## Calculus 8th Edition

series converges for all value of $x$ and interval of convergence is $R$ and radius of convergence is $\infty$
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 value of $x$ and interval of convergence is $R$ and radius of convergence is $\infty$