Thanks Professor Rivin. I have seen this theorem before. But pardon my ignorance, I don't understand how theorem 1 relates to my problem! I was asking about $\Re(s)>1$. Say, what is the sum $\sum_{n=1}^{\infty}\frac{\{\frac{n}{k}\}}{n^s}$? By the result mentioned in my question, sum is $(\log k)^{-1}$ if $s=1$. I wonder what about $s=2$, can we find a closed form?

Thank you very much for this resourceful explanation. I really didn't expect that, Crammer's conjecture could come here. Still, I am wondering whether or not it is possible to find a bound of $K(N)$ using (or not using) Sieve methods. Or does it always diverge!

I think that $|R(x,N)|\leq\tau(N)$ is a crude bound, as you are losing 'cancellation opportunity' by change of sign of $\mu(n)$. $|R(n,N)|=O(2^{\omega(N)})=O(2^{\log N})$