# Thread:

1. ## Convergence of primes

Let $p_n$ represent the n-th prime number. Determine what $\lim_{n\rightarrow \infty}\frac{p_{n+1}}{p_n}=?$. Does it even have a convergence?

Notice the following
$p_n<p_{n+1}<2p_n$ (Betrand's Conjecure)
thus,
$1<\frac{p_{n+1}}{p_n}<2$
Thus, by the limit limitation theorem,
$1\leq \lim_{n\rightarrow \infty}\frac{p_{n+1}}{p_n} \leq 2$ (I, of course, assumed the limit exists).

My guess is that the limit is one.

2. Nice problem. To my knowledge, the spacing of primes would increase as x approaches infinity, correct? If this is true it would seem that the limit would not converge.

Another cool prime problem is the sum of the reciprocals. Does $\sum_{n=1}^{\infty}\frac{1}{p_n}$ converge?

3. [QUOTE=Jameson]Nice problem. To my knowledge, the spacing of primes would increase as x approaches infinity, correct? If this is true it would seem that the limit would not converge. [qUOTE]

$p_n\sim n.\ln(n)$, so we might expect that:

$\frac{p_{n+1}}{p_n}\sim \frac{(n+1). \ln(n+1)}{n.\ln(n)}$.

Now the RHS of this relation goes to $1$ as $n \rightarrow \infty$.

This is not a proof because of the nature of the $\sim$
relation used here, but it is indicative that its not
implausible that the limit might exist and be equal to $1$.

Also, computing this ratio for all pairs of consecutive
primes $<10,000,000$ shows that this could
well be true.

RonL

4. Jameson, the problem you posted determine, $\sum^\infty_{k=1}\frac{1}{p_k}$ it diverges. It was demonstrated by Euler.

However my problem still remains unanswered. I fully believe that it converges to one. Do you are agree CaptainBlack, Jameson?

5. Oh nice post CaptainBlack. I'll have to do some research on prime number theory. At first glance the limit just doesn't seem to approach 1.

6. It is also possible that the limit does not exist at all. That the limit of this quotient fluctuates for values close to one.

The reason why I am doing this problem is to have an elegant proof that a 'polynomial prime producing phunction' (I just could not resist using 4 p's ) does not exist. The books on number theory give an elementary ugly proof. Thus, since it is true that the ratio of terms in any polynomial progression always converges to one. If we assumed that a 'polynomial prime producing phunction' does not converge to one then we have a contradiction.

Notice we can use the above concept to prove that a polynomial function for the Fibonacci sequence does not exist. Because in the Fibonacci sequence the ratio of consecutive terms must converge to 1 but in reality they converge to the Divine Proportion.

7. Originally Posted by ThePerfectHacker
However my problem still remains unanswered. I fully believe that it converges to one. Do you are agree CaptainBlack, Jameson?
Further research reveals that the required result was proven (or is a special
case of what was proven) by Erdös in a paper in 1949. As yet I haven't
found a proof online.

However it does look to me quite easy to show that if the limit exists it
must be one, as otherwise it would contradict the prime number theorem.

RonL

8. The standard reference for precise results of this kind is J.B. Rosser and L. Schoenfeld, "Approximate formulas for some functions of prime numbers", Illinois J. Maths, vol.6 (1962) 64--94.

Their (3.12) and (3.13) are $n \log n < p_n < n\log n + n\log\log n$ for all $n \ge 6$ which is enough to make CaptainBlack's argument precise.

9. Originally Posted by rgep
The standard reference for precise results of this kind is J.B. Rosser and L. Schoenfeld, "Approximate formulas for some functions of prime numbers", Illinois J. Maths, vol.6 (1962) 64--94.

Their (3.12) and (3.13) are $n \log n < p_n < n\log n + n\log\log n$ for all $n \ge 6$ which is enough to make CaptainBlack's argument precise.
Too lazy to check did you apply the squeeze theorem?

10. ## Are you sure of that?

Originally Posted by rgep
The standard reference for precise results of this kind is J.B. Rosser and L. Schoenfeld, "Approximate formulas for some functions of prime numbers", Illinois J. Maths, vol.6 (1962) 64--94.

Their (3.12) and (3.13) are $n \log n < p_n < n\log n + n\log\log n$ for all $n \ge 6$ which is enough to make CaptainBlack's argument precise.
I think that if n*log(n)<Pn where Pn is the n°ème prime number if i have understand correctly would involved that there would be no twin prime number (ie Pn=(Pn+1)+2 ) you must have forgoten a little something in your formula but indeed your formula gives sense to CapitainBlack "tilda" notation
hargh..

11. Originally Posted by SkyWatcher
I think that if n*log(n)<Pn where Pn is the n°ème prime number if i have understand correctly would involved that there would be no twin prime number (ie Pn=(Pn+1)+2 ) you must have forgoten a little something in your formula but indeed your formula gives sense to CapitainBlack "tilda" notation
hargh..
I'm not sure I understand your statement that

$n \log(n)<p_n$,

would imply that there are no twin primes.

