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    (Original post by Mladenov)
    I am too busy these days, and I considered the problem superficially.
    I need an upper bound.
    Here is one sufficient upper bound - for k>1, we have p_{k} < 2k \ln k +2. Now, it is obvious that \displaystyle \sum_{p \in \mathbb{P}} \frac{1}{p} > \sum_{k \ge 2} \frac{1}{2k \ln k +2}; the last series diverges.
    But where have you got that from? :|

    And why is it obvious that the second sum diverges?

    I've been told that, in tripos exams, you must prove any theorems you use that are not taught on the course So, if you were given a question like this, that's what you'd need to do

    I assure you that there exists an elementary solution! More than one intact! The one I case across was by Erdos and was based on "proof by contradiction"
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    (Original post by Jkn)
    But where have you got that from? :|

    And why is it obvious that the second sum diverges?

    I've been told that, in tripos exams, you must prove any theorems you use that are not taught on the course So, if you were given a question like this, that's what you'd need to do

    I assure you that there exists an elementary solution! More than one intact! The one I case across was by Erdos and was based on "proof by contradiction"
    Since he's busy I'll answer for him. The theorems he uses are certainly in the course (and are very likely to be taught quite early). From the Prime Number Theorem (specifically Dusart's inequality) one has:

    p_k < k\left(\ln k + \ln \ln k\right) for k \geq 6

    and clearly this can be relaxed to p_k < 2k \ln k + 2 for k > 1 as \ln k > \ln \ln k in that range. The series can then be shown to be divergent through say the integral test.

    Spoiler:
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    This is in fact a variation of proof 4 on Wikipedia's article on this theorem.
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    (Original post by ukdragon37)
    Note that also when in range \dfrac{1}{2k \ln k + 2} is always at least \dfrac{1}{2} [...] diverges due to the term test.
    Are you sure about this?
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    (Original post by Jkn)
    Spoiler:
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    But where have you got that from? :|

    And why is it obvious that the second sum diverges?

    I've been told that, in tripos exams, you must prove any theorems you use that are not taught on the course So, if you were given a question like this, that's what you'd need to do

    I assure you that there exists an elementary solution! More than one intact! The one I case across was by Erdos and was based on "proof by contradiction"
    The inequality is based on a result of Rosser and Schoenfeld.
    But it can be proved with the prime number theorem.

    The series diverges since the series \displaystyle \sum_{v=2}^{\infty} \frac{1}{v\ln v} diverges, which follows from the fact that \displaystyle \sum_{v=2}^{\infty} \frac{1}{v} diverges. The last is, I believe, obvious.

    More generally, we have \displaystyle \sum_{v=2}^{\infty} \frac{1}{v^{\alpha}\ln^{\beta}(v  )} converges when \alpha >1 and \alpha=1, \beta >1.
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    (Original post by ukdragon37)
    You pedant but in any case I could just take the range to be k \geq 6, or k \geq 1000000 and it wouldn't change the argument at all

    (And also the method presented was a bit round-about, proof 4 in that wiki article being a much concise version of basically the same argument.)

    (EDIT: But I suppose Mladenov wanted to massage the terms into a form that is suitable for the term test, and hence "obvious" that it diverges.)
    I honestly wasn't being pedant, I don't see how \frac{1}{2k\ln k+2} is at least \frac{1}{2} for any k>1, or how the term test is even applicable here.
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    (Original post by Lord of the Flies)
    I honestly wasn't being pedant, I don't see how \frac{1}{2k\ln k+2} is at least \frac{1}{2} for any k>1, or how the term test is even applicable here.
    EDIT: No wait, you are right. I'm sprouting gibberish at this time in the night. Meh, just use another convergence test
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    (Original post by ukdragon37)
    :confused: Are you thinking \dfrac{1}{2k\ln \left(k+2\right)} instead of \dfrac{1}{2 + 2k\ln k}? In fact I think I was right to begin with. 2k\ln k is clearly positive for k > 1.
    LOL what is going on.

    \frac{1}{2k\ln k+2}<\frac{1}{2} when k>1 so how could it possibly be at least \frac{1}{2}? Also since \frac{1}{2k\ln k+2}\to 0 how does the limit test tell us anything?

