An Analytic Overkill: $\sum_{p \leq n}\frac{1}{p} = \log \log x + A + O(\frac{1}{\log x})$ where A is a constant. Therefore $\sum_{n \leq p \leq n^4} \frac{1}{p} = \log 4 -O(\frac{1}{\log x}) < 4 $ A really poor bound tbh.

Agsh2005 wrote: Therefore $\sum_{n \leq p \leq n^4} \frac{1}{p} = \log 4 -O(\frac{1}{\log x}) < 4 $ Of course this only solves the problem for large $n$, which leaves you with quite an annoying task. No, this is really not the right way to approach the problem. But of course it follows directly from the corresponding problem in Grade 11 by applying the claim there separately to $[n,n^2)$ and $[n^2,n^4)$.

Tintarn wrote: Agsh2005 wrote: Therefore $\sum_{n \leq p \leq n^4} \frac{1}{p} = \log 4 -O(\frac{1}{\log x}) < 4 $ Of course this only solves the problem for large $n$, which leaves you with quite an annoying task. No, this is really not the right way to approach the problem. But of course it follows directly from the corresponding problem in Grade 11 by applying the claim there separately to $[n,n^2)$ and $[n^2,n^4)$. I think this is a result for all n.