} (j_{\eta+1} + B_{\eta+1})^{(m)} = \\ && \left. \sum\limits_{\eta=1}^l \sum\limits_{\xi=0}^{(2(l-\eta)+1)m+l-\eta} \sum\limits_{\xi_1=0}^m \Lambda^{(l,\eta)}_\xi \cdot (B_{l+1}-A_{l+1}-(m+1))^{(\eta-\xi_1)} \cdot \binom{m}{\xi_1} \cdot \underline{\underline{ (j_{l+1} + A_\eta -(l-\eta+1))^{(m+(l-\eta+2)+\xi)} (j_{l+1}+A_{l+1})^{(m+1+\xi_1)}}} + \right. \\ && {\mathfrak C}_l \sum\limits_{\xi_1=0}^m (B_{l+1}-A_{l+1}-(m+1))^{(\eta-\xi_1)} \cdot \binom{m}{\xi_1} \cdot \underline{(j_{l+1}+A_{l+1})^{(m+1+\xi_1)}} \tag{3b} \end{eqnarray} Now we decompose the doubly underlined term by using $(3b)$ repeatedly. We have:
\begin{eqnarray} &&\sum\limits_{1 \le j_1 < j_2 < \cdots < j_l < j_{l+1} } \prod\limits_{\eta=0}^l (j_{\eta+1} + A_{\eta+1})^{(m+1)} (j_{\eta+1} + B_{\eta+1})^{(m)} = \\ && \left. \sum\limits_{\eta=1}^l \sum\limits_{\xi=0}^{(2(l-\eta)+1)m+l-\eta} \sum\limits_{\xi_1=0}^m \sum\limits_{\xi_2=0}^{m+1+\xi_1} \Lambda^{(l,\eta)}_\xi \cdot (B_{l+1}-A_{l+1}-(m+1))^{(\eta-\xi_1)} \cdot \binom{m}{\xi_1} \cdot (A_{l+1} - (A_\eta-(l-\eta+1)) - (m+l-\eta+2+\xi))^{(m+1+\xi_1-\xi_2)} \cdot \binom{m+1+\xi_1}{\xi_2} \cdot \underline{(j_{l+1} + A_\eta-(l-\eta+1))^{((m+l-\eta+2+\xi+\xi_2)}} + \right. \\ && {\mathfrak C}_l \sum\limits_{\xi_1=0}^m (B_{l+1}-A_{l+1}-(m+1))^{(\eta-\xi_1)} \cdot \binom{m}{\xi_1} \cdot \underline{(j_{l+1}+A_{l+1})^{(m+1+\xi_1)}} \tag{3c} \end{eqnarray}
Now we do the sum over $j_{l+1} = 1,\cdots, l$ by using the identity $ \sum\limits_{j=1}^k (j + A)^{(m)} = ((k+A)^{(m+1)} - A^{(m+1)})/(m+1)$. We have:
\begin{eqnarray} &&\sum\limits_{1 \le j_1 < j_2 < \cdots < j_l < j_{l+1} \le k} \prod\limits_{\eta=0}^l (j_{\eta+1} + A_{\eta+1})^{(m+1)} (j_{\eta+1} + B_{\eta+1})^{(m)} = \\ && \left. \sum\limits_{\eta=1}^l \sum\limits_{\xi=0}^{(2(l-\eta)+1)m+l-\eta} \sum\limits_{\xi_1=0}^m \sum\limits_{\xi_2=0}^{m+1+\xi_1} \Lambda^{(l,\eta)}_\xi \cdot (B_{l+1}-A_{l+1}-(m+1))^{(\eta-\xi_1)} \cdot \binom{m}{\xi_1} \cdot (A_{l+1} - (A_\eta-(l-\eta+1)) - (m+l-\eta+2+\xi))^{(m+1+\xi_1-\xi_2)} \cdot \binom{m+1+\xi_1}{\xi_2} \cdot % %\underline{(j_{l+1} + A_\eta-(l-\eta+1))^{((m+l-\eta+2+\xi+\xi_2)}} \frac{(k+A_\eta-(l-\eta+1))^{(m+l-\eta+3+\xi+\xi_2)} - (A_\eta-(l-\eta+1))^{(m+l-\eta+3+\xi+\xi_2)}}{m+l-\eta+3+\xi+\xi_2} % + \right. \\ && {\mathfrak C}_l \sum\limits_{\xi_1=0}^m (B_{l+1}-A_{l+1}-(m+1))^{(\eta-\xi_1)} \cdot \binom{m}{\xi_1} \cdot % %\underline{(j_{l+1}+A_{l+1})^{(m+1+\xi_1)}} % \frac{(k+A_{l+1})^{(m+2+\xi_1)} - (A_{l+1})^{(m+2+\xi_1)}}{m+2+\xi_1} \tag{3d} \end{eqnarray}
Now we simplify. We firstly move the sum over $\xi_1$ to the very right (to the very bottom of the nested multiple sum) and then we substitute $\xi+\xi_2 \rightarrow \xi$. This yields:
\begin{eqnarray} &&\sum\limits_{1 \le j_1 < j_2 < \cdots < j_l < j_{l+1} \le k} \prod\limits_{\eta=0}^l (j_{\eta+1} + A_{\eta+1})^{(m+1)} (j_{\eta+1} + B_{\eta+1})^{(m)} = \\ && \left. \sum\limits_{\eta=1}^l \sum\limits_{\xi=0}^{(2(l+1-\eta)+1)m+l+1-\eta} \right. \\ && \left. \left( % \left( \sum\limits_{\xi_2=0 \vee((2(l-\eta)+1)m+l-\eta)}^{(2m+1) \wedge \xi} \sum\limits_{\xi_1=0 \vee (\xi_2-m-1)}^m \Lambda^{(l,\eta)}_{\xi-\xi_2} \cdot % (B_{l+1}-A_{l+1} -(m+1))^{(m-\xi_1)} \binom{m}{\xi_1} \cdot (A_{l+1}-(A_\eta-(l-\eta+1)) - (m+l-\eta+2+\xi-\xi_2))^{(m+1+\xi_1-\xi_2)} \binom{m+1+\xi_1}{\xi_2} \right) \cdot \frac{ (k+A_\eta-(l-\eta+1))^{(m+(l-\eta+3)+\xi)} - (A_\eta-(l-\eta+1))^{(m+(l-\eta+3)+\xi)} }{m+l-\eta+3+\xi} % % \right) +\right. \\ && {\mathfrak C}_l \sum\limits_{\xi_1=0}^m (B_{l+1}-A_{l+1}-(m+1))^{(\eta-\xi_1)} \cdot \binom{m}{\xi_1} \cdot \frac{(k+A_{l+1})^{(m+2+\xi_1)} - (A_{l+1})^{(m+2+\xi_1)}}{m+2+\xi_1} \tag{3e} \end{eqnarray}
Now we notice that the term in the last, single, sum is equal to the term in the double sum above it at $\eta = l+1$. As such we absorb the last single sum into the double sum above it at $\eta = l+1$. Then we see that the right hand side of the equality above has the same functional form as the conjecture $(2)$being evaluated at $l \rightarrow l+1$ with the new coefficients $\Lambda^{(l+1,\eta)}_\xi$ being given by $(2a)$. This completes the proof.
