How to show that the negative binomial CDF converges to the Poisson CDF? (Incomplete beta vs incomplete gamma functions) Question: Is there a straightforward proof of the following relationship between the (lower, non-regularized) incomplete beta function $\mathcal{B}(x; a ,b)$ and the (upper, non-regularized) incomplete gamma function $\Gamma(s,t)$:
$$\mathcal{B}\left(\frac{\alpha}{\alpha + \mu} ; \alpha, k + 1 \right) \sim \frac{\Gamma(\alpha) \Gamma(k+1, \mu)}{\Gamma(\alpha + k + 1)}  \quad \text{as } \alpha \to \infty ,$$
where $f(\alpha) \sim g(\alpha)$ as $\alpha \to \infty$ means $\displaystyle\lim_{\alpha \to \infty} \frac{f(\alpha)}{g(\alpha)}=1$?
Notice in particular that this would be analogous to the relationship $\mathcal{B}(\alpha, k+1) = \frac{\Gamma(\alpha) \Gamma(k+1)}{\Gamma(\alpha + k + 1)}$ that one has for the (complete) beta function and the (complete) gamma function. I believe I was able to show that the above asymptotic relationship would be equivalent to showing the convergence of the negative binomial CDF to the Poisson CDF.
Background+definitions: The negative binomial distribution with "number of successes" $\alpha > 0$, $\alpha \in \mathbb{R}$, expectation/mean $\mu > 0$, $\mu \in \mathbb{R}$, and "probability of success" $\frac{\alpha}{\alpha + \mu}$ has the following CDF for $k \ge 0$, $ k \in \mathbb{N}$:
$$\frac{\mathcal{B}\left(\frac{\alpha}{\alpha + \mu} ; \alpha, k + 1 \right)}{\mathcal{B}(\alpha, k+1)}, \quad \text{where} \quad \mathcal{B}(x; a, b) := \int_{0}^x t^{a-1} (1-t)^{b-1} \mathrm{d}t . $$
The Poisson distribution with expectation/mean $\mu > 0$, $\mu \in \mathbb{R}$ has the following CDF for $k \ge 0$, $k \in \mathbb{N}$:
$$\frac{\Gamma(k+1, \mu)}{\Gamma(k+1)}, \quad \text{where} \quad \Gamma(s,t) := \int_{t}^{\infty} e^{-u} u^{s-1} \mathrm{d}u . $$
Note: One can also show the convergence in distribution using either the PMF or MGF, and that it is easier than using the CDF. However, I want to understand the proof specifically using the CDFs. (As a sanity check that the formulas given for the CDFs are at least plausible.)
Here are some related, but distinct, questions on this website: (1) (2) (3) (4) (5) (6) (7)
As a community wiki answer I typed up an outline of what I tried so far, which either does not work or which I was not able to complete. Either way the calculations involved are not straightforward.

(community wiki) Failed attempt: What we want to show is
$$ \lim_{\alpha \to \infty} \frac{\mathcal{B}\left(\frac{\alpha}{\alpha + \mu} ; \alpha, k + 1 \right)}{\mathcal{B}(\alpha, k+1)} = \frac{\Gamma(k+1, \mu)}{\Gamma(k+1)} . \tag{$\dagger$}$$
Using $\mathcal{B}(\alpha, k+1) = \frac{\Gamma(\alpha) \Gamma(k+1)}{\Gamma(\alpha + k + 1)}$, this is the same as showing
$$\lim_{\alpha \to \infty} \frac{\mathcal{B}\left(\frac{\alpha}{\alpha + \mu} ; \alpha, k + 1 \right)\Gamma(\alpha + k + 1)}{\Gamma(\alpha) \Gamma(k+1)} = \frac{\Gamma(k+1, \mu)}{\Gamma(k+1)}  .$$
Obviously we can can cancel out the $\Gamma(k+1)$'s in the denominators to get
$$\lim_{\alpha \to \infty} \frac{\mathcal{B}\left(\frac{\alpha}{\alpha + \mu} ; \alpha, k + 1 \right)\Gamma(\alpha + k + 1)}{\Gamma(\alpha)} = \Gamma(k+1, \mu) , \tag{**}$$
which means that proving $(\dagger)$ is equivalent to showing
$$\mathcal{B}\left(\frac{\alpha}{\alpha + \mu} ; \alpha, k + 1 \right) \sim \frac{\Gamma(\alpha) \Gamma(k+1, \mu)}{\Gamma(\alpha + k + 1)}  \quad \text{as } \alpha \to \infty , \tag{*}$$
as I had claimed at the beginning of the question.
Based on how analogous $(*)$ is to the identity related the (complete) beta and gamma functions, I have to imagine that the slickest possible proof of $(\dagger)$ would prove $(*)$ directly, using some facts about special functions that I don't know about.
