I have the following summations:
If $\lambda_1 = \lambda + \delta \lambda$ with $\delta \lambda > 0$ then I know,
$$\beta_1 = \lim_{t \to \infty}\sum\limits_{n=0}^{\infty} \sum\limits_{j=0}^{[\frac n 2]} \left(\frac{e^{-\lambda_1 t}(\lambda_1 t)^j}{j!}\right)\left(\frac{e^{-\lambda t}(\lambda t)^{(n-j)}}{(n-j)!}\right)=0$$
You'll notice that the two terms in the summation are Poisson PMF's. Now, if I replace them with Negative Binomials (here $\theta_1=\theta-\delta \theta$ with $\delta \theta > 0$):
$$\beta_2 = \lim_{t \to \infty}\sum\limits_{n=0}^{\infty} \sum\limits_{j=0}^{[\frac n 2]} \left({m+j-1 \choose j}\left(\frac{\theta_1}{\theta_1+t}\right)^m \left(\frac{t}{\theta_1+t}\right)^{j}\right) \left({m+n-j-1 \choose n-j}\left(\frac{\theta}{\theta+t}\right)^m \left(\frac{t}{\theta+t}\right)^{n-j}\right)$$
I'm pretty sure that $\beta_2$ doesn't go to $0$. Is it possible to get an expression for what $\beta_2$ is (or a proof that it won't go to $0$)?
My attempt:
I only have some vague intuition that we are summing the first $[\frac n 2]$ terms of the first distribution and taking sum product with the last $[\frac n 2]$ terms of the second distribution that has its mean shifted from the first by a large amount (since $t \to \infty$). So, when the first distribution's terms are substantial, the second one's terms will be very close to $0$ and vice-versa. When we replace with negative binomials however, they have much heavier tails and so, there is always a healthy overlap between the two terms. The only thing beyond this vague intuition I have is verification through code that the first summation becomes arbitrarily small as $t$ increases and the second one, when plotted with increasing $t$ looks like the plot below (we can see that it's asymptotically approaching a finite percentage - around 3.5% in this case).
Why do I care about these summations?
We get the first summation if we apply the uniformly most powerful (UMP) test for comparing the rates of two Poisson processes to Poisson processes with slightly different rates. The false positive rate is set to 50% (which is the reason for the $[\frac n 2]$) and the summation describes the false negative rate.
The second summation comes about when we apply the UMP Poisson test to two mixed Poisson processes with prior on $\lambda$ coming from a Gamma distribution (and this results in the count of events becoming negative binomial).