Find the sum of ∑n=0∞(−1)nπ2n(2n)!\sum_{n=0}^{\infty} \frac{(-1)^n \pi^{2n}}{(2n)!}∑n=0∞(2n)!(−1)nπ2n.
cos(π)=−1\cos(\pi) = -1cos(π)=−1
sin(π)=0\sin(\pi) = 0sin(π)=0
e−πe^{-\pi}e−π
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