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I What is convergence and 1+2+3+4...... = -1/12

  1. Nov 7, 2018 #1

    bhobba

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    Ok - anyone that has done basic analysis knows the definition of convergence. The series 1-2+3-4+5..... is for example obviously divergent (alternating series test). But wait a minute lets try something tricky and perform a transform on it, (its Borel summation, but that is not really relevant to this).

    ∑an = ∑(an/n!)n! = ∑ (an/n!)∫t^n*e^-t where the sum is from n=0 to ∞ and the integral also from 0 to ∞. Now suppose ∑|(an*(xt)^n/n!)e^-t| < ∞ then dominated convergence applies and the integral and sum can be reversed, if the integral exists. It usually does for some x (in the example its |x|<1), but in the actual integral for at least x =1, which means it can be considered an analytic continuation to x=1 so ∫∑((an*t^n)/n!)*e^-t can be taken as ∑an. Or you can consider it limit x → 1- if it exists for |x|<1 because, again from dominated convergence, its continuous in x.

    Now lets apply this to 1-2+3-4 ....... where the sum is from n =1 to ∞. We have ∫∑(-n*(-t)^n*e^-t)/n! = ∫∑(-(-t)^n*e^-t)/(n-1)! = ∫∑(-(-t)^(n+1)*e^-t)/(n)! = ∫t*e^-t*e^-t = ∫t*e^-2t = 1/4.

    Are divergent series sometimes really convergent but simply written in the wrong form?

    Added Later:
    Made a goof, but fixed it. Thought I could get away without using analytic continuation - just the limit - but I was wrong. Certainly in the example you can use just the limit - but not more generally.

    Just as an aside from Wikipedia:
    Borel, then an unknown young man, discovered that his summation method gave the 'right' answer for many classical divergent series. He decided to make a pilgrimage to Stockholm to see Mittag-Leffler, who was the recognized lord of complex analysis. Mittag-Leffler listened politely to what Borel had to say and then, placing his hand upon the complete works by Weierstrass, his teacher, he said in Latin, 'The Master forbids it'.
    Mark Kac, quoted by Reed & Simon (1978, p. 38)

    Looks like Mittag-Leffler had trouble with it, but interestingly Borel wrote a book on Divergent series later, or maybe it should be superficially divergent series:-p:-p:-p:-p:-p:-p:-p:-p.

    Thanks
    Bill
     
    Last edited: Nov 7, 2018
  2. jcsd
  3. Nov 7, 2018 #2

    jedishrfu

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  4. Nov 7, 2018 #3

    bhobba

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    Thanks Jedishrfu. Didn't know that reference. I will get onto how you use this to do that strange sum. Again it's writing it in a different form that has a better range of convergence. Do not agree 100% with it - I think it really is -1/12 - it's just in the form its written it has unnecessary restrictions on the real value which you see when in the correct form. But that is something that can be discussed once we sort out my first post. I made a bit of a goof first up and thought you didn't need analytic continuation. In the case I gave you just needed to take a limit - but in general you have to use the full power of analytic continuation.

    Thanks
    Bill
     
    Last edited: Nov 7, 2018
  5. Nov 7, 2018 #4

    jedishrfu

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    It’s okay I fixed your title too as you dropped a one in the denominator.
     
  6. Nov 7, 2018 #5

    jedishrfu

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    I’m afraid my math is a bit superficial too so I won’t be able to contribute much in this thread. I liked the Borel story and the master forbids it. There are some similar ones I’ve heard with Pauli too.
     
  7. Nov 7, 2018 #6

    bhobba

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    That's fine. I suspect I will not get much interest anyway - its a bit esoteric. What I noticed is there are a number of threads here about it but all were shut down - they got out of hand. Just wanted to have a permanent record that as a mentor I will ensure remains 'calm'.

