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I want to determine whether

$${{3({p_n}-p_{n-1})}\over{p_{n-1}}}\ge\prod_{i=3}^{n-1}\Bigg(1-{2\over{p_i}}\Bigg)$$

is true for all sufficiently large $n\gt3$. (I don't know whether or not it's actually true, but it tests as true for a bunch of small $n$ that I've looked at).

My first thought was to use the lower-bound approximation for the $n$'th prime $n[ln(n)+ln(ln(n))-1]$ as $p_n$ on the left side, and the upper-bound approximation $n[ln(n)+ln(ln(n))]$ as $p_n$ on the right side. That at least gets rid of the primes, but of course the product on the right side still isn't smooth, so I don't know where to go from there. Also, I'm not sure if the inequality would remain true with those approximations, if it's indeed true with the primes. (I did test a bunch of small values and it seemed to remain true, but again I know that means nothing.)

Thanks for any help.

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Might you use $$1-\frac{2}{p_i}<(1-\frac{1}{p_i})^2$$? I vaguely recall there being plenty of estimates for $\prod_{i=1}^n(1-1/p_i)$ –  Thomas Andrews Dec 28 '12 at 17:40
    
$\prod_{i=1}^n(1-1/p_i)$, or something similar to it, shows up in Euler's phi function, which has some known bounds. Perhaps that could be useful? –  Julien Dec 28 '12 at 18:08
    
@Julien It is $\phi(N)/N$ where $N=p_1...p_n$ –  Thomas Andrews Dec 28 '12 at 18:14
1  
Maybe it's not true? My calculations for small values seem to show lhs smaller and shrinking faster than rhs. Proves nothing... –  daniel Dec 28 '12 at 18:18
    
For n = 9 I have lhs$ \approx$ 0.78 and rhs $\approx$ 0.23. For n = 100,000 I have the lhs as about 2.7(E-5) and the rhs as about 0.012. –  daniel Dec 28 '12 at 18:47

4 Answers 4

up vote 3 down vote accepted

It's not true.

I happen to have a file of the first 1400 primes sitting around so I created a spreadsheet for the LHS and the RHS. The LHS is less than the RHS for twin primes 43 and higher.

Note I started the product on the RHS at $p_n=5$ since it indicates $i=3$ not $p=3$. If you start the product with $p=3$ it takes until 229 for the first violation.

Note that merely reversing the inequality will not create a viable statement as written, since the gaps between primes vary. Thus although the twin primes violate the original statement at the number given, there are many others that continue to meet the condition given.

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Thank you so much. I hadn't yet tested out that far, and I'm glad to know. Starting the product at $p_n=5$ was indeed the intent--that's not a typo. –  Annick Dec 28 '12 at 18:43
    
I had sort of guessed this conjecture was related to twin primes. Certainly, the LHS is least in that case... –  Thomas Andrews Dec 28 '12 at 18:49
    
You did give fair warning in the statement that it might not be true. Took less time to set up the numeric test than I had already invested in the proof! Of course you should confirm for yourself in case I messed up the indexing - it was a quick and dirty test. –  half-integer fan Dec 28 '12 at 18:52
3  
This doesn't really disprove the inequality, of course, it just shows that $1400$ is not "sufficiently large." –  Thomas Andrews Dec 28 '12 at 19:14
    
From what I saw in my attempt to use induction (see other answer), once the statement is violated for a given prime gap, you can show that all further occurrences of that gap size will violate the inequality. –  half-integer fan Jan 2 '13 at 1:12

Estimate of the Product

Using the Prime Number Theorem, we get $$ \begin{align} \sum_{k=1}^n\frac1{p_k} &=\int_1^{p_n}\frac1x\,\mathrm{d}\pi(x)\\ &=\frac{\pi(p_n)}{p_n}+\int_2^{p_n}\frac{\pi(x)}{x^2}\,\mathrm{d}x\\ &=\frac{n}{p_n}+\int_2^{p_n}\frac{\frac x{\log(x)}+O\left(\frac x{\log(x)^2}\right)}{x^2}\,\mathrm{d}x\\[18pt] &=\log(\log(p_n))+O(1)\tag{1} \end{align} $$ Applying $(1)$, we have $$ \begin{align} \log\left(\prod_{i=3}^{n-1}\left(1-\frac2p_i\right)\right) &=\sum_{i=3}^{n-1}\log\left(1-\frac2p_i\right)\\ &=\sum_{i=3}^{n-1}\left(-\frac2p_i+O\left(\frac1{p_i^2}\right)\right)\\[12pt] &=-2\log(\log(p_{n-1}))+O(1)\tag{2} \end{align} $$ Therefore, $$ \prod_{i=3}^{n-1}\left(1-\frac2p_i\right)=\Theta\left(\frac1{\log(p_{n-1})^2}\right)\tag{3} $$ Estimate of the Ratio

