Necessary condition for goldbach

[kaushiks.nitt]

pi(N) > { N/(2 * pi(N/2) } + pi(N/2) is the necessary condition for goldbach and this is the proof for the inequality
Now consider two cases in case i: I am going to substitute pi(N/2) = N/2*log(N/2) in the denominator alone and take the other term in the RHS to LHS.
Thus the inequality is pi(N)- pi(N/2) > log(N/2).
This means that for any N between N and N/2 there are at least log(N/2) prime numbers. This is always true . But i haven't proved here.
Now let us come to case ii: We know that pi(N) > N/log(N) . Thus if i add an error term say e(n) on the RHS . Then pi(N) = N/log(N) + e(n). {note : It is always possible to find e(n) , however there is no such method till date to find e(n) }.
When i replace N by N/2 in the above equation then pi(N/2) = N/2*log(N/2) +e(n/2) .{ note : there is no relation between e(n) and e(n/2) }. In the denominator on the RHS i can substitute pi(N/2) by N/2*log(N/2) . The inequality is preserved .
Now the inequality is of the form,
N/log(N) + e(n) > log(N/2) + N/2*log(N/2) + e(n/2).
It is trivial that N/log(N) > log(N/2) + N/2*log(N/2).
If e(n) > e(n/2) . Then the inequality always holds good.
From observation i know e(n) > e(n/2) . Thus i claim that the inequality always holds good.
If u ask me for a proof , I need to prove that e(n) > e(n/2). Everything else is trivial.

[gmalivuk]

There are plenty of necessary conditions for goldbach, but proving any of them doesn't say anything about goldbach except that it could be true. To actually prove goldbach you need to find something sufficient.

[kaushiks.nitt]

There are plenty of necessary conditions for goldbach

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Do u know some link where i could have a look at all the necessary conditions for goldbach.
I have just started giving a try at some of the major open problems in maths .
So i haven't gone through the literature much .

[gmalivuk]

I'd say Wikipedia is usually a good place to start looking into a problem like this.

[auteur52]

I have just started giving a try at some of the major open problems in maths .

What's your background like? I would suggest that your time would be better spent learning a LOT of math. If you are serious about wanting to try to make progress towards something big like Goldbach, I would suggest you start learning Algebraic Number Theory. Mathematicians who solve these kinds of problems are standing on the shoulders of giants, spend your time trying to work yourself up there.

(If you have a PhD in Number Theory already, just ignore what I said.)

[kaushiks.nitt]

What's your background like?

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I am not a math major . I just got interested in these problems . I proved collatz conjecture can be true for a infinite case but was not able to come up with a better formula .
Then i proved the famous PNT in a single line using the concepts of relative prime.
I was able to prove goldbach till 10^8 after this i didn't had the computation power to check it's validity.
I see goldbach problem as one dealing with generating all from a few . These kinds of problem are generally tough because when we mean all it necessary mean all i.e. infinity .
I always remember the following statement even if something is proved for a infinite cases there might still be infinite cases left out .
Thus i actually believe goldbach can't be solved using any trivial techniques in number theory .
I have read all the wiki articles relating number theory besides this i have read a few other papers of erdo's ,sellberg and all.
But still my theoretical background is not that great .

[t0rajir0u]

I proved collatz conjecture can be true for a infinite case but was not able to come up with a better formula .

I have no idea what "infinite case" means. Could you elaborate? (If you mean that there exist infinitely many integers for which the Collatz conjecture is true, this is already known and trivial.)

Then i proved the famous PNT in a single line using the concepts of relative prime.

The short proofs of the PNT aren't elementary, and the elementary proofs aren't short. Again, could you elaborate?

Before you continue your excursions into open problems, you should read some advice from Terence Tao and Scott Aaronson. I don't want to discourage you, but you should try very hard to be thorough and careful and to understand how far a partial solution falls short of a full solution.

[Incompetent]

I have just started giving a try at some of the major open problems in maths .
So i haven't gone through the literature much .

Have you reflected much on why these problems might still be open, or why they are of interest?

[kaushiks.nitt]

you should read some advice from Terence Tao and Scott Aaronson.

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Actually tao sent me a really a good mail regarding these stuffs .
I am in touch with tao but because he is a really busy person he couldn't reply to me very time .

[kaushiks.nitt]

Could you elaborate? (If you mean that there exist infinitely many integers for which the Collatz conjecture is true, this is already known and trivial.)

