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Prove that \(x\) is both rational and an integer.\[\lim_{n\to\infty}\left(\ln n\frac{n}{\pi\left(n\right)}\right)=x\]The function \(\pi(n)\) counts the number of primes less than or equal to \(n\).
I've seen this equation somewhere before, but I can't remember where.
IIRC, \(x\) evaluates to \(1\). Mysterious!
 one year ago
 one year ago
Prove that \(x\) is both rational and an integer.\[\lim_{n\to\infty}\left(\ln n\frac{n}{\pi\left(n\right)}\right)=x\]The function \(\pi(n)\) counts the number of primes less than or equal to \(n\). I've seen this equation somewhere before, but I can't remember where. IIRC, \(x\) evaluates to \(1\). Mysterious!
 one year ago
 one year ago

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badreferencesBest ResponseYou've already chosen the best response.0
Furthermore, using Mathematica to algorithmically evaluate \(x\) by using arbitrarily large values of \(n\) does not converge to \(1\)!
 one year ago

mukushlaBest ResponseYou've already chosen the best response.4
* i love to know what is the answer
 one year ago

mukushlaBest ResponseYou've already chosen the best response.4
i found something similar \[\lim_{n\to\infty}\frac{\pi(n)}{\frac{n}{\ln n}}=1\] http://en.wikipedia.org/wiki/Prime_number_theorem http://mathworld.wolfram.com/PrimeNumberTheorem.html
 one year ago

badreferencesBest ResponseYou've already chosen the best response.0
It's only necessary that someone links to a proof. I know I've seen this before, somewhere.
 one year ago

eliassaabBest ResponseYou've already chosen the best response.1
This is called the prime number theorem.
 one year ago

eliassaabBest ResponseYou've already chosen the best response.1
http://www.math.columbia.edu/~goldfeld/ErdosSelbergDispute.pdf
 one year ago

badreferencesBest ResponseYou've already chosen the best response.0
Oh, wait, I found it in mathworld. Okay, thanks!
 one year ago

badreferencesBest ResponseYou've already chosen the best response.0
That's right. This is Legendre's constant. That's where I saw it before! http://en.wikipedia.org/wiki/Legendre%27s_constant
 one year ago

badreferencesBest ResponseYou've already chosen the best response.0
Thanks for hte link, @eliassaab
 one year ago
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