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I want to show the following: Let $\delta>0$ be fixed. Then for $\sigma\ge \delta, s\not=1$, $$\sum_{n\le x}n^{-s}=\frac{x^{1-s}}{1-s}+\zeta(s)+O(\tau x^{-\sigma})$$ where $s=\sigma+it,~\tau=|t|+4$ and $f(x)=O(g(x))\Leftrightarrow|f(x)|\le Cg(x)$ for a constant $C$.

Now, I have already proved the identity $$\sum_{n\le x}=\frac{x^{1-s}}{1-s}+\zeta(s)-\frac{\{x\}}{x^s}+s\int_x^\infty\{u\}u^{-s-1}du$$ and obviously $|\int_x^\infty\{u\}u^{-s-1}du|\le \frac{x^{-\sigma}}{\sigma}$. However, I don't know what to do with the prefactor $s$ and where $\tau$ comes from.

user342314
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  • All you need to conclude is $\displaystyle\frac{s}{\sigma} = \mathcal{O}(\tau)$. – reuns Nov 25 '16 at 21:35
  • @user1952009 Yes, I know but I'm not sure why this holds. I have tried the following though: $\frac{|s|}{\sigma}\le\frac{\sigma+|t|}{\sigma}\le1+\frac{|t|}{\delta}\le c_\delta(4+|t|)$ for a constant $c_\delta$ - does this look correct? – user342314 Nov 25 '16 at 21:49
  • Yes use the triangle inequality $|a+ib| \le |a|+|b|$ – reuns Nov 25 '16 at 21:50
  • That's what the first inequality is; seems like I was just thinking slowly today. Thanks! – user342314 Nov 25 '16 at 21:56
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    Also take a look at this question http://math.stackexchange.com/questions/2021975/zeta-functions-asymptotic-zeta-sigmait-mathcalot1-sigma-epsilo/2022183#2022183 – reuns Nov 25 '16 at 23:33

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