A question on the paper by Littell and Folks (1971) I have been reading this very interesting paper but have come across something that I do not quite understand, namely how the asymptotic limit of the quantity $-\frac{1}{n} \log \left(1-F_n^{F} \left(\sqrt{n}t \right) \right)$ is derived.

where $T_n^{F}$ is a test statistic. I understand that under the mull hypothesis p-values are uniformly distributed in (0,1) and hence $-2\log\left( 1-F \left( T_n^{F} \right) \right)$ follows a chi squared distribution but does that come in handy here? 
The result here must come from some approximation although none that I have tried led me to this.
All help is appreciated, thank you.
The paper is called "Asymptotic optimality of Fisher's Method of Combining Independent Tests" and I have only found in jstor:
http://www.jstor.org/stable/2284230?seq=1#page_scan_tab_contents
 A: Most probably there is a much more straightforward way, but here is what I came up with:  
According to the paper $F_n^{(F)}()$ is the cumulative distribution function of $T_n^{(F)}$. This last one is, as said, distributed as "the square root of a chi-square with $2P$ degrees of freedom". This is the chi distribution. Its cummulative distribution function is (for $2p$ degrees of freedom)
$$F_n^{(F)}(z) = \frac {\gamma (p, z^2/2)}{\Gamma (p)}$$
where $\gamma(\,,)$ is the lower incomplete gamma function, and $\Gamma ()$ is the Gamma function. Since $p$ is an integer, we have $\Gamma (p) = (p-1)!$
A series representation of the lower incomplete gamma function is
$$\gamma (p, z^2/2) = (p-1)!\cdot \left(1-e^{-z^2/2}\cdot\left[\sum_{m=0}^{p-1}\frac{(z^2/2)^m}{m!}\right]\right)$$
Then 
$$1- F_n^{(F)}(z) = 1- \frac {(p-1)!\cdot \left(1-e^{-z^2/2}\cdot\left[\sum_{m=0}^{p-1}\frac{(z^2/2)^m}{m!}\right]\right)}{(p-1)!}$$
$$=e^{-z^2/2}\cdot\left[\sum_{m=0}^{p-1}\frac{(z^2/2)^m}{m!}\right]$$
So
$$-\frac {1}{n}\ln \big[1- F_n^{(F)}(\sqrt n t)\big] = \frac 12 \frac {nt^2}{n} - \frac {1}{n}\ln\left(\sum_{m=0}^{p-1}\frac{(nt^2/2)^m}{m!}\right)$$
$$= \frac 12 t^2 - \frac {1}{n}\ln\left(\frac{t^{2p-2}}{2^{p-1}\cdot (p-1)!}n^{p-1}+ o\left(n^{(p-1)}\right)\right)$$
The leading term in the sum inside the logarithm is $O\left(n^{(p-1)}\right)$ so
$$\ln\left[O\left(n^{(p-1)}\right)\right] = O(\ln n)$$
(add and subtract to the whole expression $(1/n)\ln(n^{p-1})$ to see what happens)
Divided by $n$ it goes to zero as $n\rightarrow \infty$ (that's why the whole expression is written in the paper as $o(1)$, because it goes to zero on its own), and so 
$$-\frac {1}{n}\ln \big[1- F_n^{(F)}(\sqrt n t)\big] \rightarrow \frac 12 t^2,\;\;\; n \rightarrow \infty$$
