For all $0 <Re(s)$ and $s≠1$ with $s=σ+it$ and $1≤N$
$Re(s)$ is the real part of a complex variable, we know:
$$\zeta(s ) = \sum_{1}^N \frac {1} {n^{s}}-\frac {N^{1-s}} {1-s} \tag{1}+r_N(s)$$
$$r_N(s)=-s\int_{N}^{\infty}{\frac{u-\lfloor u \rfloor}{u^{s+1}}du}$$
And an author showed for the remainder term on http://www.maths.manchester.ac.uk/~mdc/MATH31022/2010-11/notes/Notes4Step4.pdf
as:
$$|r_N(s)| ≤ \frac {|s|} {σN^{σ}}$$
QUESTION: If we accept- $s_o$ – as one of the non-trivial zeros of the Riemann zeta function by $0 <Re(s_o)<1$ and with $s_o=σ_o+it_o$ , can we write?
$$|r_N(s_o)| ≠0$$
Best Answer
Can you show that $$r_N(s) = \sum_{n=N}^\infty \int_n^{n+1} ((n+1)^{-s}-x^{-s})dx$$ If so then what do you get from $$(n+1)^{-s}-x^{-s} =(n+1-x) sn^{-s-1}+O(s(s+1)n^{-s-2})$$ and $$\sum_{n=N}^\infty n^{-s-1} = \sum_{n=N}^\infty (s+1) \int_n^\infty t^{-s-2}dt = (s+1) \int_N^\infty\lfloor t -N+1\rfloor t^{-s-2}dt$$