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Jul 6, 2018 at 11:08 comment added juan I don't know. Our knowledge make the assertion unlikely.
Jul 6, 2018 at 10:44 comment added Đào Thanh Oai I am sorry, could you answer that the limmit is true or not true?
Jul 6, 2018 at 9:54 comment added juan If you substitute each $\pi(\cdot)$ by the corresponding $\textrm{Li}(\cdot)$, the resulting function tends to $1$. But the difference $\pi(x)-\textrm{Li}(x)$ has been shown to be $\Omega_{\pm}(\frac{x^{1/2}\log\log\log x}{\log x})$. Therefore one of the error $\pi(x^2)-\textrm{Li}(x^2)$ can be of order $\frac{x}{\log x}\log\log\log x$. Therefore the sup limit of this divided by $\pi(x)$ is $+\infty$. So only if the error in $\pi((x+1)^2)$ and $\pi(x^2)$ collaborate can the limit be $1$.
Jul 6, 2018 at 9:16 comment added Đào Thanh Oai Do You think $\lim_{n \to +\infty } \frac{\pi((n+1)^2)-\pi(n^2)}{\pi(n)}=1$ ??
Jul 6, 2018 at 8:57 history answered juan CC BY-SA 4.0