# Finding the exact sum of the series $\sum_{n=0}^\infty \frac{ (-1)^n(\pi)^{2n}}{4^n(2n)!}$

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?

• Hint: Compare the pattern to Taylor of some trigonometric functions. – Yves Daoust May 17 '19 at 18:46
• Note that $$\frac{(\pi)^{2n}}{4^n}=\left(\frac{\pi}{2}\right)^{2n}$$ – Dave May 17 '19 at 18:46

$$\sum_{n=0}^\infty \frac{ (-1)^n(\pi)^{2n}}{4^n(2n)!}=\sum_{n=0}^\infty \frac{ (-1)^n(\pi/2)^{2n}}{(2n)!}$$ then use $$\cos x=\sum_{n=0}^\infty \frac{ (-1)^n(x)^{2n}}{(2n)!}$$
This is simply $$\cos(\pi/2)$$, since $$\cos(x) =\sum_{n=0}^\infty (-1)^n\frac{x^{2n}}{(2n)!}$$
$$2\sum_{n=0}^\infty\dfrac{x^{2n}}{(2n)!}=e^x+e^{-x}$$
Here $$x=\dfrac{i\pi}2$$