The question is as follows:

Show that for any $n\ge1$, we have $$\psi(n)=\sum_{p\ge n}\left \lfloor\frac{\log n}{\log p}\right \rfloor \log p$$ where $\psi(x)=\sum_{p^m\le x}\log p$, where the sum ranges over all prime powers that are at most $x$, and $\psi(x)$ can be written as $\psi(x)=\vartheta(x)+\vartheta(x^{1/2})+\vartheta(x^{1/3})+...$ , and $\vartheta(x)=\sum_{p\le x}\log p$.

There is a section prior to this where I had to show that $\psi(n)\le 2\log 2\cdot n+2n^{1/2}\log n$ for any $n\ge1$

In essence, I think I want to find the $m$ such that $p^m\le n$ where $p$ is prime and $n\in\mathbb N$.

This is what I've done so far: $m\log p\le\log n \Rightarrow m\le\frac{\log n}{\log p}\Rightarrow m=\left\lfloor\frac{\log n}{\log p}\right \rfloor$ since $m \in\mathbb Z$, using what I think above.

Does anyone know where I go from here to show the $\psi(n)$ summation, or if I've got the complete wrong idea of the question?

Any help would be greatly appreciated.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.