Answer
$\frac{\sqrt 3}{2}$
Work Step by Step
Given: $\Sigma_{n=0}^{\infty}{(-1)^{n}}\dfrac{\pi^{2n}}{6^{2n}(2n)!}$
As we know
$cosx=\Sigma_{n=0}^{\infty}(-1)^{n}\dfrac{x^{2n}}{2n!}$
Thus,
$cos(\frac{\pi}{6})=\Sigma_{n=0}^{\infty}{(-1)^{n}}\dfrac{\pi^{2n}}{6^{2n}(2n)!}$
or
$\Sigma_{n=0}^{\infty}{(-1)^{n}}\dfrac{\pi^{2n}}{6^{2n}(2n)!}=\frac{\sqrt 3}{2}$
