Playing with some infinite products I came up with this problem, that I'm not able to figure it out by myself. Moreover in the internet it doesn't seem to appear anywhere. Maybe it is just an easy consequence of properties of characters that I'm not aware of, anyway thank you in advance for any help/answers/suggestions.

All of us know (it is fairly easy to see) that $$\lim_{n \rightarrow \infty} \frac{1 \cdot 3\cdot \dots \cdot (2n-1)}{2 \cdot 4 \cdot \dots \cdot (2n)}=0$$

Now this fact could be reformulated in this fashion: let $$\chi:\mathbb{Z}/2 \mathbb{Z} \rightarrow \mathbb{C}^*$$ the only non trivial character of $\mathbb{Z}/2 \mathbb{Z}$, then the expression $$\lim_{n \rightarrow \infty}(\prod_{i=1}^{n}[2(i-1)+1]^{\chi(2(i-1)+1 \: \text{mod} 2\mathbb{Z})}\cdot \dots \cdot [2i]^{\chi(2i \: \text{mod} s\mathbb{Z})})^{-1}$$ More generally, we can perform this construction for every $s \in \mathbb{N}$.

Indeed all we have to do is to consider a non trivial character $$\chi:\mathbb{Z}/s \mathbb{Z} \rightarrow \mathbb{C}^*$$ and consider the limit $$\lim_{n \rightarrow \infty}(\prod_{i=1}^{n}[s(i-1)+1]^{\chi(s(i-1)+1 \: \text{mod} s\mathbb{Z})}\cdot \dots \cdot [si]^{\chi(si \: \text{mod} s\mathbb{Z})})^{-1}$$

Now it is true that:

1. The value of the limit is finite for every $s$ and every non trivial character $\chi$?
2. If so the value of the limit depends only on $s$ or also on $\chi$?
3. The limit is always a real number? (Possibly $0$?)