    Even though esoteric I think that understanding exacltly whats going on with such things is important - you see it all over the place you cant do it. However good write-ups on analytic continuation, IMHO, tell a more subtle, but nonetheless different story eg:
    https://www.colorado.edu/amath/sites/default/files/attached-files/pade_analytic_continuation.pdf
    'The key point in this example is that all the three functions are identically the same. The boundaries for f1(z) and f2(z) are not natural ones, but just artifacts of the particular way we used for representing the function f3(z)'

    Divergences often are the same - a result of not writing it in a more natural way.

    Thanks
    Bill
     
  8. Nov 8, 2018 #7

    haushofer

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    Hopefully adding something: I think the whole notion of "the sum of all integers equals -1/12" is bad notation. What's actually meant, is that "if you rewrite the sum of all integers in terms of a complex function and perform an analytic continuation on it, then the value of it equals -1/12". And because the analytic continuation is unique, people sometimes forgive the lack of this subtlety.
     
  9. Nov 8, 2018 #8

    stevendaryl

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    It seems to me that the various summation techniques amount to this:
    1. You have a divergent sum: ##\sum_j a_j##.
    2. You come up with a sequence of expressions ##A_j(\lambda)## such that for each ##j##, ##lim_{\lambda \rightarrow 1} A_j(\lambda) = a_j##.
    3. Then you come up with an analytic function ##F(\lambda)## such that for some values of ##\lambda##, ##F(\lambda) = \sum_j A_j(\lambda)##
    4. Then you declare that ##\sum_j a_j = F(1)##
    If the original sum were convergent, I'm sure that this technique would give a unique answer for the sum. But if the sum is divergent, then it seems to me that there could be multiple functions ##F(\lambda)## that would work.

    Here's an example of multiple ##F##, but it actually leads to the same answer.

    The modified sum: ##1 - 2 + 3 - 4 + ...##

    We can write that as ##\lim_{x \rightarrow 1} 1 - 2x + 3x^2 - 4x^3 ...##

    That converges to the function ##F(x) = \frac{1}{(1+x)^2}##. We can immediately evaluate it to get: ##lim_{x \rightarrow 1} \frac{1}{(1+x)^2} = 1/4##

    An alternative summation is:

    ##1-2+3-4+... = lim_{x \rightarrow 1} 1^{x} - 2^{x} + 3^{x} - 4^{x} + ...##

    The function ##F'(x) = \sum_j j^{x} (-1)^{j+1}## is related to the zeta function:

    ##F'(x) = \zeta(-x) (1-2^{1+x})##

    This function is very different from ##F(x) = \frac{1}{(1+x)^2}##, but they have the same value at ##x=1##, namely 1/4.

    Off the top of my head, I can't think of an example where you get two different answers, but I also don't see any reason for it to always give the same answer, no matter which ##F## you choose
     
  10. Nov 8, 2018 #9

    bhobba

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    I will give the relation of this to that controversial 1+2+3+4...... = -1/12 soon, but would like, so people understand it better to flest out Borel summation a bit more.

    To recap about Borel summation ∑an = ∑(an/n!)*n!. n! = Γ(n+1) = ∫t^n*e^-t so ∑an = ∑an. In general you cant interchange the sum and integral but under some conditions you can so we will formally interchange them and see what happens so ∑an = ∫∑(an/n!)*t^n*e^-t. This is called the Borel sum and we will look at when it is the same as ∑an.

    If ∑an is absolutely convergent then by Tonelli's theorem since all terms are positive the integral and sum can be interchanged ∑|an| = ∑∫|an|/n!)*t^n*e^-t = ∫∑|an|/n!)*t^n*e^-t. Then by Fubini's theorem the sum and integral can be reversed and the Borel sum is the same as the normal sum. Consider the series S =1 + x + x^2 + x^3 ............ It is absolutely convergent to 1/1-x, |x| < 1. Then the Borel Sum is S = ∫∑(x^n/n!)*t^n*e^-t = ∫e^t(x-1) = 1/1-x < ∞ if x<1 - not only for Ix|<1. In other words - when S is convergent in the usual sense then it is equal to its Borel Sum but S is only valid for |x| < 1, however the Borel Sum is still the same - but true for more values of x. Borel summation has extended the values of x you get a sensible answer - in fact exactly the same answer. This is the characteristic of analytic continuation ie if a function in a smaller region it is exactly the same function in a larger region. Borel summation has extended the region the series has a finite sum. Normal summation introduces unnecessary restrictions on the sum that Boral Summation removes - at least i part. This of course works for 1+ 2x + 3x^2 +4x^3 ....... and is left as an exercise to show its Borel and Normal sum are the same ie 1/(1-x)^2.

