The Proof is Trivial!

Original post by Jkn

So, if you were given a question like this, that's what you'd need to do :tongue:

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" :smile:

Since he's busy I'll answer for him. :tongue:

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

\ln k > \ln \ln k

in that range. The series can then be shown to be divergent through say the integral test.

Spoiler

(edited 10 years ago)

Reply 942

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? :tongue:

Reply 943

Mladenov

Original post by Jkn

Spoiler

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

Reply 944

Original post by ukdragon37

You pedant :wink:

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 :tongue:

(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. :biggrin:

Reply 945

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. :biggrin:

EDIT: No wait, you are right. I'm sprouting gibberish at this time in the night. Meh, just use another convergence test :tongue:

(edited 10 years ago)

Reply 946

Original post by ukdragon37

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).

(edited 10 years ago)

Reply 947

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. :tongue:

Reply 948

Original post by ukdragon37

Since he's busy I'll answer for him. :tongue:

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

\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

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... :tongue:

Oh and btw, I certainly disapprove of your choice of Cambridge college, I've heard Emmanuel sucks... who would possibly want to go there! :s-smilie:

Spoiler

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

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:

(edited 10 years ago)

Reply 949

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. :tongue:

Mathematics is not like technology, where a knife is always more elegant than a chainsaw. :wink:

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... :tongue:

Yeah ignore that last bit, momentary brain fart due to contorting everything by jumping to conclusions at this time of the night. :tongue:

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! :s-smilie:

Spoiler

I'll be gone by the time you come so I don't have to suffer how it sucks with you. :wink:

Reply 950

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.

Reply 951

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:

Reply 952

Original post by ukdragon37

One could argue the more modern the technique the more elegant it is. :tongue:

Mathematics is not like technology, where a knife is always more elegant than a chainsaw. :wink:

Yeah ignore that last bit, momentary brain fart due to contorting everything by jumping to conclusions at this time of the night. :tongue:

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. :wink:

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? :biggrin:

Did you do maths?

Reply 953

Original post by Jkn

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. :tongue:

Original post by Jkn

What's it like? :biggrin:

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. :wink:

Nah I did CompSci undergrad as well as Part III now.

Reply 954

Original post by ukdragon37

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. :wink:

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 :colone:

I certainly will! Hahahaha :lol:

Oh awesome, where are you off to next? :smile:

Reply 955

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)

Reply 956

Original post by Jkn

Well I like elegance :cool:

So people are going to have to put up with that :colone:

I certainly will! Hahahaha :lol:

Oh awesome, where are you off to next? :smile:

I'm going to Cornell to do my PhD starting in August. Just had my US visa approved on Wednesday. :smile:

(edited 10 years ago)

Reply 957

Original post by ukdragon37

I'm going to Cornell to do my PhD starting in August. Just had my US visa approved on Wednesday. :smile:

Wow, that sounds awesome! Congrats! :biggrin:

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? :smile:

Reply 958