This is a generalization of question Another bizarre convolutional identity. .Let $\lambda^M \ge 0$, $\Lambda \ge 0$, $q \in (0,1)$ and $\theta =2,3,4,\cdots$. In addition to that let $n \in {\mathbb N}$. Now define the following coefficients: \begin{eqnarray} {\mathcal A}_{j,n} &:=& (1-q)^n \cdot \prod\limits_{\xi=1}^n \left(\frac{\frac{\Lambda}{1-q} + \lambda^M + (\xi-1+\zeta_j)_{(\theta)}}{\lambda^M + (\xi + \zeta_j)_{(\theta)}} \right) \end{eqnarray} for $j=1,\cdots,\theta$ and $n\in {\mathbb N}$ and $\left\{ \zeta_j \right\}_{j=1}^\theta$ being roots of the equation $x\cdot (x-1) \cdot (x-\theta+1) + \lambda^M=0$ in the variable $x$. Here $a^{(n)} := a\cdot (a-1) \cdot (a-\theta+1)$ the the lower Pochhammer symbol.
Having defined all that the following identity holds true. \begin{eqnarray} &&\frac{\lambda^M}{\mathcal W} \cdot \sum\limits_{n_1+n_2+\cdots+n_\theta = n} \left[ \prod\limits_{\xi=1}^\theta {\mathcal A}_{\xi,n_\xi} \right]\cdot \left( \sum\limits_{\sigma \in \Pi_\theta, \sigma_j = \theta} \mbox{sign$(\sigma)$} \frac{\prod\limits_{\xi=1,\xi\neq j}^\theta (n_\xi + \zeta_\xi)_{(\sigma_\xi-1)}}{(n - n_j - \zeta_j)} \right) =\\ &&1_{n=0} + 1_{n\ge 1} \frac{\Lambda}{\Lambda + \lambda^M \cdot (1-q)} \cdot (1-q)^n \cdot \frac{\prod\limits_{\xi=0}^{n-1} (\frac{\Lambda}{1-q}+\lambda^M +\xi_{(\theta)})}{\prod\limits_{\xi=1}^n (\lambda^M + \xi_{(\theta)})} \end{eqnarray} Here $\Pi_\theta$ is the set of all permutations of length $\theta$ and the sum in the left hand side runs over all non-negatives $\theta$-tuples of integers $(n_1,n_2,\cdots,n_\theta)$ that sum up to $n$. Finally ${\mathcal W} := \prod\limits_{1 \le \xi \le \eta \le \theta} (\zeta_\xi - \zeta_\xi) $.
The following Mathematica code verifies that identity :
In[352]:=
Clear[nn]; th = RandomInteger[{2, 5}];
{lM, q} = RandomReal[{0, 1}, 2, WorkingPrecision -> 50];
L = RandomReal[{1, 5}, WorkingPrecision -> 50];
(*Roots of the indicial equations*)
zeta = Table[Root[lM + Pochhammer[#1 - th + 1, th] &, i], {i, 1, th}];
(*Coefficients of power series expansions of solutions to the \
homogeneous ODE..*)
AA = Table[
Product[(L/(1 - q) + lM +
Pochhammer[xi - 1 + zeta[[j]] - th + 1, th])/(lM +
Pochhammer[xi + zeta[[j]] - th + 1, th]), {xi, 1, n}] (1 - q)^
n, {j, 1, th}, {n, 0, M}];
(*The Wronskian *)
WW = (-1)^(th + 1) Product[
zeta[[i]] - zeta[[j]], {i, 1, th}, {j, i + 1, th}];
sigma = Permutations[Range[1, th]];
cfs = Table[
Total[Flatten[
Table[lM/
WW With[{j =
Last[Flatten[Position[sigma[[i]], _?(# == th &)]]]},
Signature[
sigma[[i]]] (Product[
AA[[j1, 1 + n[j1] - n[j1 - 1]]] Pochhammer[
n[j1] - n[j1 - 1] + zeta[[j1]] - sigma[[i, j1]] + 2,
sigma[[i, j1]] - 1], {j1, 1,
th}]/((nn - n[j] + n[j - 1] - zeta[[j]]) Pochhammer[
n[j] - n[j - 1] + zeta[[j]] - th + 2, th - 1]))] /.
