I have following conditions and want to show that $\frac {p(n)} n \xrightarrow {n \to \infty} 0$.
\begin{align} (1)&\tilde{\gamma}(n) \xrightarrow {n \to \infty} 0, \\ (2)&\lim_{n \to \infty} n \tilde{\gamma}(n) > 0 ,\\ (3)&p(n):= \min\{q \in \mathbb{N} : q \geq \sqrt{n} \left|\log \tilde{\gamma}(q)\right|\}. \end{align}
My conclusions:
\begin{align} (1),(2) \Rightarrow \exists c,N \quad \forall n>N: 1>\tilde{\gamma}(n) > \frac{c}{n}, \\ (3) \Rightarrow p(n) < \sqrt{n} |\log \tilde{\gamma}(p(n))| + 1. \end{align}
The short proof in my book is "for large $n, p<\sqrt{n}(\log n)^2$, so that $p=o(n)$."
Clearly $ p<\sqrt{n}(\log n)^2 \Rightarrow p=o(n)$. But i couldn't proof that $ p<\sqrt{n}(\log n)^2$.
Do you have any ideas how to proof it?
Note that for $1>\tilde{\gamma}(p)> \frac c p$ it is $|\log \tilde{\gamma}(p) |< |\log \frac c p| $.
Trying to proof: $\sqrt{n} \ge \frac p {a + \log p}$ for some constant $a$.
I know from (1),(2),(3) that \begin{align} \sqrt{n} \geq \frac{p-1}{|\log c| + |\log p|} = \frac{p}{|\log c| + |\log p|} - \frac{1}{|\log c| + |\log p|}. \end{align}
And since $\frac{1}{|\log c| + |\log p|} \xrightarrow {p \to \infty} 0$, it is proofed.
Is this proof correct?