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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?

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1 Answer 1

Considering $t=\sqrt{n}$ and $x(t)=p(n)$, the hypothesis is that $x(t)\leqslant t\cdot(a+\log x(t))$ for some constant $a$ and every $t\geqslant t_0$ and the conclusion to reach is that $x(t)\leqslant t\cdot(\log t)^2$ for every $t$ large enough.

The function $u:\xi\mapsto\xi/(a+\log\xi)$ being increasing on $\xi\geqslant\mathrm e^{1-a}$, if $x(t)\geqslant t\cdot(\log t)^2$, then $$ t\geqslant u(x(t))\geqslant u(t\cdot(\log t)^2)=t\cdot(\log t)^2/(a+\log t+2\log\log t)\sim t\cdot\log t, $$ which is absurd for $t$ large enough. Hence $x(t)\leqslant t\cdot(\log t)^2$ for every $t$ large enough (and in fact $x(t)\leqslant t\cdot(\log t)^{1+\varepsilon}$ for every $t$ large enough, for every $\varepsilon\gt0$).

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Tank you for your answer. So now i have only to proof that $x(t) \le t(a+\log x(t))$ for some constant $a$. –  Runje Nov 23 '12 at 14:09
    
I tried to proof this hypothesis in my edit. –  Runje Nov 23 '12 at 14:23
    
About your edit, you should first modify the incorrect definition in (3), replacing $\geqslant$ by $\leqslant$ and adding the mention "size of" before the set defining $p(n)$. Then, as I indicate clearly in my post, $p(n)\leqslant\sqrt{n}(\log p(n)-\log c)$ by hypothesis, hence I fail to see what should be proved here. –  Did Nov 24 '12 at 9:32
    
I am sorry, i forgot an important part of the definition of $p(n)$. I hope it makes now more senses what i did there, because i want to proof it with these conditions. –  Runje Nov 26 '12 at 9:43

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