Below we verify the result numerically:
(*Checking the final result.*)
lmax = 4; kmax = 10; m = 2;(*m=1,2,3*)
Clear[Lambda, k];
Lambda = Table[
0, {l, 1, lmax}, {[Eta], 1, l}, {[Xi],
0, (2 (l - [Eta]) + 1) m + l - [Eta]}];
Lambda[[1, 1]] =
Table[(up[Subscript[B, 1] - Subscript[A, 1] - (m + 1),
m - [Xi]] Binomial[m, [Xi]])/(m + 2 + [Xi]), {[Xi], 0, m}];
Do[
Cl = !(
*UnderoverscriptBox[([Sum]), ([Eta] = 1), (l)](
*UnderoverscriptBox[([Sum]), ([Xi] =
0), (((2 ((l - [Eta])) + 1)) m +
l - [Eta])](((-1))) Lambda[([l, [Eta], \
1 + [Xi]])] up[((
*SubscriptBox[(A), ([Eta])] - ((l - [Eta])))), ((m + ((l \
- [Eta] + 2)) + [Xi]))]));
Lambda[[l + 1, [Eta], 1 + [Xi]]] = If[[Eta] <= l, (!(
*UnderoverscriptBox[([Sum]), ([Xi]2 =
Max[0, [Xi] - ((((2 ((l - [Eta])) + 1)) m +
l - [Eta]))]), (Min[((2 m + 1)), [Xi]])](
*UnderoverscriptBox[([Sum]), ([Xi]1 =
Max[0, \[Xi]2 - m - 1]\), \(m\)]Lambda[\([l, \[Eta],
1 + \[Xi] - \[Xi]2]\)] \((up[
*SubscriptBox[(B), (l + 1)] -
*SubscriptBox[(A), (l + 1)] - ((m + 1)), m - [Xi]1])) ((up[
*SubscriptBox[(A), (l + 1)] - ((
*SubscriptBox[(A), ([Eta])])) - ((m + [Xi] - [Xi]2)) -
1, ((m + 1 + [Xi]1)) - [Xi]2] Binomial[((m +
1 + [Xi]1)), [Xi]2] Binomial[m, [Xi]1]))))),
Cl up[Subscript[B, l + 1] - Subscript[A, l + 1] - (m + 1),
m - [Xi]] Binomial[m, [Xi]]] 1/(m + (l - [Eta] + 3) + [Xi]);
, {l, 1, lmax - 1}, {[Eta], 1, l + 1}, {[Xi],
0, (2 (l + 1 - [Eta]) + 1) m + l + 1 - [Eta]}];
l = lmax - 1;
Print["l,m=", {l, m}];
(Closed form expression--fast)
ex1 = Table[Expand[(!(
*UnderoverscriptBox[([Sum]), ([Eta] = 1), (l)](
*UnderoverscriptBox[([Sum]), ([Xi] =
0), (((2 ((l - [Eta])) + 1)) m +
l - [Eta])]Lambda[([l, [Eta],
1 + [Xi]])] ((up[((k + 1 +
*SubscriptBox[(A), ([Eta])] - ((l - [Eta] +
1)))), ((m + ((l - [Eta] +
2)) + [Xi]))])))) + Cl)], {k, 1, kmax}];
(From definition--slow)
ex2 = Table[Sum[!(
*UnderoverscriptBox[([Product]), ([Eta] = 0), (l - 1)](up[((
*SubscriptBox[(j), ([Eta] + 1)] +
*SubscriptBox[(A), ([Eta] + 1)])), ((m + 1))] up[((
*SubscriptBox[(j), ([Eta] + 1)] +
*SubscriptBox[(B), ([Eta] + 1)])), m])),
Evaluate[
Sequence @@
Table[{Subscript[j, [Eta]],
If[[Eta] == 1, 0, Subscript[j, [Eta] - 1]] + 1, k}, {[Eta],
1, l}]]], {k, 1, kmax}];
(ex1 - ex2) // Expand
During evaluation of In[328]:= l,m={3,2}
Out[336]= {0, 0, 0, 0, 0, 0, 0, 0, 0, 0}
Having said all this my question would be whether it is possible to derive the result $(2)$ using a combinatorial approach?
