Answer
The series converges to $-1$, that is
$\mathop \sum \limits_{n = 0}^\infty {\left( { - 1} \right)^n}\dfrac{{{{\left( \pi \right)}^{2n}}}}{{\left( {2n} \right)!}} = - 1$
Work Step by Step
From Table 2 we have
$\cos x = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^{2n}}}}{{\left( {2n} \right)!}} = 1 - \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} - \dfrac{{{x^6}}}{{6!}} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for all $x$.
Substituting $x = \pi $, we get
$\cos \pi = \mathop \sum \limits_{n = 0}^\infty {\left( { - 1} \right)^n}\dfrac{{{{\left( \pi \right)}^{2n}}}}{{\left( {2n} \right)!}}$
Since $\cos \pi = - 1$, so the series converges to $-1$, that is
$\mathop \sum \limits_{n = 0}^\infty {\left( { - 1} \right)^n}\dfrac{{{{\left( \pi \right)}^{2n}}}}{{\left( {2n} \right)!}} = - 1$
