In trying to simplify a distribution function, I stumbled upon this infinite sum involving an (upper) Gamma function. I would believe it can be simplified further, but can't find how.
$$1- \frac{(1-\rho^2)^{v/4}}{\Gamma(\frac{v}{2})} \sum_{k=0}^\infty \left( \frac{ (\frac{\rho}{2})^{2k} }{k! \;(\frac{v+2}{4})_k} \; \Gamma\Big(\frac{v}{2}+2k, \alpha \Big) \right) $$
in which $v$ is integer and a multiple of 2 (so that $v/2$ is also an integer), $\rho$ is a population correlation ($-1\le \rho\le 1$) and $(\frac{v+2}{4})_k$ is the Pochhammer function, that is, $\Gamma\Big((v+2)/4+k \Big)/ \Gamma(\Big(v+2)/4\Big)$.
I tried series expansion of the incomplete Gamma function, to no avail. I also tried replacing the Pochhammer function with various expansion, in particular $(-1)^k k! \binom{\frac{1}{4} (-v-2)}{k}$. I am pretty sure the result will involve another special function, but this is fine, I mostly desire a solution which can be expressed without an infinite sum.
As you noticed all the problem is related to the incomplete gamma function.
To simplify notation, let $x=\frac \rho 2$ and $n=\frac v2$ and write the summation as $$S_n=\sum_{k=0}^\infty \frac{x^{2 k} \,\Gamma \left(\frac{n+1}{2}\right)}{\Gamma (k+1)\, \Gamma\left(k+\frac{n+1}{2}\right)}\Gamma (2 k+n,\alpha )$$
Assuming that we could use the series expansion of $\Gamma (2 k+n,\alpha )$ around $\alpha=0$, we should have
$$S_n=\frac {\Gamma (n)}{\left(1-4 x^2\right)^{\frac n2}}+\sum_{m=0}^\infty (-1)^m \frac{\alpha^{(n+m)}}{m! \,(n+m)}\, _1F_2\left(\frac{n+m}{2};\frac{n+1}{2},\frac{n+m+2}{2};x^2 \alpha ^2\right)$$ For odd values of $m$, the summand can write in terms of Bessel functions of the first kind.
Edit
For odd values of $m=2p+1$, consider
$$G_p=\frac{\alpha ^{2 p+n+1}}{(2 p+1)! (2 p+n+1)} \, _1F_2\left(\frac{2p+n+1}{2};\frac{n+1}{2},\frac{2p+n+3}{2};x^2 \alpha ^2\right)$$
$$G_0=\frac{1}{2} \alpha ^{\frac{n+1}{2}} x^{-\frac{n+1}{2}} \Gamma \left(\frac{n+1}{2}\right) I_{\frac{n+1}{2}}(2 x \alpha )$$ $$G_1=\frac{n-1}{48} \alpha ^{\frac{n+3}{2}} x^{-\frac{n+3}{2}} \Gamma \left(\frac{n-1}{2}\right) \left((n+1) I_{\frac{n+3}{2}}(2 x \alpha )+2 \alpha x I_{\frac{n+5}{2}}(2 x \alpha )\right)$$