Now if for all $n$ sufficiently large:

$(n+1)\log(n+1)>n \log(n)+\log \log(n)+2$

then we would have but a finite number of twin prime
pairs, but I don't think this is the case.

Excercise: show that as $n \rightarrow \infty$:

$(n \log (n)+\log \log(n)) - (n+1) \log (n+1) \rightarrow \infty$

(if it does).

RonL

12. Originally Posted by CaptainBlack

then we would have but a finite number of twin prime
pairs, but I don't think this is the case.

Excercise: show that as $n \rightarrow \infty$:

$(n \log (n)+\log \log(n)) - (n+1) \log (n+1) \rightarrow \infty$

(if it does).

RonL
of course i was a litle fast on that one (milk was boiling)
but i know that i have read somewere there are an infinite number of twins primes i have a litle idea (i will check and maybe start a thread on this)

now i will try to reply your exercise : i love limits problem and its a long time i didn't did one:
here it goes (the forth the sixth...)

We have to proove that for any positive integer M we can find an integer noted here A for wich when replacing n by A in your formula we find a result greater than M (by definition of +infinity limit)

We know that log is a strictly growing up function so is her reciprocal
x->10^x keep that in mind to understand the rest (i cannot do a strictly correct demonstration on keyboard)

So we have log(n+1)>log(n) for every positive number
So for A and replacing n by A in the formula developing the formula and making a majoration will lead us to proove that for every M we can find A
for wich log(A)-log(log(A))>M

wich is equivalent to show that
A/log(A)>10^M <=> A>10^M *log(A)
we know that
log(A) < (number of digits in base 10 of A)+1

So give me an M i will give you A=10^(M^2) wich satisfy the inequality
because 10^(M^2) is greater than 10^M*((M^2)+1) ( for M>1 indeed)
Hush!

13. I do not know if there are infinite number of twin primes,
SkyWatcher.

An attempt was made by Viggo Brun which tried to show that the series:
$\sum^{\infty}_{k=1}\frac{1}{t_k}$
diverges, (where $t_k$ is k-th twin prime).
Thus, if there are finitely many twin primes then this series must for definetly converge. Thus, by contropositive, if this series diverges then there are infinitely many twin primes. But Brun demonstrated that it actually converges. Thus, the results are inconclusive. This is called Brun's Constant.

Thus, the question still remains, are there infinitely many twin prime? I once tried to prove it by assuming there are finitely many and constructing a number which would also be a prime twin, (like in Euclid's prove of infinitely many prime numbers) but no success.

14. I have been looking to paper encyclopedies and it seems that the problem of "is there an infinity of twin primes" was not conclued at the times those encyclopedies were writen.
so my memory had made two mistakes
1 i thought i had seen somewhere (in internet) that it was proven that there are an infinity of those (twin primes)
2 i had made a mistake about "is there an infinity of primes" (see jameson thread on that) i had understood (long time ago) that the product of the N first prime plus one was prime (it walks when you consider what you can calculate whithout using a computer (for the first prime)).

anyway if the formula i am doubtful about was right you would'nt had to squeeze your brain trying to find if there is an infinity of that stuff because it's an evident coloraly of such a théorème that's there is not!
what made me looking about the thrueness of this formula is that it's using decimal logarithm (as far as we have the same notation in english and in french) and i could'nt see why the number ten would had to do with the distribution of prime number.

15. Originally Posted by SkyWatcher
I have been looking to paper encyclopedies and it seems that the problem of "is there an infinity of twin primes" was not conclued at the times those encyclopedies were writen.
so my memory had made two mistakes
1 i thought i had seen somewhere (in internet) that it was proven that there are an infinity of those (twin primes)
2 i had made a mistake about "is there an infinity of primes" (see jameson thread on that) i had understood (long time ago) that the product of the N first prime plus one was prime (it walks when you consider what you can calculate whithout using a computer (for the first prime)).

anyway if the formula i am doubtful about was right you would'nt had to squeeze your brain trying to find if there is an infinity of that stuff because it's an evident coloraly of such a théorème that's there is not!
what made me looking about the thrueness of this formula is that it's using decimal logarithm (as far as we have the same notation in english and in french) and i could'nt see why the number ten would had to do with the distribution of prime number.
Just because your found on the internet it said that they had a proof does not mean anything. I once searched for Fermat's Last Theorem, then I get hunders of links that promise they have "proofs". Thus, I enter some on them and I see that the people that made the site do not know what they are talking about. For example, someone gives a false proof on one of these website and in the end he writes, "...Thus, it does not work for integral,rational and irrational numbers." WHAT! it certainly does work for irrational numbers. If it said that on Wikipedia, or Mathworld then you can trust it otherwise people try to make themselves look smart by tricking people in making them think they got proofs.
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