Answer
$\dfrac{\pi }{2} - \dfrac{{{\pi ^3}}}{{{2^3}\cdot 3!}} + \dfrac{{{\pi ^5}}}{{{2^5}\cdot 5!}} - \dfrac{{{\pi ^7}}}{{{2^7}\cdot 7!}} + \cdot\cdot\cdot = 1$
Work Step by Step
From Table 2 of Section 11.7, we know that
$\sin x = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^{2n + 1}}}}{{\left( {2n + 1} \right)!}} = x - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^5}}}{{5!}} - \dfrac{{{x^7}}}{{7!}} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for all $x$.
Substituting $\dfrac{x}{2}$ for $x$ in the series above gives
$\sin \dfrac{x}{2} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^{2n + 1}}}}{{{2^{2n + 1}}\left( {2n + 1} \right)!}}$
Setting $x = \pi $, we obtain
$\sin \dfrac{\pi }{2} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{\pi ^{2n + 1}}}}{{{2^{2n + 1}}\left( {2n + 1} \right)!}} = \dfrac{\pi }{2} - \dfrac{{{\pi ^3}}}{{{2^3}\cdot3!}} + \dfrac{{{\pi ^5}}}{{{2^5}\cdot5!}} - \dfrac{{{\pi ^7}}}{{{2^7}\cdot7!}} + \cdot\cdot\cdot$
Since $\sin \dfrac{\pi }{2} = 1$, so
$\dfrac{\pi }{2} - \dfrac{{{\pi ^3}}}{{{2^3}\cdot 3!}} + \dfrac{{{\pi ^5}}}{{{2^5}\cdot 5!}} - \dfrac{{{\pi ^7}}}{{{2^7}\cdot 7!}} + \cdot\cdot\cdot = 1$
