I have with the benefit of context the following result:
$$\sum\limits_{n=0}^{\infty} \sum\limits_{j=0}^{n} \left({m+j-1 \choose j}\left(\frac{\theta_1}{\theta_1+t}\right)^m \left(\frac{t}{\theta_1+t}\right)^{j}\right) \left({m+n-j-1 \choose n-j}\left(\frac{\theta}{\theta+t}\right)^m \left(\frac{t}{\theta+t}\right)^{n-j}\right)=1$$
I have verified this also numerically. Now, how do I go about proving it?
My attempt:
I basically collected all terms together and "simplified" this to (calling the summation $\beta$ since it's related to the false negative rate of the Binomial test applied to the negative binomial distribution):
$$\beta = \sum\limits_{n=0}^{\infty} \sum\limits_{j=0}^{n} \frac{n \choose j}{2m+n-2 \choose m+j-1}{2m-2\choose m-1} \left(\frac{\theta_1\theta}{(\theta_1+t)(\theta+t)}\right)^m \frac{t^n}{(\theta_1+t)^j(\theta+t)^{n-j}}$$
Taking as many terms out of the summations as possible we get:
$$\beta ={2m-2\choose m-1} \left(\frac{\theta_1\theta}{(\theta_1+t)(\theta+t)}\right)^m \sum\limits_{n=0}^{\infty} {2m+n-2\choose n} t^n\sum\limits_{j=0}^{n} \frac{n \choose j}{2m+n-2 \choose m+j-1}\frac{1}{(\theta_1+t)^j(\theta+t)^{n-j}}$$
Now, the inner summation over $n$ becomes a thorn since it involves a ratio of binomial terms. This has not provided any simplification over the original expression.
Note: I secretly don't care so much about the summation in the question since I know it's $1$ (it's basically the sum over entire domain of the PMF of the density of a sum of two negative binomial distributions). The summation I really care about is:
$$\sum\limits_{n=0}^{\infty} \sum\limits_{j=0}^{[\frac{n}{2}]} \left({m+j-1 \choose j}\left(\frac{\theta_1}{\theta_1+t}\right)^m \left(\frac{t}{\theta_1+t}\right)^{j}\right) \left({m+n-j-1 \choose n-j}\left(\frac{\theta}{\theta+t}\right)^m \left(\frac{t}{\theta+t}\right)^{n-j}\right)$$
But hoping working out the mechanics of the original one (which should be easier and I know the answer) will provide insight into this one.
Best Answer
Recall the formula \begin{eqnarray*} \sum_{i=0}^{\infty} \binom{K+i-1}{i} x^{i} =\frac{1}{(1-x)^K}. \end{eqnarray*} You also need to invert the order of the sums \begin{eqnarray*} \sum_{n=0}^{\infty} \sum_{j=0}^{n} \cdots = \sum_{j=0}^{\infty} \sum_{n=j}^{\infty} \cdots \end{eqnarray*} The sums will then decouple into the following $2$ products \begin{eqnarray*} \sum_{j=0}^{\infty} \binom{m+j-1}{j} \left( \frac{\theta_1}{\theta_1+t} \right)^m \left( \frac{t}{\theta_1+t} \right)^j \end{eqnarray*} and if you then let $p=n-j$ \begin{eqnarray*} \sum_{p=0}^{\infty} \binom{m+p-1}{p} \left( \frac{\theta}{\theta+t} \right)^m \left( \frac{t}{\theta+t} \right)^p. \end{eqnarray*} Both of these are easily seen to equal $1$.