    There is also another way of looking at this by introducing whats called the Boral exponential sum. Personally I don't use it much but sometimes its of dome use. It is defined as limit t → ∞ e^-t*S(t). Here Sn = a0 + a1 +a3 +.....+an so limit n → ∞ Sn = S, and S(t) = ∑Sn*t^n/n!. Its not hard to see if Σan converges normally to S then its exponential sum also converges to S. limit t → ∞ e^-t*S(t) = Sn limit t → ∞ e^-t*t^n/n! = 0 after using L-Hopital n times on e^-t*t^n. So let K be large then Sk is close to S with it getting closer and closer as K gets larger. limit t → ∞ = e^-t (S0 + S1*t + S2*t^2/2! ++++++++ e^-t*t^k-1/(k-1)! + ∑S*t^n/n! (from K to infinity - and we use Sk. Sk+1, Sk+2 ..... = S for all practical purposes since K is large and gets better as K gets larger). Hence the exponential sum is ∑S*t^n/n! (from K to infinity). But since the terms preceding it are all zero we have the exponential sum is limit t → ∞ e^-t*S*∑t^n/n! = S. Thus if ∑an converges in the usual sense to S then S is also the exponential sum

    Now we will show something interesting - if the exponential sum S exists then the Borel Sum exists and is also S. However the reverse is not true.

    Let B(t) = ∑an*t^n/n! = a0 + (S1-S0)*t + (S2-S1)*t^2/2! + (S3-S2)*t^3/3! ++++++. Hence B(t)' = S(t)' - S(t) .
    S - a0 = e^-t*S(t)| from 0 to ∞ = ∫d/dt [e^-t*S(t)] = ∫e^-t*(S(t)' - S(t)) = ∫B(t)'*e^-t = e^-t*B(t)| (o to ∞) + ∫e^-t*B(t) = -a0 + ∫e^-t*B(t). On cancelling a0 we end up with what was claimed S = ∫e^-t*B(t) which is the Borel Sum. We also have shown if ∑an normaly converges to S then the Borel Sum is also S. You can also see this intuitively using an argument to the exponential sum giving S. If K is very large all the aK beyond K are for all practical purposes 0, with the approximation getting better and better as K gets larger. But the Boral Sum S then becomes the sum to k as ∫e^-t*t^n/n! =1 so it is S.

    What this is saying is the Boral Sum is exactly the same for normally convergent sums. But if it not normally convergent it can still give an answer. Not only this but if ∑an*x^n has any non zero radius of convergence the Borel Sum is exactly the same as the normal sum in the radius of convergence but is an analytic continuation for all x it exists. And the view of the paper I linked to on analytic continuation is its simply removing an unnatural restriction in the way the sum is written. So, one way of viewing Borel summation is simply removing an unnatural restriction in the way a series is written so it be expressed in a more natural way.

    Anyway I have another example of this in deriving 1 +2 +3 +4 ...... = -1/12. But will wait for comments now.

    Thanks
    Bill
     
    Last edited: Nov 8, 2018
  11. Nov 8, 2018 #10
    I have never understood this equivalence. As stevendaryl said:

    I mean, isn't it possible to find another function that, when s=0 gives "the same" as this sum and when you analytically continue it, you get something different than -1/12????

    Thanks
     
  12. Nov 8, 2018 #11

    stevendaryl

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    Following your definitions (to the extent that I understand them), I don't get a finite answer.

