$1+\frac{1}{4}+\frac{1.3}{4.8}+\frac{\mathrm{1.3.5}}{\mathrm{4.8.12}}+\mathrm{....}=$

$\begin{array}{l}1+\frac{1}{4}+\frac{1.3}{4.8}+\frac{\mathrm{1.3.5}}{\mathrm{4.8.12}}+\cdots \\ =1+\frac{1}{4}+\frac{1(1+2)}{4.8}+\frac{1.(1+2)(1+2\left(2\right))}{\mathrm{4.8.12}}+\cdots \\ p=1,q=2\\ \frac{p}{1!}\left(\frac{x}{q}\right)=\frac{1}{4}\end{array}$

$\begin{array}{l}\Rightarrow \frac{x}{2}=\frac{1}{4}\\ \Rightarrow x=\frac{1}{2}\\ {(1-x)}^{\frac{-p}{q}}={(1-\frac{1}{2})}^{\frac{-1}{2}}\\ ={\left(\frac{1}{2}\right)}^{\frac{-1}{2}}\\ ={\left(2\right)}^{\frac{1}{2}}\\ =\sqrt{2}\end{array}$