Answer
$\dfrac{1}{2}$
Work Step by Step
Since, we know that the Maclaurin Series for $\cos x$ is defined as:
$ \cos x=\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n}}{(2n)!}=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-...$
Thus, we have the form of the given series as $\cos x$ series.
Here, in the given problem $x=\dfrac{\pi}{3}$
Thus, the sum of the series is:
$\cos \dfrac{\pi}{3}=\dfrac{1}{2}$
