I've been searching for a while now, in several directions, so I end up asking here. My question is: Does $\left|\sum\limits_{k=1}^{2n}k^{-s}\right|-\left|2^{1-s}\sum\limits_{k=1}^{n}k^{-s}\right|$ diverge, if $1/2<\Re(s)<1$?
In my research, it is assumed that both $\left|\sum\limits_{k=1}^{2n}k^{-s}\right|$ and $\left|\sum\limits_{k=1}^{n}k^{-s}\right|$ diverge, and that $\left|\sum\limits_{k=1}^{2n}k^{-s}\right|^2-\left|2^{1-s}\sum\limits_{k=1}^{n}k^{-s}\right|^2$ converges to $0$. We also know that $\left|\sum\limits_{k=1}^{2n}k^{-s}-2^{1-s}\sum\limits_{k=1}^{n}k^{-s}\right|$ converges to $0$ but it may not help that much.
Any ideas?
A comment that got too long.
Note that one has the asymptotic formula $\sum\limits_{k=1}^{n}k^{-s}=\frac{n^{1-s}}{1-s}+O(n^{-\sigma}), n \ge |t|/(2\pi), s=\sigma+it, \sigma \ge 1/2$ (uniform in $\sigma \ge 1/2$ say); see Titchmarsh.
From this one can easily see $\left|\sum\limits_{k=1}^{2n}k^{-s}\right|-\left|2^{1-s}\sum\limits_{k=1}^{n}k^{-s}\right|=O(n^{-\sigma}), n \ge t/(2\pi)$