    Edit: phew :lol: (I believe Cauchy works fine).
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    (Original post by Lord of the Flies)
    Edit: phew :lol: (I think Cauchy works fine).
    It's a typical "I think this is what he wanted" and then crafting a whole "explanation" based around that wrong assumption.
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    (Original post by ukdragon37)
    Since he's busy I'll answer for him. The theorems he uses are certainly in the course (and are very likely to be taught quite early). From the Prime Number Theorem (specifically Dusart's inequality) one has:

    p_k < k\left(\ln k + \ln \ln k\right) for k \geq 6

    and clearly this can be relaxed to p_k < 2k \ln k + 2 for k > 1 as \ln k > \ln \ln k in that range. Note that also (eventually) \dfrac{1}{2k \ln k + 2} is always at least \dfrac{1}{2}, so it is obvious that \sum_{k\geq 2}\dfrac{1}{2k \ln k + 2} diverges due to the term test.

    Spoiler:
    Show
    This is in fact a variation of proof 4 on Wikipedia's article on this theorem.
    Still not as nice as the Erdos proof though! Or Euler's! It relies on such sophisticated modern techniques that it seems, to me, inelegant (it's overkill!) :eek:

    I've read the Wikipedia article but thanks for the explanation. Not sure about the "eventually greater than a half" business though...

    Oh and btw, I certainly disapprove of your choice of Cambridge college, I've heard Emmanuel sucks... who would possibly want to go there!
    Spoiler:
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    Edit: I see I am not the only one that was lost by this so called "obvious" statement :lol:
    (Original post by Lord of the Flies)
    x
    It doesn't seem too hard to find a closed for for \int_{-\infty}^{\infty}\left(\frac{\sin  {x}}{x}\right)^ndx (yet!) so I'm going to naively have a go! (And probably drown in a sea of funny-looking Ls) :lol:
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    (Original post by Jkn)
    Still not as nice as the Erdos proof though! Or Euler's! It relies on such sophisticated modern techniques that it seems, to me, inelegant (it's overkill!) :eek:
    One could argue the more modern the technique the more elegant it is. Mathematics is not like technology, where a knife is always more elegant than a chainsaw.

    (Original post by Jkn)
    I've read the Wikipedia article but thanks for the explanation. Not sure about the "eventually greater than a half" business though...
    Yeah ignore that last bit, momentary brain fart due to contorting everything by jumping to conclusions at this time of the night. It is still easy to show it diverges though.

    (Original post by Jkn)
    Oh and btw, I certainly disapprove of your choice of Cambridge college, I've heard Emmanuel sucks... who would possibly want to go there!
    Spoiler:
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    I'll be gone by the time you come so I don't have to suffer how it sucks with you.
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    (Original post by Jkn)
    It doesn't seem too hard to find a closed for for \displaystyle\int_{-\infty}^{\infty}\left(\frac{\sin  {x}}{x}\right)^ndx (yet!) so I'm going to naively have a go! (And probably drown in a sea of funny-looking Ls) :lol:
    I wouldn't be so sure. I had a quick try after seeing your integral and could not see anything obvious. I shall try it again when I am more awake but it looks tough.
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    (Original post by Lord of the Flies)
    I wouldn't be so sure. I had a quick try after seeing your integral and could not see anything obvious. I shall try it again when I am more awake but it looks tough.
    It feels like I'm nearing a solution... I've got some insane stuff written down! I'm a Laplace transform newbie so I may have gone wrong... (I will trudge on!) :lol:
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    (Original post by ukdragon37)
    One could argue the more modern the technique the more elegant it is. Mathematics is not like technology, where a knife is always more elegant than a chainsaw.

    Yeah ignore that last bit, momentary brain fart due to contorting everything by jumping to conclusions at this time of the night. It is still easy to show it diverges though.

    I'll be gone by the time you come so I don't have to suffer how it sucks with you.
    Perhaps... though, when a two star problem is posted, it is not appropriate to leave one line solutions that rely on sophisticated theories most people on the thread haven't heard of.. and most likely won't for years to come :lol:

    What's it like? Did you do maths?
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    (Original post by Jkn)
    Perhaps... though, when a two star problem is posted, it is not appropriate to leave one line solutions that rely on sophisticated theories most people on the thread haven't heard of.. and most likely won't for years to come :lol:
    Nowhere does there say there is that rule - the stars just means a solution using simpler techniques are possible (which does not necessarily imply conciseness or elegance) but that doesn't mean a solver can't use a large hammer to crack the small nut if he wants to.

    (Original post by Jkn)
    What's it like? Did you do maths?
    It's great, will be very sad to leave. Even now there are students in other colleges who don't know that we get our laundry done for us so make sure you rub it in their face.