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I was trying to prove collatz using induction .
Thus if it is true till all numbers till N then it is true for N+1. This proof holds for a large values of N . But if N = 2^k then proving collatz is true for N = 2^k +1 is not trivial.
Also i observed this too has been proved by someone else that if N = 2^k*p +1(odd) then after k times the N = 3^k*p +1 this is even. Thus the function becomes odd and even repeatedly and it never grows unbounded.
So it must converge to one .
I said that after carrying out the operation infinite times it approaches one but there was an obvious error and i withdrew the proof which i had submitted to ams.

[kaushiks.nitt]

The short proofs of the PNT aren't elementary, and the elementary proofs aren't short. Again, could you elaborate?

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Consider any N .
I have this formula for finding pi(N). Let us say there are k prime numbers till √N .
Then pi(N) ~ N*(p1-1)*(p2-1)*....(pk-1)/p1*p2*....pk.
Now consider N tends to infinity then √N also tends to infinity.
Then pi(N) ~ N*(p1-1)*(p2-1)*....(pk-1)*....(pinf-1)/p1*p2*....pk*....p(inf).
and this is nothing but N* Π( 1 -1/p) = N/log(N).
Π( 1 -1/p) = 1/log(N) this has been proved by euler .
I hope there is no mis communication.
Also pi(N) ~ N*(p1-1)*(p2-1)*....(pk-1)/p1*p2*....pk. Is a better approx than N/log(N).
For N =100 pi(N) calculated by my formula is 23 while my N/logN it is around 21.
For N= 1000 pi(N) calculated by my formula is 153 while my N/logN it is around 144.
Infact using the above formula i can also find the error between the actual and computed value . while the PNT never deals with the error.

[kaushiks.nitt]

Have you reflected much on why these problems might still be open, or why they are of interest?

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I don't get you these are open problems because no one has solved it till now and each problem have their own applications .
So what are you asking me .

[achan1058]

Have you reflected much on why these problems might still be open, or why they are of interest?

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I don't get you these are open problems because no one has solved it till now and each problem have their own applications .
So what are you asking me .

What he means is, the only reason these problems are open is because they are very, very hard. If it is easy and well known, someone would have done it already. My suggestion is to work on some not famous open problem. They tend to be easier. (since they are open because nobody worked on it, instead of being open despite the fact that over 9000 geniuses have worked on it)

[mike-l]

Consider any N .
I have this formula for finding pi(N). Let us say there are k prime numbers till √N .
Then pi(N) ~ N*(p1-1)*(p2-1)*....(pk-1)/p1*p2*....pk.
Now consider N tends to infinity then √N also tends to infinity.
Then pi(N) ~ N*(p1-1)*(p2-1)*....(pk-1)*....(pinf-1)/p1*p2*....pk*....p(inf).
and this is nothing but N* Π( 1 -1/p) = N/log(N).
Π( 1 -1/p) = 1/log(N) this has been proved by euler .
I hope there is no mis communication.
Also pi(N) ~ N*(p1-1)*(p2-1)*....(pk-1)/p1*p2*....pk. Is a better approx than N/log(N).
For N =100 pi(N) calculated by my formula is 23 while my N/logN it is around 21.
For N= 1000 pi(N) calculated by my formula is 153 while my N/logN it is around 144.
Infact using the above formula i can also find the error between the actual and computed value . while the PNT never deals with the error.

I haven't looked too deeply into this (see why below), but what is this product over?

Π( 1 -1/p) = 1/log(N)

And, are you sure that that statement isn't equivalent to the PNT to begin with?


If you want to be taken seriously, you need to work on your presentation skills. Quite frankly, when I see a non-math person claiming to have proved a number of difficult problems and presenting horribly formatted proofs, it immediately goes into the 'crank' file, especially when they know nothing of the literature on the subject. While it's not impossible you've come up with some new proofs, people claiming these things are a dime a dozen, and most of them are wrong.

[kaushiks.nitt]

I remember this famous line from shawshank redemption.
"Hope is a good thing , it might be even the best of things and no good thing ever dies".
Perhaps there is nothin wrong in being optimistic an working on this and even wiles wasn't known before he solved fermat's last theorem and maybe when he was 20 no one would have ever expected him to solve.
History just inspires
It might take a long time or may be i never ever solve one of them .
But there is always a pleasure in doing and i do it for that . M just well settled in life with a government job and a hefty pay so i do this only in my leisure time .
After all i love maths and ramanujam ( I hope u might be knowing) is my distant relative

[t0rajir0u]

Infact using the above formula i can also find the error between the actual and computed value . while the PNT never deals with the error.

