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

[bhobba]

I have been dealing with this problem for two years and proposed my solution in the article Y.N. Zayko, The Geometric Interpretation of Some Mathematical Expressions Containing the Riemann zeta-Function, Mathematics Letters, 2016; 2 (6): 42-46.

Very very interesting stuff. It emboldened me to explain, as promised, Ramanujan Summation - which I do below. The conclusion I have reached is its simply writing, in a reasonable way, the series so it can be summed.

It's based on the Euler-Maclauren series, a not what I would call rigorous, but short, derivation I will now give. Define the linear operator Ef(x) = f(x+1). Define Df(x) = f'(x). Also f(x+1) = f(0) + f'(x) + f''(x)/2! + f'''(x)/3! ... (the Maclauren expansion), or Ef(x) = f(x) +Df(x) + D^2f(x)/2! + D^3f(x)/3! ... = e^Df(x). We Also need the Bernoulli numbers which are defined by the expansion of x/(e^x - 1) = ∑B(k)*x^k/k! - you can look up the values with a simple internet search.

f(0) + f(1) + f(2) ... f(n-1) = (1 + E + E^2 ++++++E^(n-1))f(0) = (E^n -1/E-1)f(0) = (E^n -1)(1/e^D -1)f(0) = D^-1*(D/e^D -1)f(x)|0 to n = D^-1f(x)|0 to n + ∑ B(k)*D^(k-1)f(x)/k!| 0 to n = 0 to n ∫f(x) + ∑B(k)*D^(k-1)f(n)/k! - ∑B(k)*D^(k-1)f(0)/k!. Now the sum is from 1. Let n → ∞ and most of the time Sn = ∑B(k)*D^(k-1)f(n)/k! → 0 so we will assume that - certainly its the case for convergent series since → 0. So you end up with ∫f(x) - ∑B(k)*D^(k-1)f(0)/k!. So f(0) + f(1) + f2) +f(3) ...= ∫f(x) - ∑B(k)*D^(k-1)f(0)/k!. Notice regardless of n (- ∑B(k)*D^(k-1)f(0)/k!) does not depend on n and is called the Ramanujan sum. Again note the sum is from 1. We would like the Ramanujan Sum to be the same for concergent series. This is done by defining it as ∫f(x) + C where C is the Ramanujan sum. If it is finite then it is ∫f(x) + C. If it is infinite it is defined as C.

I won't go any further into it - you can look it up and see what various series give in various cases - it gives the good old -1/12 for 1+2+3+4... but with an interesting twist.

Thanks
Bill

 

[bhobba]

What does it mean that a sequence is Stable?

Take S = 1+1+1+1... . It is not stable because S = 1+1+1+1... = 1+S . It's inherent in the series. The interesting thing is by transforming that series to the Eta function you get S = (1-1+1-1... )/-1. The thing is in that form it is stable. 1-1+1-1... (called the Grande Series) is stable. One way to show a series is stable is to sum it by a methods that are known to only work with stable series - one such is Borel Exponentiation summation. Borel summation (which can be used to sum some series that are not stable - so is not a stable method) is usually easier to use in practice than the exponential method, but to help here, there is this theorem that says Borel Summation gives the same answer as Borel Exponential summation (hence must be stable) if limit t → ∞ e^-t*an = 0. Now applying that to Grande's series you see it's true. So if Borel summation gives an answer the series is stable ∫∑(-t)^n*e^-t/n! = ∫ e^-2t = 1/2. Also for stable series one can apply generic summation that uses that stability to figure out the sum. For the Grande Series this gives S = 1 - S or S=1/2. This means 1+1+1+1... = -1/2 despite the series not being stable. How this happens is you make reasonable assumptions about converting it to the Grande Series, for example we could have shown 1 + S = 1 -1/2 = 1/2 - that's why its not stable, its inconsistent etc. Knowing this we do make use of stability properties in calculating the sum so you can easily convert it to the Grande Series. Unless you do something like that you can't sum it. To be clear if a series is not stable when summing it you can't use stability ie you can't use S = a0 + S' when you sum it. So you do not do it.

One other thing I want to mention is where does the infinity in 1+1+1+1... etc go in the summation. Well I showed C(k) (1 - 2*2^k) = E(k) - where E(k) is stable and summable to a finite number. Its that (1- 2*2^k) term - the infinities in C(k) are canceled by the infinities in (2*2^k)*C(k) to give finite answers. I have done, in a smart way of course, that dreaded shuffling of infinities done in non-Wilsonian re-normalization that I so detest - and so did Wilson. He vowed and declared to sort it out - and did - for which he got a Nobel. The real answer in physics of course is you must have a cutoff so there is nothing infinite to begin with - that's the basis of Wilsonian Re-normalization - every theory must come with a natural cutoff to avoid infinities if they can occur:
http://scipp.ucsc.edu/~dine/ph295/wilsonian_renormalization.pdf

See equation 1.

The thing is this. A re-normalization technique called zeta function re-normalization exists - it's a bit of generalization of dimensional re-normalization and has some nice properties such as no counter-terms needed. I at first thought - this may be the clue that Wilson may not be right - on investigation - bummer - he was. Well he was a Putman Fellow twice at 19/20 or so (he entered Harvard at about 16)- I can't compete with that.

The reason analytic continuation works is now simple from converting it to the Eta function - that is always finite summable by say Borel Summation. If C(k) is the same when k<0 it must be the same in the entire place - except k=-1 of course which is a non-removable singularity. This is more easily seen using Ramanujan Summation which I will post a bit later - have to go to lunch now (I eat a lot of lunches don't I)

Thanks
Bill

 

[Svein]

I think this thread should mention cardinal numbers (https://en.wikipedia.org/wiki/Cardinal_number) where addition, subtraction etc. works somewhat differently from standard numbers.