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The Bertrand postulate is that $p_{n+1} < 2p_{n}$ but it is far to be enough to conclude.

I made a program in $R$, to see the convergence of this sequence for the first $500$ prime numbers and it seems to converge to 1. Is this true, and how to prove it ?

reuns
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Dr Richard Clare
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    Technically, Bertrand shows that $$\limsup \frac{p_{n+1}}{p_n}\leq 2$$ It doesn't show a limit exists. – Thomas Andrews Apr 15 '17 at 01:59
  • @ThomasAndrews I didn't write that Bertrand prove those limits, I just take the limits playing with the results of $p_{n+1} < 2p_n$. Thanks anyway. – Dr Richard Clare Apr 15 '17 at 04:57
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    No, my point is, you can't even write anything about $\lim_{n\to\infty}\frac{p_{n+1}}{p_n}$ until you can show it exists, and that the only thing you can conclude from Bertrand's postulate is the $\limsup$ is at most $2$. – Thomas Andrews Apr 15 '17 at 11:58

5 Answers5

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Hint: Use the fact that $p_{n} /n\log n \to 1$ via Prime Number Theorem and your desired limit gets converted to $$\lim_{n\to\infty}\frac{(n+1)\log(n+1)}{n\log n}$$ which you can easily prove to be equal to $1$. Bounding the prime gap $p_{n+1}-p_{n}$ is a difficult problem and is not really needed here.

Paramanand Singh
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  • Do we really need the PNT ? – reuns Apr 15 '17 at 02:45
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    @user1952009: The formula $p_n/n\log n\to 1$ is equivalent to PNT. – Paramanand Singh Apr 15 '17 at 02:47
  • Yes of course but the target is to show that $p_n / p_{n+1} \to 1$ – reuns Apr 15 '17 at 02:48
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    @user1952009: The wiki link on prime gap (given in one of the answers here) indicates the same idea. Bounding the prime gap is essentially done by using PNT as far as the wiki link says. I don't know of any other approach to deal with ratio $p_{n+1}/p_{n}$. – Paramanand Singh Apr 15 '17 at 02:52
  • $\log \zeta(s) \approx \sum_p p^{-s}= s\int_1^\infty \pi(x) x^{-s-1}dx \approx s \sum_{m=1}^\infty m, p_m^{-s-1} (p_{m+1}-p_m) $ $= s \sum_{m=1}^\infty m, p_m^{-s} (\frac{p_{m+1}}{p_m}-1)$ which we know converges absolutely for $Re(s) > 1$. – reuns Apr 15 '17 at 03:07
  • I wonder if it reduces to showing $\sum_p \frac{1}{p\ln^a p}$ converges for some $a$ – reuns Apr 15 '17 at 03:12
  • @user1952009: how do you proceed further to the limit of $p_{m+1}/p_{m}$? The convergence of your series above is granted by the bounded nature of the ratio $p_{m+1}/p_{m}$. – Paramanand Singh Apr 15 '17 at 03:13
  • It would improve your Answer to detail that $((n+1)\log(n+1))/(n\log n) = 1$. – hardmath Apr 15 '17 at 05:15
  • @hardmath: it was a hint so I omitted that simple limit. But as you suggest adding this will improve the answer so there it goes. – Paramanand Singh Apr 15 '17 at 05:30
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Yes, this is true. The current "state-of-the-art" estimate for prime gaps is $$p_{n+1} - p_n = O(p_n^{0.525}) $$ (due to Baker, Harman, Pintz), which guarantees that your limit is 1.

