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Let $n$ be a positive integer and let $p(n)$ be the $n$th prime. Let $$f(n) = \dfrac{1}{30} \prod_{3<i<n+1} \left(\dfrac{p(i)- \left( \dfrac{2i}{\ln(p(i))}\right) + 1}{p(i)} \right).$$

How does $f(n)$ behave asymptotically? Does $\lim_{n\to oo} (n+7)^2 f(n)$ exist and what value is it? Can the limit be given in closed form?

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Careful observation of human reactions suggests that people tend to show more interest in problems that seem relevant to them. The lack of responses to your question may reflect the fact that it is singularly devoid of such motivation. – joriki Oct 31 '12 at 17:01
@joriki : On chat ppl said they find it intresting but hard. – mick Oct 31 '12 at 17:17
I'd find it quite interesting if I knew how it came up and why you're interested in the answer. – joriki Oct 31 '12 at 17:55
The limit is probably 0. reason is the product over primes $(P_i - k)/P_i = O ( 1/log(n)^k )$. – mick Oct 31 '12 at 19:07
The limit then appear to be like $x^2 / log(x)^{O(log(x))}$ which is $0$. – mick Oct 31 '12 at 19:09
up vote 1 down vote accepted

An infinite product is said to converge if the limit exists, and it is not zero. This is because the log of a product is a sum of logs, and $\ln0=-\infty$. Furthermore, there's a theorem about the convergence or divergence of $\prod_i(1-a_i)$ being the same that of $\sum_ia_i$. In this case, since $\sum_i\frac{2\,i}{p(i)\ln p(i)}$ diverges then so does the product. Since each term is clearly $<1$, then it diverges towards $0$. QED.

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Yes. I new it already by this time. (its an old question and my comment suggested 0 already ). Thanks anyway. Its nice now. :) – mick Nov 27 '13 at 23:24
I'm glad to hear that ! :-) However, $(1)$ unanswered questions merely clog up the Unanswered Questions queue (and there are over $100$ pages of these only on my labels alone), and $(2)$ questions are to be answered publicly for the benefit of the entire community (e.g., interested third parties). – Lucian Nov 27 '13 at 23:43
Agreed Lucian ! – mick Nov 27 '13 at 23:47

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