Incomplete gamma function

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.

• If any of the answeres below were useful to you, then you should upvote all answers you find useful and accept the one that was most useful to you. It is a way to show that you have found the answer to your question and it shows your appreciation. Now it seems like you still need help. If answers are not useful to you, then it helps if you say why not. This helps others to help you. For more information read this. – gebruiker Mar 15 '16 at 12:27

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.
• Thanks. Actually I was interested in an analytic formula, we're not going to do numerical work as fas as I know. I found that $$y^{8}\sum_{n=0}^{\infty}\frac{y^{n}}{(n+8)!}=e^{y}-\sum_{n=0}^{7}\frac{y^{n}}{n!}$$ – Jorge Lavín Sep 28 '13 at 8:48