Show that for $\displaystyle s\in \mathbb{C}$ and $\displaystyle Re(s) > 0$ the integral $\displaystyle \int_1^\infty \lfloor x\rfloor^{-s} - x^{-s}dx$ converges.
If we denote s =u+iv we have $\displaystyle \lfloor x\rfloor^{-s}-x^{-s} =\int_x^{\lfloor x\rfloor}(-s)t^{-1-s}dt =
\int_{\lfloor x\rfloor}^xst^{-1-s}dt$ hence $\displaystyle |\lfloor x\rfloor^{-s}-x^{-s}|\leq
\int_{\lfloor x\rfloor}^x|s||t^{-1-s}|dt = \int_{\lfloor x\rfloor}^x|s||t^{-1-s}|dt
=\int_{\lfloor x\rfloor}^x|s||t^{-1-u-iv}|dt=\int_{\lfloor x\rfloor}^x|s|t^{-1-u}dt =|s|\frac{-1}u(x^{-u}-\lfloor x\rfloor^{-u})$ hence we only have to show the result if $\displaystyle s$ is a positive real number.
It's true if $\displaystyle s>1$ because $\displaystyle \sum_{n=1}^{+\infty}n^{-s}$ and $\displaystyle \int_1^{+\infty}t^{-s}dt$ are convergent. If $\displaystyle s=1$ use the fact that the sequence $\displaystyle \left\{\sum_{k=1}^n\frac 1k-\ln n\right\}$ is convergent. For $\displaystyle s<1$, use inequalities involving $\displaystyle \int_1^n t^{-s}dt$ and $\displaystyle \sum_{k=1}^nk^{-s}$.
The function $\displaystyle t\mapsto t^{-1-u}$ is decreasing hence $\displaystyle |\lfloor x\rfloor^{-s}-x^{-s}|\leq \frac{|s|}u(\lfloor x\rfloor^{-u}-x^{-u}|)=|s|\int_{\lfloor x\rfloor}^xt^{-1-u}dt\leq |s|(x-\lfloor x\rfloor ){\lfloor x\rfloor}^{-1-u}$. We have
$\displaystyle \begin{align*}
\int_1^{n+1}|\lfloor x\rfloor^{-s}-x^{-s}|dx&\leq |s|\sum_{k=1}^n\int_k^{k+1}
\left(x\lfloor x\rfloor^{-1-u}-\lfloor x\rfloor^{-u}\right)dx\\
&=|s|\sum_{k=1}^n\int_k^{k+1}
\left(xk^{-1-u}-k^{-u}\right)dx\\
&=|s|\left(\sum_{k=1}^nk^{-1-u}\left(\frac{(k+1)^2}2-\frac{k^2}2\right)-k^{-u}\right)\\
&= |s|\left(\sum_{k=1}^nk^{-1-u}\left(k+\frac 12\right)-k^{-u}\right)\\
&= \frac{|s|}2\sum_{k=1}^nk^{-1-u}
\end{align*}$
which is convergent (we don't have to distinguish the case $\displaystyle s>1$ or not, so it's better than my previous work).