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.