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The solution says that $\sum_{n=0}^\infty \frac{ (-1)^n(\pi)^{2n}}{4^n(2n)!} = 0$. I only know how to prove this converges. What can I use to calculate the sum or where should I start?

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  • $\begingroup$ Hint: Compare the pattern to Taylor of some trigonometric functions. $\endgroup$ – Yves Daoust May 17 '19 at 18:46
  • $\begingroup$ Note that $$\frac{(\pi)^{2n}}{4^n}=\left(\frac{\pi}{2}\right)^{2n}$$ $\endgroup$ – Dave May 17 '19 at 18:46
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$$\sum_{n=0}^\infty \frac{ (-1)^n(\pi)^{2n}}{4^n(2n)!}=\sum_{n=0}^\infty \frac{ (-1)^n(\pi/2)^{2n}}{(2n)!}$$ then use $$\cos x=\sum_{n=0}^\infty \frac{ (-1)^n(x)^{2n}}{(2n)!}$$

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This is simply $\cos(\pi/2)$, since $$\cos(x) =\sum_{n=0}^\infty (-1)^n\frac{x^{2n}}{(2n)!}$$

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$$2\sum_{n=0}^\infty\dfrac{x^{2n}}{(2n)!}=e^x+e^{-x}$$

Here $x=\dfrac{i\pi}2$

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