Alex
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  • I have an idea , I hope it's correct : it comes from this fact if $f$ is a bijection of $\mathbb{N} \to \mathbb{N}$ then $\lim_{n \to \infty} \frac{f(n)}{n}=1$ define $f$ such that $f(p_{n})=p_{n+1}$ and keep non prime numbers (i.e that doesn't appear in the sequence $(p_{n}){n\geq 1}$ as they are , $f$ is clearly a bijection , and since $(\frac{f(p{n})}{p_{n}}){n\geq 1}$ is a subsequence of the convergent sequence $(\frac{f(n)}{n}){n\geq 1}$ then it's limit is also $1$ . – uvw Apr 15 '17 at 02:11
  • @oty Where are you ever using that this sequence is of primes? If your argument was correct, you'd be showing the same result for $a_n=2^n$ instead, which is false. So find your mistake :) – Arkady Apr 15 '17 at 02:20
  • @oty Consider this bijection: $2^{2k}\to2^{2k+1}$, $2^{2k+1}\to2^{2k}$, all other integers (that are not powers of 2) stay in place. With this bijection, the limit does not exist, does it? – Alex Apr 15 '17 at 02:23
  • @Ravi , i am afraid that I don't understand your point of vue , I have a bad english , I apologize if it is not what you meant , but if $a_{n}=2^{n}$ we also have $\frac{f(a_{n})}{a_{n}}=1+\frac{1}{2^{n}}$ which converge to one . in my prove the function defined in my previous post is indeed bijective. I don't see where the argument break can you elaborate , Thank you . – uvw Apr 15 '17 at 02:33
  • @Alex ok let me prove the statement : suppose that there is $N$ such that forall $n\ge N$ , $f(n)>n$ then this doesn't let enough value for $f$ to complete the ${0,...,N}$ indeed only ${f(0),...,f(N-1)}$ can Map this set , a contradiction with $f$ is a bijection , and so $\forall N$ there is $n\geq N$ such that $f(n)\leq n$ hence $\lim inf \frac{f(n)}{n} \leq 1$ , by a similar raisonning supposing the is $N$ such that $f(n)< n$ for all $n\geq N$ we can show that $\lim sup \frac{f(n)}{n} \geq 1$ .... – uvw Apr 15 '17 at 02:43
  • @Alex , Ravi , you are right , oh thank you a lot , actually the statement is if $f$ is a bjiection as stated in my first post and if $\lim \frac{f(n)}{n}$ exist then this limit is $1$ , I apologize I need to prove that the limit exist if I want to make my argument valid Thank you again for correcting me . – uvw Apr 15 '17 at 02:49
  • For the existence of the limit : first by bertrand we have that $p_{n+1} \in (p_{n} , 2p_{n})$ , so if we can prove that $p_{n+1}^{2} < p_{n}p_{n+2}$ we can be done we will get $(\frac{f(n)}{n})_{n\geq 1}$ is increasing and bounded by $2$ then the limit exist , any idea to prove the inequality ? – uvw Apr 15 '17 at 03:18
  • Why it is important to you that $p_{n+1}^2 < p_n p_{n+2}$ ?? I can prove that the asymptotically this inequality holds. – Dr Richard Clare Apr 16 '17 at 22:39
  • @RichardClare , can you elaborate please , if the inequality does hold that mean that $\frac{f(p_{n})}{p_{n}} <\frac{f(p_{n+1})}{p_{n+1}} wich mean that the sequence $\frac{f(n)}{n}$ is increasing , the fact we needed to prove the existence of the limit . Thank you – uvw Apr 16 '17 at 23:32
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Here's a different proof of that fact, which doesn't make a reference to the prime number theorem assuming the limit exists. We'll use the fact that the series $$\sum_{n=1}^\infty \frac{1}{p_n}$$ diverges, which has a nice elementary proof due to Erdos. Now suppose that the limit is more than $1$. Then $p_{n+1}/p_n>1+a$ for some $a>0$ and all $n\geq N$ for some $N$. Thus $$\sum_{n=1}^\infty \frac{1}{p_n}\leq \sum_{n=1}^{N-1} \frac{1}{p_n} + \frac{1}{p_n}\sum_{i=0}^\infty \frac{1}{(1+a)^i}$$ which converges.

It would be interesting to give an elementary (not using the prime number theorem) proof that the limit exists.

amakelov
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  • Very elegant proof! – Dr Richard Clare Apr 15 '17 at 04:58
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    There is a gap here, in that the negation of the limit of $p_{n+1}/p_n$ being one is not "that the limit is more than $1$", but rather should encompass the alternative possibility that the limit fails to exist. – hardmath Apr 15 '17 at 05:13
  • yes, that's why I said "... assuming the limit exists" – amakelov Apr 15 '17 at 05:13
  • This reminds me of the fact that using very elementary arguments one can prove that if limit $\lim_{x\to\infty} \pi(x) /x$ exists then it must be $1$. It is rather difficult to prove that the limit exists and this is the whole point of Prime Number Theorem. – Paramanand Singh Apr 15 '17 at 05:38
  • Interesting, but should $a > 1$ not read $a > 0$, and the right-most $1/p_n$ in your second equation not read $1/p_N$ ? – Ross Ure Anderson Feb 15 '23 at 21:54
  • @RossUreAnderson you are absolutely right, I shall fix it. Thank you for pointing this out. – amakelov Apr 28 '23 at 12:11
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yes, for sufficiently large prime $p,$ the next prime is no larger than $$ p + p^{0.7} $$ Let me get link and correct exponent.

https://en.wikipedia.org/wiki/Prime_gap#Upper_bounds

Will Jagy
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Let $p_{n+1}=(1+x_n)p_n .$ From the Prime Number Theorem we have, as $n\to \infty,$ $$1+o(1)=\frac {\pi(p_{n+1})}{p_{n+1}/\log p_{n+1}}=\frac {n(\log p_n+\log (1+x_n))}{(1+x_n)p_n}$$ $$\text {and }\quad 1+o(1)=\frac {\pi (p_n)}{p_n/\log p_n}=\frac {(n-1)\log p_n}{p_n}.$$

Taking the ratio of these two formulas, since $x_n>0$ we have $$1+o(1)=\frac {n}{n-1}\cdot \frac {1+(\log (1+x_n))/\log p_n}{1+x_n}<$$ $$<\frac {n}{n-1}\cdot \frac {1+x_n/\log p_n}{1+x_n}=$$ $$=\frac {n}{n-1}\left(1+\frac {x_n}{1+x_n}(-1+1/\log p_n)\right)=V(n).$$

Now $x_n>0,$ while $1/\log p_n\to 0$ as $n\to \infty$. So if $k>0$ and $x_n>k$ for infinitely many $k,$ we would have $$\lim_{m\to \infty}\inf_{n\geq m}V(n) \leq 1-\frac{k}{1+k}<1$$ contrary to $V(n)=1+o(1).$

Therefore $\lim_{n\to \infty}x_n=0.$

DanielWainfleet
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