    You start with ##A = \sum_n n##.
    You get a related series: ##B(t) = \sum_n n t^n/n!##
    Then you can write: ##A = \int_0^\infty B(t) e^{-t} dt = \int_0^\infty \sum_n n t^n/n! e^{-t} dt = \sum_n n/n! \int_0^\infty t^n e^{-t} dt = \sum_n n## (since the inner integral gives ##n!##)

    But ##B(t) = t e^t##, which gives ##\infty## when you integrate ##A = \int_0^\infty B(t) e^{-t} dt##
     
  13. Nov 8, 2018 #12
    Noone answered my question but I will state it in a more formal way. I saw different derivations of the "-1/12" result, all of them look like this:

    There is, at least, a function F(n,s) such that:
    1) F(n,-1)=1+2+3+4+...n (for finite "n"). I'm using here s=-1 for the place in which F is equal to 1+2+3...+n in order to emulate the demostration that uses the z function. Clearly, it does not matter which value of the s variable we use for this equality to happen.
    2) The limit of that function where "n" tends to infinite is well defined in some region of the "s" variable, but not in the point s=-1
    3) Nevertheless, you can analytically continue this function in the s variable so that we can asign a value to F(infinite;-1).

    4) It seems to me that, no matter which function F we use, F(infinite;-1)=-1/12.

    This would be an astonishing result, wouldn't it? ... but... Is 4 proved somewhere?


    Ps: To me it looks like using this procedure with different F could give different results, but every other example that I find always results in -1/12!!!!!!!!
     
  14. Nov 9, 2018 #13

    bhobba

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    Ok here is the issue in clear form. Now 1^k + 2^k+ 3^k ........... = ζ(-k) where ζ is the standard zeta function 1 + 1/2^s + 1/3^s ........... = ζ(s). This function is only convergent for s>1. But the Riemann Hypothesis says the primes are the zeroes on the imaginary axis through s =1/2. But it only exists if s>1. Wow I have solved one of the million dollar Clay prizes - its undefined - yippi. But let us not get up our hopes up. You see there is another function called the Eta function defied by η(s) = 1 - 1/2^s + 1/3^s - 1/4^s ....... Now one can easily derive a simple relation between the two:
    https://proofwiki.org/wiki/Riemann_Zeta_Function_in_terms_of_Dirichlet_Eta_Function

    We have ζ(s) = η(s)/(1-2^(1-s)). Note we have just done some algebraic manipulations to arrive at this equation. Also note n(s) is convergent for s >0 (alternating series test) - we have extended the vales of s so make the Riemann Hypothesis well formulated (it still has issues at s =1 ie the harmonic series - it has a singularity there). Also using Borel Summation the Eta function is summable.

    We can apply it easily - we have 1 + 2^k + 3^k +++++++ = (1 - 2^k + 3^k - 4^k.......)/(1- 2^(1+k)) and calculate 1+1+1+1....... and 1 + 2 + 3 +4 ....

    If K=0 we have 1+1+1+1....... = (1-1+1-1+1.....)/-1 = -1/2
    If K=1 we have 1+2+3+4....... = (1-2+3-4........)/-3 = -1/12.

    I have used Borel Summation but Steven used Abel summation which is based on Abel's Theorem:
    https://sites.math.washington.edu/~morrow/335_16/AbelLaplace.pdf

    Notice the second bit about the Laplace Transform. The Borel Sum is the Laplace Transform of B(t) with S=1. Let S = 1/x and then take the limit x→1- just like Abel Summation.

    Lets look at (1/x)*∫B(t)*e^-t/x. Do a change of variable to t' = t/x. We get ∫Σ(an*(x*t)^n)/n!)*e^-t. In a similar way to Abel Σan*x^n is absolutely convergent if |x|<1 (hence is exactly the same as the Borel Sum) and Abel says take the limit x→1-, so we can take limit x→1- of ∫Σ(an*(x*t)^n)/n!)*e^-t to give the Borel Sum ∫Σ(an*t^n)/n!)*e^-t. So Abel and Borel summtion are related. By taking the limit x → 1- you get both - but Boral is more general.

