Chapter 11 - Section 11.10 - Taylor and Maclaurin Series - 11.10 Exercises - Page 809: 84

$\frac{\sqrt 3}{2}$

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 $\Sigma_{n=0}^{\infty}{(-1)^{n}}\dfrac{\pi^{2n}}{6^{2n}(2n)!}=\frac{\sqrt 3}{2}$