analysis - 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 ? $\left\lfloor e^{a(m+1)} - e^{am} \right\rfloor$.