The PNT doesn't deal with error because a precise bound on the error in the PNT is equivalent to the Riemann Hypothesis. (There are less precise error bounds known.)

And you don't have a proof. What you have is a heuristic calculation which suggests that the result "should" be true, but you haven't proven either of the asymptotics you've cited. Making your argument rigorous is more difficult than you think.

[NathanielJ]

even wiles wasn't known before he solved fermat's last theorem and maybe when he was 20 no one would have ever expected him to solve.

Wiles was a professor at Princeton and a Guggenheim Fellow before he proved Fermat's Last Theorem. And his proof was over 100 pages long and took seven years to complete.

[Hackfleischkannibale]

Have you reflected much on why these problems might still be open, or why they are of interest?

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I don't get you these are open problems because no one has solved it till now and each problem have their own applications .
So what are you asking me .

For example, you could have found out the barriers to a proof. I don't know the situation with Goldbach, but for P=NP there are precise reasons why a proof would have to look very different from "usual" proofs in that area, and I read this has been the case for Fermat, too.

[achan1058]

Wiles was a professor at Princeton and a Guggenheim Fellow before he proved Fermat's Last Theorem. And his proof was over 100 pages long and took seven years to complete.

Not to mention that he wasn't proving the FLT, but something that implies the FLT. Also, there are some highly overpowered mathematics in his proof.

[kaushiks.nitt]

While it's not impossible you've come up with some new proofs, people claiming these things are a dime a dozen, and most of them are wrong.

This is an arbit statement highly irrelevant to the context .May be your math is so very poor that you cant understand even simple statements and want to be spoon feeded ..
Also i never said it's something very different from the PNT itself . I had a look at sellberg's proof and i knew he never proved it like i did . That is why i mentioned.

[kaushiks.nitt]

And you don't have a proof. What you have is a heuristic calculation which suggests that the result "should" be true, but you haven't proven either of the asymptotics you've cited. Making your argument rigorous is more difficult than you think.

I guess u will accept that the sieve of Eratosthenes can be used to compute / find all prime numbers till N by knowing all the primes till √N. This is what i have used.
Just consider there are just two primes say 2,3 till √N then the number of primes till N would be N- [N/2] - [N/3] +[N/6] + 2( note even 2 and 3 are cancelled but these have to be included ) Using my formula the result would be N -N/2 -N/3 + N/6 .
Thus the error is 2*O(1) - O(1) + 2( pi(√N) . So u could just generalize the result if there are k primes till √N , the coefficients of O(1) are nothing but those which we encounter in the expansion of (1-1)^pi(√N) .
U might be tempted to say that 1-1 = 0 . but O(1) - O(1) != 0 .
Thus the average error that i get is √N/log(N) . But this is not the worst case error .
I don't understand by what u mean when u say that i don't have a proof .
It is very clear about the computation of pi(N) and also as N -> inf the zeta function is what i have on the RHS thus clearly proving pi(N) = N/log(N) . which is nothing but your PNT. So it can be regarded as the proof to PNT as well as an method to calculate the exact error which is nothing but your RH

[kaushiks.nitt]

Wiles was a professor at Princeton and a Guggenheim Fellow before he proved Fermat's Last Theorem. And his proof was over 100 pages long and took seven years to complete.

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This is what i also said . I never claimed proving anything i just started .I also knew that wiles didn't became a prof at 20 perhaps no one could . So to say u can't do things is ridiculous and actually a baseless statement .

[kaushiks.nitt]

I don't know the situation with Goldbach, but for P=NP there are precise reasons why a proof would have to look very different from "usual" proofs in that area, and I read this has been the case for Fermat, too.

I never claimed to have proved goldbach i just found a necessary condition and i know it is no where closer to the proof .
I myself know that it is impossible to prove goldbach using trivial methods . Just like wiles may it can be stated that something else implies goldbach . Simple additive number theory is in sufficient for this problem . This is beyond doubt .
I was just checking if all the necessary condition's are being satisfied and i found they are which simply gives support to the fact that goldbach is true .

