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Any help is appreciated in proving/disproving the following inequality $$ \frac{\ln{p_{n+1}}}{\ln{p_{n}}} < \frac{n+1}{n} $$

user65008
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  • What is $p_n$? Some sequence? – fixedp Mar 04 '13 at 21:57
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    @fixedp: Given the title and tags, I think it's reasonable to assume that $p_n$ denotes the $n$th prime number. – Zev Chonoles Mar 04 '13 at 21:59
  • An equivalent formulation is to show that $p_n^{1/n}$ is a decreasing function of the (integer) variable $n$. I thought this might be in Guy's Unsolved Problems In Number Theory, but didn't find it there. – Gerry Myerson Mar 04 '13 at 22:39
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    The claim ist that $\frac{\ln p_n}n$ is strictly decreasing. With $x=p_n$ the prime number theorem suggests that this is approximately $\frac{\ln^2x}{x}$. While this approximation is clearly decreasing, the exact behaviour is ... no tthateasy. – Hagen von Eitzen Mar 04 '13 at 22:46

2 Answers2

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This is Firoozbakht’s conjecture. According to the link, it has been verified for primes up to $4\times10^{18}$, but is believed to be false, as it contradicts the Cramér–Granville heuristic.

Gerry Myerson
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  • I found this inequality trying to make a generalization of the well-known $p_n<2^n$. Thank you very much for your answer – user65008 Mar 05 '13 at 07:39
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Not so much an answer as adding another level to the same question.

The Firoozbakht's conjecture (1982) is equal to:

$$(p_{n+1})^{n} < (p_n)^{n+1}.$$

Then the natural log is: $$n \ln(p_{n+1}) < (n+1)\ln(p_n).$$

Now, $$\ln(p_n) \leq \ln(n) + \ln(\ln(n)) + 1\text{, for $n \geq 2$. (*)}$$ With $n = n+1$ we have: $$\ln(p_{n+1}) \leq \ln(n+1) + \ln(\ln(n+1)) + 1, \text{for $n \geq 2$.}$$

And, because $$p_n \geq n*\ln(n)\text{, for $n \geq 2$; (**)}$$

the natural log of $p_n$ is: $$\ln(n) + \ln(\ln(n)) \leq \ln(p_n), \text{for $n \geq 2$}.$$

With $n = n+1$ we have: $$\ln(n+1) + \ln(\ln(n+1)) \leq \ln(p_{n+1}), \text{for $n \geq 2$}.$$

So, if $$n \ln(n+1) + \ln(\ln(n+1)) \leq n \ln(p_{n+1}) < n(\ln(n+1) + \ln(\ln(n+1)) + 1) < (n+1)(\ln(n) + \ln(\ln(n))) < (n+1)\ln(p_n) < (n+1) \ln(n) + \ln(\ln(n)) + 1$$ holds then the conjecture is true for all terms. If only the outer terms only hold then the primes with largest gap maximal primes will hold true. Primes with smaller prime gaps require sharper bounds on the inner terms.

With the outer terms, dividing by $n(\ln(n) + \ln(\ln(n)))$ we have: $$\frac{\ln(n+1) + \ln(\ln(n+1))}{\ln(n) + \ln(\ln(n)) + 1} < \frac{n+1}{n} \text{, for $n \geq 2$}.$$

This inequality is true because the left-side increases slower than the right-side.

With the inner terms, dividing by $n(\ln(n) + \ln(\ln(n)))$ we have: $$\frac{\ln(n+1) + \ln(\ln(n+1)) + 1}{\ln(n) + \ln(\ln(n))} < \frac{n+1}{n} \text{, for $n \geq 2$}.$$

This inequality is false for every $n$ value tested.

Is there a problem with any of these statements?

References:

$(*)$ Proved in Dusart 2010, ESTIMATES OF SOME FUNCTIONS OVER PRIMES WITHOUT R.H.,section 4. Useful Bounds)

$(**)$ Used in Dusart 1999, THE k th PRIME IS GREATER THAN k(ln k + ln ln k − 1) FOR k ≥ 2 Lemma 1. p. 413

user160140
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