However the closest I've come so far to attempting to prove $(\dagger)$ is by attacking $(**)$ via brute force and not using any facts about special functions (because I don't know many).
The first observation is that the ratio of the $\Gamma$ functions in $(**)$ is the Pochhammer symbol a.k.a. rising factorial :
$$\frac{\Gamma(\alpha + k + 1)}{\Gamma(\alpha)}  = \prod_{i=0}^k (\alpha + i) .$$
This follows using induction from the functional equation that shows that the gamma function interpolates the factorials, $\Gamma(x+1)=x \Gamma(x)$. This is also related to how $\frac{\Gamma(\alpha + k + 1)}{\Gamma(\alpha)} \sim \alpha^{k+1}$ as $\alpha \to \infty$. Therefore we can rewrite $(**)$ as
$$ \lim_{\alpha \to \infty} \left[\mathcal{B}\left(\frac{\alpha}{\alpha + \mu} ; \alpha, k + 1 \right)  \left( \prod_{i=0}^k (\alpha + i) \right) \right] = \Gamma(k+1, \mu) \tag{***} .$$
Then I did the following brute-force change of variables on the integral defining $\mathcal{B}\left(\frac{\alpha}{\alpha + \mu} ; \alpha, k + 1 \right)$ to force its bounds of integration to match those of the integral defining $\Gamma(k+1, \mu)$:
$$ u = \frac{\alpha}{\left( \frac{\alpha}{\alpha + \mu} \right) -t}  - \alpha , \quad \mathrm{d}t = \frac{\alpha}{(\alpha + u)^2} \mathrm{d}u , \quad t = \frac{\alpha(u - \mu)}{(\alpha + \mu)(\alpha + u)}, \quad 1 - t = \frac{\alpha^2 + 2\alpha\mu + \mu u}{(\alpha+\mu)(\alpha + u)} .$$
So in other words I think I showed that
$$\mathcal{B}\left(\frac{\alpha}{\alpha + \mu} ; \alpha, k + 1 \right) = \int_{\mu}^{\infty} \left( \frac{\alpha(u - \mu)}{(\alpha + \mu)(\alpha + u)} \right)^{\alpha - 1} \left( \frac{\alpha^2 + 2\alpha\mu + \mu u}{(\alpha+\mu)(\alpha + u)} \right)^k \frac{\alpha}{(\alpha + u)^2} \mathrm{d}u . $$
Then, using that $\frac{\Gamma(\alpha + k + 1)}{\Gamma(\alpha)} = \prod_{i=0}^k (\alpha + i) \sim \alpha^{k+1}$ as $\alpha \to \infty$, pulling $\frac{\Gamma(\alpha + k + 1)}{\Gamma(\alpha)} \sim \alpha^{k+1}$ inside of the integral, and then simplifying and grouping terms, I think I was able to show that
$$\lim_{\alpha \to \infty} \left[\mathcal{B}\left(\frac{\alpha}{\alpha + \mu} ; \alpha, k + 1 \right)  \left( \prod_{i=0}^k (\alpha + i) \right) \right] = \lim_{\alpha \to \infty} \int_{\mu}^{\infty} \left( \frac{\alpha}{\alpha + u}  \right)^{\alpha + k + 1} \cdot \frac{(\alpha^2 + 2 \alpha \mu + \mu u)^k (u-\mu)^{\alpha - 1}}{(\alpha + \mu)^{\alpha + k -1}} \mathrm{d}u . $$
Now it's straightforward enough to show that
$$ \lim_{\alpha \to \infty} \left( \frac{\alpha}{\alpha + u}  \right)^{\alpha + k + 1} = \lim_{\alpha \to \infty} \left( \frac{\alpha + u}{\alpha}  \right)^{-\alpha} = \left(\lim_{\alpha \to \infty}\left( 1 + \frac{u}{\alpha} \right)^{\alpha} \right)^{-1} = e^{-u} ,$$
so we would actually be done now if it is possible to at least show that
$$ \frac{(\alpha^2 + 2 \alpha \mu + \mu u)^k (u-\mu)^{\alpha - 1}}{(\alpha + \mu)^{\alpha + k -1}} \sim u^k \quad \text{as }\alpha \to \infty \tag{****} .$$
However, that seems to be more wishful thinking on my part than anything else, and I probably made an algebra or other error on the way there, so the result $(****)$ probably is not true. And even if it is, the calculations involved to get there are so tedious and not straightforward and not intuitive that it would not be a very straightforward proof.
On the other hand, if there were some fact about special functions that would allow one to show $(*)$ in one fell swoop, that in my opinion would be a much "slicker" argument. So that's why I phrased the question as asking about $(*)$.