[mike-l]

While it's not impossible you've come up with some new proofs, people claiming these things are a dime a dozen, and most of them are wrong.

This is an arbit statement highly irrelevant to the context .May be your math is so very poor that you cant understand even simple statements and want to be spoon feeded ..
Also i never said it's something very different from the PNT itself . I had a look at sellberg's proof and i knew he never proved it like i did . That is why i mentioned.

My point was that we encounter posts like yours all the time, and most of the time they are a complete waste of time to read, so if you want to have any chance of people considering what you have to say, you need to present it much much better.

[skeptical scientist]

After all i love maths and ramanujam ( I hope u might be knowing) is my distant relative

Even Ramanujan, with all his genius, might never have accomplished anything important if Hardy hadn't brought him to England where he had access to modern journals, books, and a mathematical community with an enormous body of known results that he could learn, important (but attackable) open questions that he could work on, and other clever people that he could collaborate with.

Even if you are a mathematical genius, any attempt to answer such long-standing and well-studied questions as these will almost certainly require both an encyclopedic knowledge of known techniques/results and substantial new techniques.

To use your example of Wiles - he was successful both because he was smart and creative enough to develop new techniques and because there was a large body of mathematics that he could build on, and he knew of an important conjecture in his area (superficially unrelated to Fermat) that had been shown to imply Fermat's last theorem.

I'm not suggesting you stop looking at these problems - far from it. After all, independently rediscovering results of Euler is nothing to be ashamed of. I'm simply suggesting that your time would be more fruitfully spent if you spent some more time reading books and papers (many of which are available online) to get a better understanding of the current state of the art, and spread your research between these (important but unlikely to be solved any time soon) open questions and other open questions which are less well-known, but more attackable.

[kaushiks.nitt]

My point was that we encounter posts like yours all the time, and most of the time they are a complete waste of time to read, so if you want to have any chance of people considering what you have to say, you need to present it much much better.

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May be presentation plays an important role but i was just posting it in one of the forum and i wasn't submitting it ti some journal where things have to be crystal clear .
Besides i presume that everyone who is replying to my posts is atleast as good as me and infact better .
So whatever seems trivial to me should appear the same to them .
I just need not keep mentioning the obvious things . I don't mind doing it would just lengthen my posts and the reader might not be interested in having a look at it . That's why i keep things short .
Also i am ready to answer any doubt's that u have in my posts .
N moreover in forums we just discuss ideas and ways of thinking .
A person who is producing or even capable of producing great results would never post his stuffs in the forum.
Have ever heard wiles posting results in forums . Or for that matter tao or anyone else . They never do.
So it is not always possible to expect results which are worth of a paper in the forums .
How many have you encountered till date in this very forum ???

[kaushiks.nitt]

important (but attackable) open questions that he could work on, and other clever people that he could collaborate with.

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I would first like to thank skeptical scientist for his advice. This was pretty much reasonable than many of the stupid comments passed before which simply stated that one shouldn't try these problems which literally made no sense as in if no one is to try this then will these stuffs be left unsolved for ever . I do accept a strong foundation of the basics is necessary which just helps you prevent wasting time . Just like i quoted it is to interact with clever and better people being the reason for me joining the forum not to just waste my time and waste others .
As i am not a math major only through forums i keep learning existing but new things to me . Everyone mentions few papers or links which deal with these stuffs so i have a look at it and then get back to the problem. Googling isn't of much help because as you could see when i first searched for math forums i never found this link . Only after a long time i got to know about this forum . So it doesn't always help and it is always better to have it from the people who had already studied stuffs because they exactly know which paper is more relevant .

[kaushiks.nitt]

independently rediscovering results of Euler is nothing to be ashamed of.

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I do this because i could always say had it not been for euler i would have proved the results nevertheless. Or may be if i had been before euler's time the results would have been known by my name .
That's all nothing else . Ofcourse i would accept that i would have never discovered A.P or sum upto n terms which gauss did at class 3 . That is mathematical genius for you

[gmalivuk]

Please stop posting two or three times in a row. You can edit your most recent post if no one has replied.

[kaushiks.nitt]

Please stop posting two or three times in a row. You can edit your most recent post if no one has replied.

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Ok . But how do i answer to different people's question in a single post . Wont it become cumbersome.

