Answer
$\frac{x-sinx}{x^{3}}=\frac{1}{6}-\frac{1}{120}x^{2}
+\frac{1}{5040}x^{4}-...=\Sigma_{n=1}^{\infty}(-1)^{n+1}\frac{{(x)}^{(2n-2)}}{(2n+1)!}$, for all values of $x$
$R=\infty$
Work Step by Step
$\frac{x-sinx}{x^{3}}=\frac{1}{x^{2}}-\frac{1}{x^{3}} \cdot sinx$
$=\frac{1}{x^{2}}-\frac{1}{x^{3}} \Sigma_{n=0}^{\infty}(-1)^{n}\frac{{(x)}^{2n+1}}{(2n+1)!}$
$=\frac{1}{x^{2}}+\Sigma_{n=0}^{\infty}(-1)^{n+1}\frac{{(x)}^{(2n+1)-3}}{(2n+1)!}$
$=\frac{1}{x^{2}}+\Sigma_{n=0}^{\infty}(-1)^{n+1}\frac{{(x)}^{(2n-2)}}{(2n+1)!}$
$=\Sigma_{n=1}^{\infty}(-1)^{n+1}\frac{{(x)}^{(2n-2)}}{(2n+1)!}$
$R=\infty$
