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anonymous
 4 years 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!
anonymous
 4 years 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!

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0* i love to know what is the answer

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i 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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This is called the prime number theorem.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0http://www.math.columbia.edu/~goldfeld/ErdosSelbergDispute.pdf

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh, wait, I found it in mathworld. Okay, thanks!

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks for hte link, @eliassaab
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