# 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 :

$$F(z) = \zeta(-\ln z) = \sum_{n=1}^\infty z^{\ln n}$$

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

• Sorry, didn't see the floors around it when I wrote my answer. But why do you take the big leap to say: $$\tilde{F}_a(z) = \sum_{n=0}^\infty z^{a \left\lfloor \textstyle \frac{\ln n}{a} \right\rfloor } = \sum_{n=0}^\infty z^{a n} \left\lfloor e^{a(n+1)} - e^{a n} \right\rfloor$$ Is there a justification for that step? – Thomas Andrews Jun 5 '15 at 4:38
• Hi, I had the justification above. – reuns Jun 5 '15 at 6:11
• and I edited your answer, of course I saved it if you want it back. sorry that the floor function wasn't discussed more. – reuns Jun 5 '15 at 6:17
• I tried verifying your formula numerically. The sum seems to tend to $0$, or thereabouts. – Lucian Jun 5 '15 at 8:45
• yes it's normal : it's because the first non-zero term comes farther and farther each time your $a$ decreases to $0$. you should use the formula $\tilde{F}_a(z) = \sum_{n=1}^\infty z^{a \lfloor (\ln n)/ a \rfloor}$ instead to get a convergence to $\zeta(-\ln z)$ that you could verify numerically. or maybe you could use a mix of the two : find which $\lfloor e^{a(m+1)} - e^{an}\rfloor$ are non-zero and compute the first 100 non-zero ones. – reuns Jun 5 '15 at 10:11

\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 $a\to 0$.
Also, why is $$\sum z^{a \left\lfloor \textstyle \frac{\ln n}{a} \right\rfloor } = \sum z^{a n} \left\lfloor e^{a(n+1)} - e^{a n} \right\rfloor?$$