Answer
$x \cos \ x =\Sigma_{n=0}^{\infty} (-1)^{n} \dfrac{(x)^{2n+1}}{(2n)!}$
Work Step by Step
We know that the Maclaurin series for $\cos x=\Sigma_{n=0}^{\infty} (-1)^{n} \dfrac{(x)^{2n}}{(2n)!}=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+..... $
Multiply the above series by $x$.
$x \cos x=x [1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+.....] $
Thus, our required series is:
$x \cos \ x =\Sigma_{n=0}^{\infty} (-1)^{n} \dfrac{(x)^{2n+1}}{(2n)!}$
