$$\zeta(2) = \sum_{n>0} \frac{(-1)^{n+1} (2n+1) \pi^{2n+2}} { (2n+3)!} $$
How to show this ?
In particular I am intrested in showing it without complex numbers or complex analysis.
Do similar identities exist for $\zeta(3)$, $\zeta(4)$ or $\zeta(5)$ ?
$$\zeta(2) = \sum_{n>0} \frac{(-1)^{n+1} (2n+1) \pi^{2n+2}} { (2n+3)!} $$
We can show (with the taylor series for cos and sin) that
$$\sum_{n=0}^{\infty} \frac{(-1)^{n+1} (2n+1) x^{2n+3}} { (2n+3)!} = x - 2 \sin(x) + x \cos(x)$$
Plugging in $\pi = x$ we get
$$\sum_{n=0}^{\infty} \frac{(-1)^{n+1} (2n+1) \pi^{2n+3}} { (2n+3)!} = \pi - 2 \sin(\pi) + \pi \cos(\pi) = 0$$
and likewise by dividing both sides by $\pi$
$$\sum_{n=0}^{\infty} \frac{(-1)^{n+1} (2n+1) \pi^{2n+2}} { (2n+3)!} = 0$$
Now
$$\sum_{n=0}^{\infty} \frac{(-1)^{n+1} (2n+1) \pi^{2n+2}} { (2n+3)!} - \sum_{n>0} \frac{(-1)^{n+1} (2n+1) \pi^{2n+2}} { (2n+3)!} = [\frac{(-1)^{n+1} (2n+1) \pi^{2n+2}} { (2n+3)!}]_{n=0} = - \frac{\pi^2}{6} = - \zeta(2)$$
hence the proposed value is correct.
QED
