Answer
$\frac{\sqrt 2}{2}$
Work Step by Step
Given: $\Sigma_{n=0}^{\infty}(-1)^{n}\dfrac{\pi^{2n+1}}{4^{2n+1}(2n+1)!}$
$\Sigma_{n=0}^{\infty}(-1)^{n}\dfrac{\pi^{2n+1}}{4^{2n+1}(2n+1)!}=sin(\frac{\pi}{4})$
$=\frac{\sqrt 2}{2}$
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