Calculus 8th Edition

$1-\frac{1}{4}x-\Sigma_{n=2}^{\infty}\frac{3.7.......(4n-5)x^{n}}{4^{n}n!}$ and $R=1$
$\sqrt[4] {1-x}=(1-x)^{1/4}=(1+(-x))^{1/4}$ $=1-\frac{1}{4}x-\Sigma_{n=2}^{\infty}\frac{3.7.......(4n-5)x^{n}}{4^{n}n!}$ $\lim\limits_{n \to \infty}|\dfrac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\dfrac{\dfrac{3.7.......(4n-5)(4(n+1)-5)x^{n}}{4^{n+1}(n+1)!}}{\dfrac{3.7.......(4n-5)x^{n}}{4^{n}n!}}|$ $=\lim\limits_{n \to\infty}|\frac{(4n-1)x}{4(n+1)}|$ $=|x|$ The series will converge when $|x|\lt 1$, so $R=1$.