For each integer $n>2$, we define $p(n)$ to be the smallest prime number that does not divide $n$. Prove that $$\lim_{n\to \infty}\frac{p(n)}{n}=0.$$
My argument is: We only need to prove $$\lim_{n\to \infty,\; n\; \text{is square-free}}\frac{p(n)}{n}=0$$
since $\frac{p(n)}{n}\leq \frac{p(n)}{Rad(n)}=\frac{p(Rad(n))}{Rad(n)}$. Let $p_k$ is the $k-$th prime number, for each square-free number $p_1\dots p_{k-1}<n<p_1\dots p_{k}$, $p(n)$ must be less than $p_{k+1}$ otherwise $p_1\dots p_k|n$. So $\frac{p(n)}{n}\leq \frac{p_k}{p_1\dots p_{k-1}}$. To finish, we must prove
$$\lim_{k\to \infty}\frac{p_k}{p_1\dots p_{k-1}}=0$$
For a moment, I thought this is obvious however I couldn't prove the above limit equals $0$. How can I continue?
Best Answer
Armed with Bertrand's postulate, we know that $p_k\leq2p_{k-1}$, and we have $$ \frac{p_k}{p_1\dots p_{k-1}}\leq \frac{2}{p_1\cdots p_{k-2}} $$ and it is not difficult to see that this goes to $0$ as $k$ increases.