We can derive from the Prime Number Theorem that $p_n=n\log(n)+O(n\log(\log(n)))$. This shows that for any $\alpha\gt0$ $$ \lim_{n\to\infty}\frac{p_{\alpha n}}{p_n}=\alpha\tag{4} $$ Furthermore, $(4)$ shows that $\lim\limits_{n\to\infty}\frac{p_n}{p_{n-1}}=1$, therefore, $\frac{p_k-p_{k-1}}{p_{k-1}}\sim\log\left(\frac{p_k}{p_{k-1}}\right)$ Thus, for any $\epsilon>0$, there is an $N$ so that for $m\ge N$, $$ \begin{align} m &=\pi(p_m)\\ &\ge(1-\epsilon)\frac{p_m}{\log(p_m)}\tag{5} \end{align} $$ and $$ \begin{align} \sum_{k=m+1}^n\frac{p_k-p_{k-1}}{p_{k-1}} &\le(1+\epsilon)\sum_{k=m+1}^n\log\left(\frac{p_k}{p_{k-1}}\right)\\ &=(1+\epsilon)\log\left(\frac{p_n}{p_m}\right)\tag{6} \end{align} $$ Thus, for any $n\ge2N$, there must be a $\frac n2\lt k\le n$ so that $$ \begin{align} \frac{p_k-p_{k-1}}{p_{k-1}} &\le(1+\epsilon)\frac{2\log(2)}{n}\\ &\le(1+\epsilon)\frac{2\log(2)}{k-1}\\ &\le\frac{1+\epsilon}{1-\epsilon}2\log(2)\frac{\log(p_{k-1})}{p_{k-1}}\tag{7} \end{align} $$ Comparing $(3)$ and $(7)$, there is no constant of proportionality that would make the given inequality true for all $n$.

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This is just an elaboration of my comment above, definitely not a solution.

It is fairly easy to show that $$\prod_{i=3}^{n}\frac{1-2/p_i}{(1-1/p_i)^2}=\prod_{i=3}^n\left(1-\frac{1}{(p_i-1)^2}\right)$$ is decreasing and has a non-zero limit as $n\to\infty$.

Let $\alpha$ be that limit. It looks like $\alpha \approx 0.880$.

For any $\epsilon>0$, then, for large enough $n$, $$9\alpha a_n^2<\prod_{i=3}^n (1-2/p_i) <9(\alpha+\epsilon)a_n^2$$

This might make it easier, since $\{a_i\}$ is a fairly well-known sequence.

where $a_n=\prod_{i=1}^n (1-1/p_i)$. (The factor of $9$ comes because we've added the cases of $i=1,2$.)

So, at minimum, you need:

$$a_n^2 \leq \frac{2}{3\alpha}\frac{p_{n+1}-p_n}{p_n}$$

for sufficiently large $n$.

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First, I replaced $p_n - p_{n-1}$ with $2$ since there are infinitely many twin primes.

Then, by incrementing $n$ and rearranging, I get $$\frac6{p_{n-1}} \cdot \frac{p_{n-1}}{p_n} \ge \left( 1-\frac2{p_n} \right) \cdot \prod_{i=3}^{n-1} \left( 1-\frac2{p_i} \right)$$

Hmm, just saw a problem with the next step - I'll come back and edit after working on it.

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We don't know that there are infinitely many twin primes, but it doesn't matter, since $p_n-p_{n-1}\geq 2$, so if you can prove $\leq 6/p_{n-1}$ you are done. –  Thomas Andrews Dec 28 '12 at 18:00
    
Conjecture, proof - who can keep track :-). Though the induction approach is not panning out so far. –  half-integer fan Dec 28 '12 at 18:23
    
It's a neat problem. I wonder if the sense of the inequality might be the other way. –  daniel Dec 28 '12 at 18:31

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