    Note, and this is the key point, we have have not done anything except putting divergent series in different forms, and in those forms apparent divergences are gone. Basically it's as the article on analytic continuation says - many divergences occur simply because it is written in an unnatural form. When written in a better form the divergence disappears. Not all divergent series are like that eg the divergence in the Harmonic series 1 + 1/2 + 1/3 .......... can't be removed. This is where analytic continuation comes in - the Harmonic series is a singularity that cant be removed.

    Thanks
    Bill
     
    Last edited: Nov 9, 2018
  15. Nov 9, 2018 #14

    bhobba

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    No - from the theory of analytic continuation it is unique. It is easily seen if f1 and f2 are analytic continuations then on the region where they are the same f1-f2 =0. Analytic continue 0 by any method and it's continuation is zero. This means f1 -f2 = 0 in the region of continuation and f1 =f2 everywhere

    This is what leads to so called generic summation. The analytic continuation of a series S = a0 + a1x +a2x^2 + ...... that has some radius of convergence not = 0 is zero in that region. Since it is unique it has values at all x and you have the normal sumability properties of linearity and stability - the last one is simply S = a0 + ( a1x +a2x^2 + ......).

    So lets take S = 1 + x + x^2 .............

    We have Sx = x + x^2 + x^3 ........

    S - Sx = 1 so S 1/1-x regardless of x - except of course x=1 which is a non removable singularity - you need a more powerful method of summation for 1+1+1+1......

    While many methods extend convergence, when looked at they are just forms of analytic continuation and we can use generic summation to often get values for all x - not just those where it is naturally convergent. Again we have written it in a form with unnecessary restrictions and analytic continuation allows you to remove them - often anyway.

    Here is another way of looking at it. Suppose someone has cut a circle out of the complex plane and then asks you what is the function in the whole plane. Giving you that circle is an unnatural restriction on the function - like the divergent series. But that small circle is enough to reconstruct the full function and remove that restriction that is artificial anyway.

    That's why I say 1+2+ 3+4....... really equals -1/12. The way its written obscures artificially its real value which can be recovered by some perfectly valid manipulations. I know many say its not true - I know Micromass for example did a very scathing appraisal of it years ago, saying it made him really mad saying its -1/12. I cant prove him wrong - but what I hope I have shown is you can also view it as simply removing unnatural restrictions and it really has that value.

    Thanks
    Bill
     
    Last edited: Nov 9, 2018
  16. Nov 9, 2018 #15

    stevendaryl

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    But a sum is not an analytic function. The expression ##1+2+3+4+...## has no variables in it, so you can't analytically continue it. What you can do is to introduce a variable, for example: It is the limit as ##t \rightarrow 1## of ##1^t + 2^t + 3^t + ...##. That can be analytically continued.

    So in order to apply the theorem that analytic continuations are unique, you first have to modify the sum to parametrize it. There isn't a unique way of doing that. For example, instead of the above approach, you could also use:

    limit as ##t \rightarrow 1## of ##1 + 2t + 3t^2 + 4 t^3 + ...##.

    That has a different analytic continuation--it's a different function. So I don't understand how you can guarantee that two different ways to "enhance" the original series by introducing a parameter ##t## will give you the same analytic continuation at ##t=1##.
     
  17. Nov 9, 2018 #16
    I was going to ask the exact same thing as stevendaryl!!!!
     
  18. Nov 9, 2018 #17

    bhobba

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    That is a good point - how to parameterize the function that you analytically continue. What I have shown is that Borel Summation - which requires no 'direct' parameterization ie ∫Σan*t^n e^-t/n! (the t comes from writing n! = Γ(n+1) = ∫t^n*e^-t) is a natural extension of ordinary summation. Power series are easily and uniquely analytically continued. In fact any sum ∑an can be written as ∑an*x^n where x is one. Thar is unique with a unique analytic continuation providing its radius of convergence is not zero. I have transformed the zeta function into the Eta function which is also analytic (though not as easy to see). But it's true pathological functions exists that do not have a unique analytic continuation, or even any analytic continuation at all.

