## University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{1}{2}$
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: $\cos \dfrac{\pi}{3}=\dfrac{1}{2}$