Chapter 11 - Section 11.8 - Power Series - 11.8 Exercises - Page 752: 42

Radius of convergence is $\sqrt R$

Let $x^{2}=y$, then $\sum_{n=1}^{\infty}c_{n}x^{2n}=\sum_{n=1}^{\infty}c_{n}y^{n}$ So it is convergent for $-R\lt y\lt R$. Therefore, for $x$ it is convergent if $-\sqrt R\lt x\lt \sqrt R$ Hence, the radius of convergence is $\sqrt R$.