[z4lis]

Ok . But how do i answer to different people's question in a single post . Wont it become cumbersome.

Doing it like this is just fine. Just answer one question under one quote.

Please stop posting two or three times in a row. You can edit your most recent post if no one has replied.

And then make a new quote underneath it like this and answer the next question. Having a long post with several quotes is preferable to three posts by the same person in a row. =P

Hopefully gmalivuk won't hurt me for this. :3

[stephentyrone]

To echo what others have said, work on your writing. Even little things like not putting a space before a period at the end of a sentence -- though seemingly inconsequential -- will make your writing tremendously easier to read, which will make it much easier for other people to take what you write seriously. Even the greatest genius will go undiscovered if she cannot write well enough to express her ideas clearly.

The fact that you're "not publishing this in a journal" is inconsequential. Presumably you are posting here because you want people to respond and have an exchange of ideas (if not, why bother?). The more clearly you express yourself, the better the quality of feedback you receive.

[mike-l]

So I've actually given some thought to your PNT 'proof'. I'm inclined to agree with t0rajir0u that you have a good heuristic, but it needs work to become a proof. If you do manage to find a short elementary proof of the PNT, well, you'll get noticed .

Here's what you need to do to convince me (and I believe at least one of these tasks to be non-trivial).

(disclaimer, math mode doesn't work on my work computer so I can't preview the markup below )
1) You've given the error between your estimate and \pi(N) for small N (call this e(N) say). Generalize this and prove that

\lim_{N\rightarrow \infty} e(N)/\pi(N) = 0

.

2) Prove that (correct me if I have the bounds wrong, but they seem to be the ones that fit in)

\lim_{N\rightarrow \infty} \ln{N} \prod_{p<\sqrt n} (1-p^{-1}) = 1

My gut is that 1 is probably pretty doable, but you are the one claiming the statement, so I'll leave it to you to get your hands dirty on it.

For 2, I get that you are writing 1 - p^{-1} = (1+p^{-1}+p^{-2}+...)^{-1} and using prime factorization to rewrite this as a partial sum of the harmonic series, and comparing that to ln(N). But you get a lot of terms bigger than N, and you miss a lot of terms smaller than N, so I need convincing that they somewhat balance and asymptotically we still have ln(N).

[kaushiks.nitt]

2) Prove that (correct me if I have the bounds wrong, but they seem to be the ones that fit in)
limNlnNpn(1−1p)=1

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Euler proved that sigma 1/n = Π (1/ 1 - p^-1). ( the infinite product in the right hand side extends over all prime numbers p )
i.e sigma 1/n = Π ( p/(p-1)) and Π ( (p-1)/p) = 1/( sigma 1/n ) .
Thus Π ( 1 - 1/p ) = 1/( sigma 1/n ) = 1/log(N) .
In fact the error between sigma (1/n) and log (N) is given by the euler constant y . So you could use that also.
That is why i use this is formula to calculate pi(N) only as N -> infinity . In this case √N will also tend to infinity and the euler products would still hold good.
This what is done in the sieve theory to calculate all primes till N by just knowing all the primes till √N .
If there is something wrong with the proof let me know

[mike-l]

Euler proved that sigma 1/n = Π (1/ 1 - p^-1). ( the infinite product in the right hand side extends over all prime numbers p )
i.e sigma 1/n = Π ( p/(p-1)) and Π ( (p-1)/p) = 1/( sigma 1/n ) .

I agree up to here. This is what I was refering to in my last post.

Thus Π ( 1 - 1/p ) = 1/( sigma 1/n ) = 1/log(N) .

You need to state what your bounds are. The first "equality" only holds for (all p) and (all n), where the second "equality" only holds for (n <= N). (quotes because neither is really an equality, one is a rewriting of an infinite divergent sum, the other is an approximation)

In fact the error between sigma (1/n) and log (N) is given by the euler constant y . So you could use that also.
That is why i use this is formula to calculate pi(N) only as N -> infinity . In this case √N will also tend to infinity and the euler products would still hold good.
This what is done in the sieve theory to calculate all primes till N by just knowing all the primes till √N .
If there is something wrong with the proof let me know

You need to be more careful. While proving the PNT is about asymptotic behaviour, every value of N we look at is itself finite. You can assume it's as large as you need it to be, but it's still finite. There is no N 'at infinity', but your proof seems to hinge on an equality that's not true until infinity.

