Answer
$0$
Work Step by Step
Since, we know that the Maclaurin Series for $\sin x$ is defined as:
$ \sin x=\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n+1}}{(2n+1)!}=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+...$
Thus, we have the form of the given series as a $\sin x$ series.
Here, in the given problem $x=\pi$
Thus, the sum of the series is:
$\sin \pi =0$
