Riemann zeta function, representation as a limit is it true that $$\displaystyle \zeta(s) = \ \lim_{\scriptstyle a \to 0^+}\ 1 + \sum_{m=1}^\infty e^{\textstyle -s m a } \left\lfloor e^{\textstyle(m+1)a} - e^{\textstyle m a} \right\rfloor$$ my proof :
\begin{equation}F(z) = \zeta(-\ln z) = \sum_{n=1}^\infty z^{\ln n}\end{equation}
which is convergent for $|z| < \frac{1}{e}$. now I consider the functions :
$$\tilde{F}_a(z) = \sum_{n=1}^\infty z^{a \left\lfloor \textstyle \frac{\ln n}{a} \right\rfloor } = 1 + \sum_{m=0}^\infty z^{a n} \left\lfloor e^{a(m+1)} - e^{a m} \right\rfloor $$
because $\displaystyle\lim_{ a \to 0^+} a \left\lfloor \textstyle \frac{\ln n}{a} \right\rfloor = \ln n$, we get that :
$$\lim_{\scriptstyle a \to 0^+} \ \tilde{F}_a(z) = \sum_{n=1}^\infty z^{\ln n} = \zeta(-\ln z)$$

(details)
$\displaystyle\sum_{m=0}^\infty z^{a m} \left\lfloor e^{a(m+1)} - e^{a m} \right\rfloor $
is also convergent for $z < \frac{1}{e}$ because $\displaystyle\sum_{m=0}^\infty (z^a e^a)^{m}$ is convergent for $z < \frac{1}{e}$ and $\displaystyle\sum_{m=0}^\infty z^{am} \left\{e^{a(m+1)} - e^{a m}  \right\} $ is convergent for $z < 1$.
to justify $\displaystyle\sum_{n=1}^\infty z^{a \left\lfloor \textstyle \frac{\ln n}{a} \right\rfloor } = 1 + \sum_{m=1}^\infty z^{a m} \left\lfloor e^{a(m+1)} - e^{a m} \right\rfloor $ : if $\left\lfloor \frac{\ln n}{a}  \right\rfloor = m \ne 0$ then
  $\displaystyle\frac{\ln n}{a} \in [m, m+1[ \implies n \in [ e^{am}, e^{a(m+1)}[ $ . how many different $n$'s is that ? $\left\lfloor e^{a(m+1)} - e^{am}  \right\rfloor $.

$$\begin{align}\left|e^{-nas}\left\lfloor e^{(n+1)a}-e^{na}\right\rfloor\right|&\leq e^{-nas}\left(e^{(n+1)a}-e^{na}\right)\\&=e^{-nas}e^{na}\left( e^{a}-1\right)\\&=\frac{e^a-1}{e^{na(s-1)}}
\end{align}$$
As $a\to 0$, $e^a-1\to 0$. and $e^{na(s-1)}\to 1$. So each term goes to zero as $a\to 0$.
That seems to contradict your conclusion.
Also, why is $$\sum z^{a \left\lfloor \textstyle \frac{\ln n}{a} \right\rfloor } =  \sum z^{a n} \left\lfloor e^{a(n+1)} - e^{a n} \right\rfloor?$$
That seems like a huge leap.
