Enhance your knowledge!

(This is just the worthless standard solution) :

$\quad\displaystyle\sum_{n=0}^{\infty}\frac{\left(-\frac{\pi^2}4\right)^n}{(2n+1)!}$

$=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n\left(\frac{\pi}2\right)^{2n}}{(2n+1)!}$

$=\displaystyle\frac2\pi\sum_{n=0}^{\infty}\frac{(-1)^n\left(\frac{\pi}2\right)^{2n+1}}{(2n+1)!}$

$=\displaystyle\frac2\pi\sin\left(\frac\pi2\right)$

$=\displaystyle\frac2\pi$

$=\displaystyle\left(\frac2\pi\right)^1$

This problem was to enhance the knowledge of cardinal sine function.

$\sum _{ n=0 }^{ \infty }{ \frac { { \left( -{ x }^{ 2 } \right) }^{ n } }{ \left( 2n+1 \right) ! } } =sinc\left( x \right) \\ =\frac { sin(x) }{ x } \\$

This is not a solution.