Chapter 11 - Section 11.10 - Taylor and Maclaurin Series - 11.10 Exercises - Page 771: 39

$\Sigma_{n=0}^{\infty}(-1)^{n}\frac{x^{4n+1}}{2^{2n}(2n)!}$ $R=\infty$

$xcos(\frac{1}{2}x^{2})=x\Sigma_{n=0}^{\infty}(-1)^{n}\frac{(\frac{1}{2}x^{2})^{2n}}{2n!}$ $=x\Sigma_{n=0}^{\infty}(-1)^{n}\frac{(\frac{1}{2})^{2n}x^{4n}}{2n!}$ $=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{x^{4n+1}}{2^{2n}(2n)!}$ $R=\infty$