n[0] :> 0 /. n[th] :> nn,
Evaluate[
Sequence @@
Table[{n[j1], If[j1 == 1, 0, n[j1 - 1]], nn}, {j1, 1,
th - 1}]], {i, 1, Length[sigma]}]]], {nn, 0, 4}];
If[cfs[[1]] == -1, cfs = -cfs];
cfs1 = Join[{1},
L/(L + lM (1 - q)) Table[(1 - q)^
nn Product[(L/(1 - q) + lM + Pochhammer[j1 - th + 1, th])/(lM +
Pochhammer[j1 - th + 2, th]), {j1, 0, nn - 1}], {nn, 1, 4}]]
cfs - cfs1
During evaluation of In[352]:= Part::pkspec1: The expression 1-n[0] cannot be used as a part specification.
During evaluation of In[352]:= Part::pkspec1: The expression 1+n[4] cannot be used as a part specification.
During evaluation of In[352]:= Part::pkspec1: The expression 1-n[0] cannot be used as a part specification.
During evaluation of In[352]:= General::stop: Further output of Part::pkspec1 will be suppressed during this calculation.
Out[361]= {1, 1.552571781271098723342828740166072576457571141824, \
2.858204502867962173959497265074121316357015883823, \
5.26180694429884504388721410448191744303577055378, \
0.3061559870934861729788929398768477083156310159048}
Out[362]= {0.*10^-45, 0.*10^-45, 0.*10^-44, 0.*10^-43, 0.*10^-44}
The identity in question has been obtained in the following way. It is the coefficient at $z^n$ in the power series expansion of the solution to the following ordinary differential equation. \begin{equation} z^{\theta-1} \frac{d^\theta}{d z^\theta} X(z) + \left( \frac{\lambda^M}{z} - \frac{\Lambda}{1-(1-q) \cdot z} \right) \cdot X(z) = \frac{\lambda^M}{z} \end{equation} subject to $X(0)=1$. We can solve the ODE above in two different ways, firstly using the power series expansion method and in doing so we obtain the right hand side of the identity in question. On the other hand we can find the solutions to the homogeneous ODE and out of them construct the solution to the in-homogeneous ODE using the Green's function method. The later method gives the left hand side of the identity in question.
How would you go about proving this identity for every integer value of $n$ without resorting to the apparatus of ordinary differential equations?
This is by no means a full answer to this question but at least step in the right direction. Firstly the left hand side can be written as follows: \begin{eqnarray} lhs[n] = \frac{\lambda^M}{{\mathcal W}} \cdot \sum\limits_{n_1+n_2+\cdot+n_\theta=n} \left(\prod\limits_{\xi=1}^\theta {\mathcal A}_{\xi,n_\xi} \right) \cdot \left| \begin{array}{rrrr} 1 & 1 & \cdots & 1 \\ (n_1+\zeta_1)_{(1)} & (n_2+\zeta_2)_{(1)} & \cdots & (n_\theta+\zeta_\theta)_{(1)} \\ \vdots & \vdots & \vdots \\ (n_1+\zeta_1)_{(\theta-2)} & (n_2+\zeta_2)_{(\theta-2)} & \cdots & (n_\theta+\zeta_\theta)_{(\theta-2)} \\ \frac{1}{(n-n_1-\zeta_1)} & \frac{1}{(n-n_2-\zeta_2)} & \cdots & \frac{1}{(n-n_\theta-\zeta_\theta)} \end{array} \right|\quad (i) \end{eqnarray} Now since $\lambda^M = \prod\limits_{\xi=1}^\theta (-\zeta_\xi)$ we immediately see that $lhs[0]=1$ as it should be. Otherwise we have: \begin{eqnarray} lhs[n] = \frac{\lambda^M}{\mathcal W} \sum\limits_{1 \le j_1,\le j_2 \le \cdots \le j_n \le \theta} \left(\prod\limits_{\xi=1}^s {\mathcal A}_{{\tilde j}_\xi,{\tilde d}_\xi} \right) \cdot \frac{\prod\limits_{1 \le \xi \le \eta \le \theta} (\zeta_\xi - \zeta_\eta + \sum\limits_{\lambda=1}^n (-1_{\eta=j_\lambda} + 1_{\xi=j_\lambda})}{\prod\limits_{\xi=1}^n(\zeta_\xi- \sum\limits_{\lambda=1}^n 1_{\xi \neq j_\lambda})} \quad (ii) \end{eqnarray} for $n\ge 1$. Here we have: \begin{equation} \left(j_1,j_2,\cdots,j_\theta\right) = \left( \underbrace{{\tilde j}_1,\cdots,{\tilde j}_1}_{{\tilde d}_1}, \underbrace{{\tilde j}_2,\cdots,{\tilde j}_2}_{{\tilde d}_2}, \cdots, \underbrace{{\tilde j}_s,\cdots,{\tilde j}_s}_{{\tilde d}_s} \right) \end{equation} and the indices on the right hand side $\left\{ {\tilde j}_\xi \right\}_{\xi=1}^s $ above are all distinct.
Now expression $(ii)$ would be the starting point for completing the proof. Let us do the case of $n=1$ in here. We have: \begin{eqnarray} lhs[1] &=& \sum\limits_{j=1}^{\theta} {\mathcal A}_{j,1} \cdot \frac{\zeta_j-1}{\zeta_j} \cdot \prod\limits_{\xi\neq j} \left(\frac{1+\zeta_j - \zeta_\xi}{\zeta_j - \zeta_\xi} \right) \\ &=& \sum\limits_{j=1}^{\theta} {\mathcal A}_{j,1} \cdot \frac{\zeta_j-1}{\zeta_j} \cdot \frac{\left[ (\zeta_j + 1)_{(\theta)} + \lambda^M\right]} {\prod\limits_{\xi\neq j}(\zeta_j - \zeta_\xi) } \\ &=& \Lambda \cdot \sum\limits_{j=1}^\theta \left( 1 - \frac{1}{\zeta_j} \right) \cdot \frac{1}{\prod\limits_{\xi\neq j} (\zeta_j - \zeta_\xi)}\\ &=& \frac{\Lambda}{\lambda^M} \end{eqnarray} as it should be.
Now, let us do the case $n=2$. We have: \begin{eqnarray} %First line. &&lhs[2]=\frac{\lambda^M}{(\lambda^M + 2_{(\theta)})} \cdot \sum\limits_{1 \le j_1 \le j_2 \le \theta} \left[ \right. \\ && \left. 1_{j_1 \neq j_2} {\mathcal A}_{j_1,1} {\mathcal A}_{j_2,1} \frac{-(\zeta_{j_1} - \zeta_{j_2})^2}{1-(\zeta_{j_1} - \zeta_{j_2})^2} \frac{(\zeta_{j_1} - 2)}{(\zeta_{j_1}-1)} \frac{(\zeta_{j_2} - 2)}{(\zeta_{j_2}-1)} \cdot \prod\limits_{\lambda=1}^2 \left( \prod\limits_{\xi\neq j_\lambda} \left( \frac{1+\zeta_{j_\lambda}-\zeta_\xi}{\zeta_{j_\lambda} - \zeta_\xi}\right) \right) + \right. \\ && \left. +1_{j_1=j_2} {\mathcal A}_{j_1,2} \cdot \frac{\zeta_{j_1} - 2}{\zeta_{j_1}} \cdot \prod\limits_{\xi \neq j_1} \left( \frac{2+\zeta_{j_1} - \zeta_\xi}{\zeta_{j_1} - \zeta_\xi}\right) \cdot \frac{1}{2} \right. \\ &&\left. \right] = \\ %Second line &&\frac{\lambda^M}{(\lambda^M + 2_{(\theta)})} \cdot \sum\limits_{1 \le j_1 \le j_2 \le \theta} \left[ \right. \\ && \left. 1_{j_1 \neq j_2} \Lambda^2 \frac{-(\zeta_{j_1} - \zeta_{j_2})^2}{1-(\zeta_{j_1} - \zeta_{j_2})^2} \frac{(\zeta_{j_1} - 2)}{(\zeta_{j_1}-1)} \frac{(\zeta_{j_2} - 2)}{(\zeta_{j_2}-1)} \cdot \prod\limits_{\lambda=1}^2 \left( \prod\limits_{\xi\neq j_\lambda} \left( \frac{1}{\zeta_{j_\lambda} - \zeta_\xi}\right) \right) + \right. \\ && \left. +1_{j_1=j_2} \Lambda \left( \frac{\Lambda}{(1+\zeta_{j_1})_{(\theta)} + \lambda^M } +(1-q) \right) \cdot \frac{\zeta_{j_1} - 2}{\zeta_{j_1}} \cdot \prod\limits_{\xi \neq j_1} \left( \frac{1}{\zeta_{j_1} - \zeta_\xi}\right) \cdot \frac{1}{2} \right. \\ &&\left. \right] = \\ %Third line %Fourth line &&\frac{\lambda^M}{(\lambda^M + 2_{(\theta)})} \cdot \sum\limits_{1 \le j_1 \le j_2 \le \theta} \left[ \right. \\ && \left. 1_{j_1 \neq j_2} \Lambda^2 \frac{-(\zeta_{j_1} - \zeta_{j_2})^2}{1-(\zeta_{j_1} - \zeta_{j_2})^2} \frac{(\zeta_{j_1} - 2)}{(\zeta_{j_1}-1)} \frac{(\zeta_{j_2} - 2)}{(\zeta_{j_2}-1)} \cdot \prod\limits_{\xi\neq j_1} \left( \frac{1}{\zeta_{j_1} - \zeta_\xi}\right) \prod\limits_{\xi\neq j_2} \left( \frac{1}{\zeta_{j_2} - \zeta_\xi} \right) + \right. \\ && \left. +1_{j_1=j_2} \Lambda \left( \frac{\Lambda}{(1+\zeta_{j_1})_{(\theta)} + \lambda^M }\right) \cdot \frac{( - 2)}{\zeta_{j_1}} \cdot \prod\limits_{\xi \neq j_1} \left( \frac{1}{\zeta_{j_1} - \zeta_\xi}\right) \cdot \frac{1}{2} \right. \\ &&\left. \right] + \frac{\Lambda(1-q)}{(\lambda^M + 2_{(\theta)})} = \\ &&\left. \frac{\Lambda (\Lambda + (1-q)(\lambda^M + 1_{(\theta)}))}{(\lambda^M+1_{(\theta)})(\lambda^M + 2_{(\theta)})} \right. \end{eqnarray} In the first line we used the identity $\prod\limits_{\xi=1}^\theta (2- \zeta_\xi) = (\lambda^M + 2_{(\theta)})$ and we cancelled out common factors in the numerator and in the denominator. In the second line we used the fact that : \begin{eqnarray} {\mathcal A}_{j,1} &=& \frac{\Lambda}{[\lambda^M + (1+\zeta_j)_{(\theta)}]} \\ {\mathcal A}_{j,2} &=& \Lambda \cdot \frac{\Lambda + (1-q) (\lambda^M + (1+\zeta_j)_{(\theta)} )}{(\lambda^M + (1+\zeta_j)_{(\theta)}) (\lambda^M + (2+\zeta_j)_{(\theta)})} \end{eqnarray} In the third line we carried out some partial summation and in the last line we used a summation identity How do I prove that a double sum has a neat closed form? .
We endeavor to deal with the cases $n\ge 3$ accomplish this in the future.