Zeros of Riemann Zeta Function

The functional equation for RZC is given by
$$\zeta(s)=2^{s}\pi^{s-1}\sin\left(\frac{\pi s}{2}\right)\Gamma(1-s)\zeta(1-s)$$
By this equation, its easy to see that $\zeta(-2n)=0$ for all $n\in\mathbb{N}$. But if we substitute $s=2n$ for any $n\in\mathbb{N}$, by the functional equation, we got $\zeta(2n)=0$, since $\sin(n\pi)=0$. This contradicts the fact that for $\mathrm{Re}(s)>1$, $\zeta(s)=\sum_{n=1}^{\infty}n^{-s}$, because $\sum_{n=1}^{\infty}n^{-s}>0$ for all $s$ with $\mathrm{Re}(s)>1$.

What am I misunderstanding?

Answer

You are forgetting that $\Gamma(s)$ has poles at all negative integers. Thus you should interpret functional equation literally only where each factor is defined, and may take limits near the poles of $\Gamma(s)$.

These poles are simple poles, so they cancel with the zeros of $\sin(\frac{\pi s}{2})$.