Why doesn't the Riemann Zeta Function have zeroes at positive even integers?

Question

According to the Wikipedia entry on "Riemann Functional Equation", the Zeta Function is equal to itself multiplied by a bunch of stuff, including the term $$\sin(πs/2)$$ This sine term means that for negative even integers, the Zeta Function is 0. But why doesn't this apply to positive even integers? Is it because the negative even integer values of s create a divergent series, but the positive even integer values are convergent?

Answer 1

$$ \zeta(s)=2^s\pi^{s-1}\sin\Big(\frac{\pi s}{2}\Big)\Gamma(1-s)\zeta(1-s) $$

At positive odd integers, $\Gamma(1-s)$ has a pole, but none of the terms to the left of it in the equation have a zero or a pole, which forces $\zeta(1-s)$ to have a zero to "cancel out" the pole. Notice demonstrating $\zeta$ vanishes at negative even integers uses the fact $\zeta$ is nonzero for $s>1$. If we plug in positive even integers, $\Gamma(1-s)$ still has a pole but it is cancelled out by the sine term having a zero of the same order, which forces $\zeta(1-s)$ to actually be nonzero.