Questions about the prime counting function

[riemannian]

greetings . i have a couple of questions about the prime counting function .
when $\pi _{0}(x)$ changes by 1, then it's logical to assume that it should happen at a prime argument . meaning :

$$\lim_{\xi\rightarrow 0}\pi _{0}(x+\xi )-\pi _{0}(x-\xi )=1$$

implies that x is a prime .
is this a true assumption ?

according to the literature, we can expand $\pi_{0}(x)$ using the riemann R function .

$$R(x) = 1+\sum_{k=1}^{\infty}\frac{(ln x)^{k}}{k!k\zeta (k+1)}$$

$$\pi_{0}(x)= R(x)-\sum_{\rho} R(x^{\rho })-\frac{1}{lnx}-\frac{1}{\pi}tan^{-1}\left( \frac{\pi}{lnx}\right) = \sum_{k=1}^{\infty}\frac{(lnx)^{k}[1-\sum \rho^{k} ]}{k!k\zeta(k+1)}-\frac{1}{lnx}-\frac{1}{\pi}tan^{-1}\left( \frac{\pi}{lnx}\right)$$$\rho$ being the nontrivial zeros of the zeta function .
is this correct ?? i mean , is the expansion correct ??

 

[riemannian]

i think i got my answer .

$$\Pi _{0}(x)=\sum_{n=1}^{\infty}\frac{1}{n}\pi_{0}(x^{1/n})$$

and $\Pi_{0}(x)$ does change by 1 at primes .

now i am intrigued by the terms $\sum \rho^{k}$ , it seems to me we don't need to know $\rho$ themselves to evaluate $\Pi_{0}(x)$ , we just need to evaluate $\sum \rho^{k}$ . is this correct ??