Answer
$x \sin \ x =\Sigma_{n=0}^{\infty} (-1)^{n} \dfrac{(x)^{2n+2}}{(2n+1)!}$
Work Step by Step
We know that the Maclaurin series for $\sin x=\Sigma_{n=0}^{\infty} (-1)^{n} \dfrac{(x)^{2n+1}}{(2n+1)!}=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+..... $
Multiply the above series by $x$.
$x \sin x=x [x-\dfrac{(x)^3}{3!}+\dfrac{(x)^5}{5!}+.....] $
Thus, our required series is:
$x \sin \ x =\Sigma_{n=0}^{\infty} (-1)^{n} \dfrac{(x)^{2n+2}}{(2n+1)!}$
