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 $y$.


2 Answers 2


$$\int_x^\infty \{ u\} u^{-s-1} du = \int_x^\infty \left( \{ u\}-\frac{1}{2}\right) u^{-s-1} du + \frac{1}{2} \frac{x^{-s}}{s}$$ 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.


If $s=\sigma\in(0,\infty)$, then the answer is positive because

$$\begin{split} \int_{x}^{\infty} \{u\}u^{-\sigma-1}du &\ge \sum_{N\ge x} \int_{N+1/2}^{N+1} \{u\}u^{-\sigma-1}du \\ &\ge\frac{1}{2\sigma}\sum_{N\ge x}\Big( \frac{1}{(N+1/2)^\sigma}-\frac{1}{(N+1)^{\sigma}}\Big) \\ &\ge \frac{1}{4}\sum_{N\ge x} \frac{1}{(N+1)^{1+\sigma}} \\ &\gg x^{-\sigma}. \end{split}$$

  • 3
    $\begingroup$ Indeed, the question is quite easy for real $s$, unlike for $s\in \mathbb{C}$. $\endgroup$
    – user140392
    Jul 1, 2019 at 1:52
  • 2
    $\begingroup$ @OneTwoOne Yes, I just wanted to point this out, because maybe it is still possible to split the integral in some way to get a lower bound for $s\in \mathbb{C}$. The formula was too big for a comment. $\endgroup$ Jul 1, 2019 at 2:14

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