Answer
The series converges for all values of $x$ and the interval of convergence is $R$. Radius of convergence = $\infty$.
Work Step by Step
Power series for $sinx^{4}=\Sigma_{n=0}^\infty(-1)^{n}\frac{x^{8n+4}}{2n+1!}$
$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{(-1)^{n+1}\frac{x^{8(n+1)+4}}{2(n+1)+1!}}{(-1)^{n}\frac{x^{8n+4}}{2n+1!}}|$
$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{(-1)^{n+1}\frac{x^{8n+12}}{(2n+3)!}}{(-1)^{n}\frac{x^{8n+4}}{2n+1!}}|$
$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|-\frac{x^{8}}{(2n+2)(2n+3)}|$
$=0 \lt 1$
Thus, the series converges for all values of $x$ and the interval of convergence is $R$.