    This means I will have to limit my remarks to those functions that can be uniquely analytically continued - most are eg Boral summation gives a unique result and analitic continuation, and simple algebraic manipulations on such functions like the zeta function to the Eta function are true.

    So we will have to impose the rule generic summation will work on it - that means its sum is unique by any method. For 1 +2 +3 ....... Note it is equal to 1^1 +2^1 +3^1 ....... - its a direct equality - and the relation to 1^1 - 2^1 + 3^ .... is a direct equality so beyond doubt. Now one can use generic summation on 1 -2 + 3 - 4 to show its sum must be 1/4. Its unique - any summation method must give it.

    That said - you are correct - it must be done with care to show the answer is unique.

    Thanks
    Bill
     
    Last edited: Nov 10, 2018
  19. Nov 10, 2018 #18

    haushofer

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    That's a good point; so to rephrase it, I guess my quote ""if you rewrite the sum of all integers in terms of a complex function" can be done in different ways, and I can't think of a reason why all these different ways still would give the same answer after analytical continuation; after all, you're analyically continuing different functions.

    I don't have a clear-cut answer to that.
     
  20. Nov 10, 2018 #19

    bhobba

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    Its the standard transformation between the Zeta and Eta functions. Even though its an equality doing this has a number of advantages - first the Eta function converges for S>0 but the Zeta function for S>1. It simply is an advantage of writing it in a different but equal form. The second advantage is a bit more subtle - and shows there is an unstated assumption being made - The Zeta function is not stable - this is easily seen considering s=0 so you have 1+1+1+1........... This is the same as 1 + (1+1+1+1........). So you cant use generic summation. I think it was Hardy that showed because of that when written in that form you cant assign any value to it - you really must make further assumptions.

    This is the real key - what assumption was made in deriving this equality? Each an in Σan must be associated with a n^s so you have an implied order making the series stable. In 1 + 2 + 3 + 4..... the ordering is obvious ie you generalize to 1^k + 2^k + 3^k ......... This also means when you have Σan*n^k + Σbn*n^k it equals Σ(an+bn)*n^k. Although obvious, and easily provable for finite and conventionally convergent sums, it is something that is assumed - the n^k is associated by the order of the series - this can break down if you have zeroes - you might think removing them would make no difference but because its not stable it can.

    Assuming the above the transformation from the Zeta to Eta function is rigorous - we now have a series that has had it's region of convergence extended and is stable so generic summation can now be applied.

    Its an assumption, but I think most would accept it.

    Thanks
    Bill
     
  21. Nov 10, 2018 #20

    stevendaryl

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    I agree that Borel summation seems like a reasonable way to extend the set of summable series.

    However, here's the abstract description of the issue.

    Mathematically, an infinite sequence is an infinite-dimensional vector, an element of the vector space ##V = R^\omega##. Let's make up a term, an "evaluation", which is a partial linear map from ##V## into ##R##. An evaluation has a domain, which is the set of infinite sequences that it evaluates.

    There is one particular evaluation, which I'll call ##\Sigma##, which is the usual notion of the limit of the finite sums of terms in a sequence. Its domain is sum subset of ##R^\omega##.

    The goal in summation techniques is to come up with an evaluation ##E## such that
    1. ##dom(E) \supset dom(\Sigma)##: It declares more sequences to be summable.
    2. If a sequence ##s## is in ##dom(\Sigma)##, then ##E(s) = \Sigma(s)##. It agrees with ##\Sigma## on all convergent series.
    Examples of such evaluations include Borel summation and Abel summation and Zeta regularization, etc.

    My concern is the possibility that you might have two different evaluations ##E_1## and ##E_2## that give different answers for a particular series.
     
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