Is there any approach to solve:
$$\sum\limits_{n=1}^{N}{n^s}$$
where $s \in \mathbb{C}$ and $ 0<\Re(s) < 1$?
Actually I don't need an exact result but just coefficient of the higher exponent, or more explicitly to calculate the next limit:
$$\lim_{N->\infty}{\frac{ \sum\limits_{n=1}^{N\over 2}{{(n+\frac{N}{2})}^s} }{\sum\limits_{n=1}^{N\over 2}{n^s}}}$$
I have an idea on how to proceed for natural exponents, but I don't know how to proceed for $s$ complex.
I expect that the Euler—Maclaurin formula still works, although I'm not totally sure about the estimate of the remainder term: $$f(N)\stackrel{\text{def}}{=}\sum_{k=1}^N k^s= \tfrac{N^{1+s}-1^{1+s}}{1+s} +\tfrac{N^s+1^s}{2} +O(1)$$ Then $$\frac{f(2N)-f(N)}{f(N)}=2^{1+s}-1+o(1)\text{.}$$