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Is there a way to show that $\sqrt{p_{n}} < n$?

In this article, I show that $f_{2}(x)=\frac{x}{ln(x)} - \sqrt{x}$ is ascending, for $\forall x\geq e^{2}$. As a result, $\forall n \geq 3$ $\frac{p_{n}}{ln(p_{n})} - \sqrt{p_{n}}\leq \frac{p_{n+1}}{ln(p_{n+1})} - \sqrt{p_{n+1}}$ Also (and as a result), $\forall n \geq 3$ $ \frac{p_{n}}{ln(p_{n})} - \sqrt{p_{n}} > 0$ Or $ \frac{\pi (p_{n})}{p_{n}/ln(p_{n})} < \frac{\pi (p_{n})}{\sqrt{p_{n}}}$

According to PNT $\displaystyle\smash{\lim_{n \to \infty }}\frac{\pi (p_{n})}{p_{n}/ln(p_{n})}=1$ Or, $\forall \varepsilon >0$, $\exists N(\varepsilon )$: $\forall n>N(\varepsilon )$ $1- \varepsilon < \frac{\pi (p_{n})}{p_{n}/ln(p_{n})} < 1+ \varepsilon$ Or $1- \varepsilon < \frac{\pi (p_{n})}{p_{n}/ln(p_{n})} < \frac{\pi (p_{n})}{\sqrt{p_{n}}}$ As a result $\forall \varepsilon >0$, $\exists N(\varepsilon )$: $\forall n>N(\varepsilon )$ $(1 - \varepsilon ) \cdot \sqrt{p_{n}} < \pi (p_{n}) = n$

But this is not enough.

Interestingly, Andrica's conjecture is true iff function $f_{4}(x)=\pi (x) - \sqrt{x}$ is strictly ascending ($x < y \Rightarrow f(x) < f(y)$) for prime arguments.

If $f_{4}(p_{n}) < f_{4}(p_{n+1})$ then $\pi (p_{n}) - \sqrt{p_{n}} < \pi (p_{n+1}) - \sqrt{p_{n+1}}$ Or $\sqrt{p_{n+1}} - \sqrt{p_{n}} < \pi (p_{n+1}) - \pi (p_{n}) =1$

And vice-versa, if $\sqrt{p_{n+1}} - \sqrt{p_{n}} < 1$ Then $-\sqrt{p_{n}} < -\sqrt{p_{n+1}} + 1$ Or $\pi (p_{n})-\sqrt{p_{n}} < \pi (p_{n}) + 1 -\sqrt{p_{n+1}} = \pi (p_{n+1}) -\sqrt{p_{n+1}}$

So, if Andrica's conjecture is true then $\forall n \geq 3$ $\pi (p_{n})-\sqrt{p_{n}} > 0$ Or $\sqrt{p_{n}} < \pi (p_{n})= n$

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    Even relatively crude bounds on $\pi(x)$, such as $\pi(x)\gt \frac{x}{\log x}$ are enough. Or are you looking for an elementary argument?2012-10-03

2 Answers 2

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The following upper bound for $p_{n}$ holds for $n\ge 6$: $ \frac{p_{n}}{n} < \ln n + \ln \ln n=\ln(n\ln n) < n, $ so $p_n < n^2$ in those cases. It clearly also holds for $p_2=3<4$, $p_3=5<9$, $p_4=7<16$, and $p_5=11<25$ (though it fails for $p_1=2\not<1$).

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    Nice ... I have missed this part http://en.wikipedia.org/wiki/Prime_number_theorem#Approximations_for_the_nth_prime_number2012-10-03
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This follows from Rosser's result that except at the beginning, $\pi(x)\gt \frac{x}{\log x}.$ Just put $x=p_n$. We need to do hand calculation until $\log x>\sqrt{x}$, which happens early.

Remark: We used Rosser's not so easy result only for convenience. The lower bound on $\pi(x)$ obtained by Chebyshev in the middle of the $19$th century, using an "elementary" argument, is already enough.

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    I've missed this too http://en.wikipedia.org/wiki/Prime_number_theorem#Bounds_on_the_prime-counting_function, the Vallée-Poussin proof's ... so, basically \pi (x) > \frac{x}{ln(x) - (1-\varepsilon )} > \frac{x}{ln(x)} > \sqrt{x} from some $x$ onwards2012-10-03