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I observed the following limit empirically. Let $p_n$ be the $n$-th prime and $c_n$ be the $n$-th composite number then,

$$ \lim_{n \to \infty}\frac{1}{n}\sum_{i=1}^{n}\frac{p_n c_n}{p_n c_n + p_i c_i} = \frac{\pi}{4}. $$

I am looking for a proof.

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How "empirically"? Have you developed a computer program so as to check what happens with $\;n=100\,,\,1000\,,\,....$ ? –  DonAntonio May 30 '13 at 10:25
    
Yes, for the first 1 million primes. –  user60930 May 30 '13 at 11:00
    
That's a very nice check. Read below my answer the comment by Shreevatsa: I think he's got a point there. –  DonAntonio May 30 '13 at 11:01

2 Answers 2

up vote 32 down vote accepted

It turns out that

  1. The limit is correct, but
  2. It's not saying anything that's very special to primes and composites.

Note that (inspired by DonAntonio's answer) $$ \lim_{n\to\infty} \frac1n \sum_{k=1}^n \frac{n^2}{n^2 + k^2} = \lim_{n\to\infty} \frac1n \sum_{k=1}^n \frac{1}{1 + \left(\frac{k}{n}\right)^2} = \int_{0}^{1} \frac{dx}{1 + x^2} = \frac{\pi}{4} $$

It just so happens that $p_n$ and $c_n$ are (at a very loose level of approximation) on the order of $n$ each, so that $p_n c_n$ is of the order of $n^2$, and therefore your sum $$ \frac{1}{n}\sum_{k=1}^{n}\frac{p_n c_n}{p_n c_n + p_k c_k} \approx \frac{1}{n}\sum_{k=1}^{n}\frac{n^2}{n^2 + k^2}, $$ the approximation turning exact in the limit.

To prove this rigorously, we have from the prime number theorem, that $p_n \sim n\ln n$, or to be precise $$p_n = n\left(\ln n + \ln\ln n - 1 + O\left(\frac{\ln\ln n}{\ln n}\right)\right) = n\left(\ln n + o(\ln n)\right).$$ Similarly for the $n$th composite number $c_n$, we have $$c_n = n\left(1 + \frac1{\ln n} + O\left(\frac{1}{\ln^2 n}\right)\right) = n\left(1 + o(1)\right).$$

So $$p_nc_n = n^2\left( \ln n + o(\ln n) \right).$$

Consider a particular value of $\frac{k}{n}$ (say $\alpha$) so that $k = \alpha n$. Then $$ \frac{p_kc_k}{p_nc_n} = \frac{k^2 (\ln k + o(\ln k))}{n^2(\ln n + o(\ln n))} = \frac{k^2}{n^2}\frac{\ln n + \ln \alpha + o(\ln n)}{\ln n + o(\ln n)} = \frac{k^2}{n^2}(1 + o(1)) $$

Therefore $$ \lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^{n}\frac{p_n c_n}{p_n c_n + p_k c_k} = \lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^{n}\frac{1}{1 + \frac{p_k c_k}{p_nc_n}} = \lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^{n}\frac{1}{1 + \frac{k^2}{n^2}(1 + o(1))} = \int_{0}^{1} \frac{dx}{1 + x^2} = \frac{\pi}{4}. $$

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4  
+1 Beautiful analysis –  cyclochaotic May 30 '13 at 13:26
2  
Actually we can probably get away with an even looser analysis. It's an interesting reverse question: what is the loosest "nice" condition on a sequence of numbers $a_n$, which guarantees that $\displaystyle \lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^{n} \frac{a_n}{a_n + a_k} = \frac{\pi}{4}$? –  ShreevatsaR May 30 '13 at 13:38
1  
I thank you for mentioning in your overwhelming beautiful answer but that's way too modest from you. Too bad one can upvote only once. +1 –  DonAntonio May 30 '13 at 17:16
    
@ShreevatsaR : Partial answers to the general sequence $a_n$ is present in the following question. Can you have a shot at this bounty question? math.stackexchange.com/questions/403679/… –  user60930 May 31 '13 at 5:35

We have that

$$\lim_{n\to\infty}\,\frac1n\,\sum_{k=1}^n\frac n{n+k}=\lim_{n\to\infty}\,\frac1n\,\sum_{k=1}^n\frac1{1+\frac kn}=\int\limits_0^1\frac{dx}{1+x}=\log 2$$

Your sum is not, of course, the above one, but it ressembles it a little, so I'd expect its sum to be closer to $\,\log 2\,$ than to $\,\pi/4\,$ , in particular since the difference between these two numbers is less than $\,1/100\,$ , yet I cannot tell for sure.

A simple program that can add automatically can make some checkings...

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4  
Note that $\displaystyle \lim_{n\to\infty} \frac1n \sum_{i=1}^n \frac{n^2}{n^2 + i^2} = \int_{0}^{1} \frac{dx}{1 + x^2} = \frac{\pi}{4}$. I think this limit may be more relevant. :-) –  ShreevatsaR May 30 '13 at 10:44
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Oh, I think you're right, @ShreevatsaR . Why won't you post this as an answer? I would upvote it... –  DonAntonio May 30 '13 at 10:55
4  
Still I think this was the germ of the idea, that the answer was an artifact of summation rules. +1. –  daniel May 30 '13 at 15:34

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