    Nah I did CompSci undergrad as well as Part III now.
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    (Original post by ukdragon37)
    Nowhere does there say there is that rule - the stars just means a solution using simpler techniques are possible (which does not necessarily imply conciseness or elegance) but that doesn't mean a solver can't use a large hammer to crack the small nut if he wants to.

    It's great, will be very sad to leave. Even now there are students in other colleges who don't know that we get our laundry done for us so make sure you rub it in their face.

    Nah I did CompSci undergrad as well as Part III now.
    Well I like elegance :cool: So people are going to have to put up with that

    I certainly will! Hahahaha :lol:

    Oh awesome, where are you off to next?
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    (Original post by Lord of the Flies)
    I wouldn't be so sure. I had a quick try after seeing your integral and could not see anything obvious. I shall try it again when I am more awake but it looks tough.
    Hmm... I've simplified it so that the Laplace transform relies on a really simple-looking yet awkward integral (I'm not sure I have tools to evaluate it, hmmm... :s)
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    (Original post by Jkn)
    Well I like elegance :cool: So people are going to have to put up with that

    I certainly will! Hahahaha :lol:

    Oh awesome, where are you off to next?
    I'm going to Cornell to do my PhD starting in August. Just had my US visa approved on Wednesday.
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    (Original post by ukdragon37)
    I'm going to Cornell to do my PhD starting in August. Just had my US visa approved on Wednesday.
    Wow, that sounds awesome! Congrats!

    That's what I'm thinking of doing! Parts I-III of the Cambridge tripos and then a PhD in the US (MIT or Caltech sound fun :rolleyes: but are probably hard to get into). Obviously I don't know if I'm going to end up doing this... but it's the only good idea I've come up with to explore/have-a-real-further-in my subject whilst attending a world-leading university and not being confined to one city all at the same time (i.e. the closest thing a mathmo will ever have a gap year :lol:)

    Is that what made you want to do it?
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    (Original post by Jkn)
    Wow, that sounds awesome! Congrats!
    Thanks!

    (Original post by Jkn)
    That's what I'm thinking of doing! Parts I-III of the Cambridge tripos and then a PhD in the US (MIT or Caltech sound fun :rolleyes: but are probably hard to get into).
    You can't possibly know at this moment where you would want to go, unless you know exactly what you want to research already and the strengths of each university in that research. The most important criteria for choosing your PhD is the potential advisors/actual advisor who admitted you at the university, and also its research in the specific field you want to go into. For example while MIT, Berkeley and Stanford commonly seem high-and-mighty and all that for CompSci, they were actually quite low on my list compared to Carnegie Mellon or Cornell as they did not have a good specialisation in the research direction I wanted to do. (Although they would have been nice to get into for pride points But that's another thing, universities tend not to take students either if they pursue directions that are uninteresting to their faculty.)

    (Original post by Jkn)
    Obviously I don't know if I'm going to end up doing this... but it's the only good idea I've come up with to explore/have-a-real-further-in my subject whilst attending a world-leading university and not being confined to one city all at the same time (i.e. the closest thing a mathmo will ever have a gap year :lol:)

    Is that what made you want to do it?
    Well I was being a bit bored of the UK, and the funding situation is much better in the US, so that's why I went instead of staying. It's not really a gap "year" though, in the US PhDs tend to be five years or more, averaging six, whereas in the UK pretty much you are kicked out after four years maximum (you are expected to finish in three) if you still haven't graduated. If you ever decide that this is too much of a slog, doing Part I - II in Cambridge and then a Master's in the US would achieve what you are saying
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    A fun one, for people who like solutions to historically notorious problems, that are today, within the reach of an A-level student and a wikipedia page:

    Problem 149 */**

    Part 1) Take  f(x) = \pi x - x^2, 0 \leq x \leq\pi . Find its periodic extension (ie, calculate its Fourier Series).

    Part 2) Solve the Basel Problem.

    Edit: The hope are for literally anyone who comes across this to give it a try to see some particularly nice mathematics. I just realised, this does need more thought than simply applying the definitions to construct a fourier series, since upon first glance you can't apply it directly (as f is not explicitly defined over an interval of the form -L to L).. if you get stuck here, that's when this spoiler will be useful

    Spoiler:
    Show
    f(x) here is an even function, and as such you don't even need to calculate any of the b_k, as obviously an even function has no odd parts! Now for the a_n, by the definition of the range x can take, you find limits of -L to L to be impossible.. well, since f is even, the integral can be doubled, and instead go from 0 to L. (where L will be \pi / 2 )
 
 
 
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