I have to calculate the following quantity
$$\gamma(8,(R/\alpha)^{1/4})$$
where $\gamma(s,x)$ is the lower incomplete gamma function. Choosing the series representation I'm trying to actually sum the series, but I can't find a compact answer. Let $x=R/\alpha$
$$\gamma(8,x^{1/4})=\Gamma(8)x^{2}e^{-x^{1/4}}\sum_{n=0}^{\infty}\frac{x^{n/4}}{\Gamma(9+n)}$$
I was thinking in manipulations like multiplying the series by the factor of $x^2$ but the point will be in some convenient way of writing
$$\frac{\Gamma(8)}{\Gamma(9+n)} $$
to get the $1/n!$ for the exponential series. So far
$$\frac{\Gamma(8)}{\Gamma(9+n)}=\frac{7!}{(8+n)!}=\frac{7!}{7!8(8+1)...(8+n)} $$
but I can't give the number the way I want. Of course I'm assuming that what I want to do is possible, which may not be the case.
Thanks for your time.
There isn't a "compact" answer.
Since the series is the sum of positive terms, it is quite suitable for a grind-it-out summation.
Letting $y = x^{1/4}$, the series is $\sum_{n=0}^{\infty} \dfrac{y^n}{(n+8)!}$.
The first term is $\dfrac1{8!}$, and the ratio of consecutive terms is $\dfrac{\dfrac{y^{n+1}}{(n+9)!}}{\dfrac{y^n}{(n+8)!}} =\dfrac{y}{n+9} $.
Note that the ratio is greater than $1$ until $n+9 \ge y$, so you have to accumulate at least $y-9$ terms. After you get to this many terms, all subsequent terms decrease, so you can stop when $\dfrac{term}{total} < \epsilon$, where $\epsilon$ determins how accurate you want the answer to be. $\epsilon < 10^{-10}$ is probably good enough.