Denote by $\zeta$ the Riemann zeta function. For $\Re(s)=\sigma>0$, it is well known that


$$\sum\_{n\leq x} n^{-s} = \zeta(s) + \frac{x^{1-s}}{1-s}+ O(x^{-\sigma}).$$


But do there exist infinitely many $x$ such that


$$ \Bigg|\sum\_{n\leq x} n^{-s} - \zeta(s) - \frac{x^{1-s}}{1-s} \Bigg| \gg x^{-\sigma} ?$$


**ADDENDUM:** Since the left-hand side is equal to $\Big|s\int\_{x}^{\infty} \lbrace u \rbrace u^{-s-1}\mathrm{d}u\Big|$, the problem amounts to finding a lower bound for this integral, where $\lbrace y \rbrace$ denotes the fractional part of $y$.



On some analytic property of the Riemann zeta function
$$\int\_x^\infty \{ u\} u^{-s-1} du = \int\_x^\infty \left( \{ u\}-\frac{1}{2}\right) u^{-s-1} du + \frac{1}{2} \frac{x^{-s}}{s}$$ and by integration by parts,


$$\int\_x^\infty \left( \{ u\}-\frac{1}{2}\right) u^{-s-1} du= - \int\_x^\infty 
 \left(\int\_x^u \left( \{ t\}-\frac{1}{2}\right)dt \right) (-s-1) u^{-s-2} du=$$


$$= O\left( |s+1| \int\_x^\infty u^{- \sigma -2 } \right) = O\left( |s+1| x^{-\sigma-1} / |\sigma +1| \right)$$


so the $\frac{1}{2} \frac{x^{-s}}{s}$ term dominates and you get a lower bound.