[kaushiks.nitt]

kaushiks.nitt wrote:Thus Π ( 1 - 1/p ) = 1/( sigma 1/n ) = 1/log(N) .


You need to state what your bounds are. The first "equality" only holds for (all p) and (all n), where the second "equality" only holds for (n <= N). (quotes because neither is really an equality, one is a rewriting of an infinite divergent sum, the other is an approximation)

.

This was a typing error perhaps i meant Π ( 1 - 1/p ) = 1/( sigma 1/n ) = 1/log(n).

While proving the PNT is about asymptotic behaviour, every value of N we look at is itself finite. There is no N 'at infinity

But pi(N) * log(N) / N =1 only as N tends to infinity this is PNT . Thus i will of course take N -> infinity.
I don't understand your statement no N at infinity.

[mike-l]

Do you understand what a limit is?

You have to show that for any \epsilon > 0, there is some M such that for any N>M, we have |\pi(N)*\ln(N)*N^{-1} - 1| < \epsilon.

N -> infinity does not mean take N=infinity, it means that you can get arbitrarily close to your value by taking sufficiently large N (this is just a paraphrasing of the above)

[PM 2Ring]

I'm sorry, kaushiks.nitt, I find your posts too difficult to read thoroughly. How does your prime counting function compare to this one found by Riemann, which uses the Zeta function?

Estimating the number of primes in C.

PS. It's very cool that you are a distant relative of Ramanujan. I always thought he was interesting, even before I read his biography. As you probably know, Ramanujan was also fond of using divergent series in his work. The first English mathematicians to see these results dismissed him as a madman, but Hardie realized he was onto something. However, it took a lot of careful mathematics to transform these results into something provable and useful.

[kaushiks.nitt]

How does your prime counting function compare to this one found by Riemann, which uses the Zeta function?

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I just presume that one can find all primes till N by just knowing all the primes till √N . For this we use sieve theory .
For example consider there are just two primes 2 and 3 till √N then the number of primes till N would be actually N - [N/2] - [N/3] +[ N/6] +2( 2 and 3 are also sieved and hence must be included and [N] denotes the greatest integer function) . This can be approximated to N - N/2 - N/3 +N/6 which is N ( 1 - 1/2)*(1-1/3) . which is nothing but N* (1/2)*(2/3) .
Like wise if say we have k primes less than √N the formula for calculating pi(N) = N* ( p1 -1/p1)*( p2-1/p2)* .... ( pk-1/pk).
This is trivial and of course this has an error because i should have been considering the greatest integer function.
Euler just said that all numbers are nothing but products of prime so when u consider all prime numbers then (p1/p1-1)*(p2/p2-1)*... would be nothing but the sum of the harmonic series of all natural numbers this is also trivial.
sigma(1/n) can be approximated to log(n) . Thus i thought when N -> infinity pi(N) = N/log(N) .
I considered this to be the proof of the prime number theorem . But then mike-1 pointed out that this is valid only when you consider all primes . That is also obvious . But i thought as √N tends to infinity almost all primes are included and the argument might still hold good.
I hope this is clear . Maybe i don't know to phrase it in better words than in this posts and most of the results are just trivial .

[mike-l]

sigma(1/n) can be approximated to log(n)

The n on the left hand side and right hand side are different. One is the index of the summation, the other is the upper bound of the summation.

But i thought as √N tends to infinity almost all primes are included and the argument might still hold good.

The issue is how fast √N goes to infinity compared to N. (Hey, this is related to the other thread!)

Your contention is that

\prod_{p<\sqrt N} 1-p^{-1} \sim (\ln N)^{-1}

I believe this to be true, but only because the PNT is true. I don't see an elementary proof of it.

Your argument is that if N were infinity, then the LHS would be equal to the harmonic series, whose partials are approximated by ln N. But to see why this argument fails, consider another function in place of √N as a bound. Your argument would imply that

\prod_{p<f(N)} 1-p^{-1} \sim (\ln N)^{-1}

for any f that goes to infinity. But I can choose an f to make the LHS grow at any speed I want. Say I want it to be like N^{-2}. Then I define

f(N) = \sup (i | \prod_{p<i} 1-p^{-1} > N^{-2})

We certainly have f(N) -> infinity since N^{-2} -> 0, so your argument should apply, but we constructed the function specifically to fail the conclusion.