## Divergent Series

For the discussion of math. Duh.

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Voekoevaka
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### Divergent Series

Most of the mathematicians claim that divergent series don't have a result, or tait their result is ∞.

But Euler himself (happy 306th birthday) has published himself an article entitled De Seriebus Divergentibus, where he explains that we can find results for divergent series. In this essay, he described some methods to sum divergent series, then he tried to compute one of them (1-1+2-6+24-120+...=Σn=0n!), using four differents methods that leads him to the same result. Studies about divergent series were made by Ramanujan too.

But today, in traditionnal context, finding a sum for divergent series is absurd, and publications about there are rare. I have a discussion with a math teacher about such results but he says it is useless and clearly false.

I consider myself that summing divergent series is legitimate, and infinity should be reconsiderated. Here is a simple proof about divergent series, which is legitimate with p-adic numbers, but not with ordinary numbers :

S=1+2+4+8+16+...
S=1+2.1+2.2+2.4+2.8+...
S=1+2.(1+2+4+8+...
S=1+2S
S=-1

The same result can be found with the geometric series formula :

S=1+2+4+8+...=1/(1-2)=-1

In classical mathematics, this formula (S=1/(1-r)) must be used with |r|<1.

When I says 1+2+4+8+...=-1, I don't claim that when we add the powers of two, the result will magically converge to -1. I think that, seen through operations, the series have the behaviour of -1. And I also think that most divergent series could have the behaviour of a finite number or expression.

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jestingrabbit
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### Re: Divergent Series

So long as you keep the two concepts clearly separate ie have on the one hand what you call "classical" convergence, and on the other hand have these other possible notions of convergence as a separate thing, I don't really think they can do much damage.

But I do have a question. You can show, with a particular algebraic manipulation, that the limit of the sum of powers of 2 is -1. Can you show that there are no algebraic manipulations that show it is something else?
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Voekoevaka
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### Re: Divergent Series

jestingrabbit wrote:You can show, with a particular algebraic manipulation, that the limit of the sum of powers of 2 is -1.

I wasn't talking about a limit. The definition of a limit doesn't make sense in this case. I was talking about the properties of the series.

jestingrabbit wrote:Can you show that there are no algebraic manipulations that show it is something else?

When you are at the step S=1-2S, there are two roots of the equation : -1 and ∞. But most of the equations does have ∞ as a solution on the Riemann sphere. As ∞ is a result of almost equations, it doesn't makes the other root not legitimate. It is not known if it exists a summation operator for divergent series which is an extension of the classical one and which gives one and only one result for each series. But as S is geometric series, there is only one analytic continuation of the generalized geometric series.

When you define the function f(x)=1+x+x2+x3+..., you can write f(x)=1/(1-x) assuming |x|<1. This function does have an unique analytic continuation which gives f(2)=-1.

But I don't think a generalized series summation operator would give an unique result in all cases. This operator could be like a multivalued function : there can be more than one result correspounding to one series.
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Dark Avorian
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### Re: Divergent Series

Well actually, it is the limit under the p-adic ultrametric, as you mentioned.
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z4lis
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### Re: Divergent Series

Try thinking about the reverse problem. Begin by a nice linear relation, say:

S = 2 + 5S

Now, start expanding, substituting S in for itself each time:

S = 2 + 5(2 + 3S) = 2 + 10 + 25S

= 2 + 10 + 25(2 + 3S) = 2 + 10 + 50 + 125S

and we can spot how the formula unwinds itself. At the nth iteration, it appears as if we see

$S = 2 + 2*5 + 2*25 + \cdots + 5^n S$

and so this would give the series $\sum_{i=0}^\infty 2 5^i,$ whatever that might mean. Now, if you aren't worried about convergence, it's perfectly fine to talk about formal power series like this, as long as your rules of manipulation are set up properly.What might an algebraic object that satisfies S = 2 + 5S in some ring look like? If 4 is a unit in the ring, then we can solve for S. But if 4 is not a unit in the ring... perhaps we want to add such an object. How might we do it? Would we want to use a power series like that? I've just learned about some possibly related things in a class. Look at http://en.wikipedia.org/wiki/I-adic_topology and see if that might give some of the tools you want. It's basically a method of formally adding something like certain power series to a ring, just like the p-adics.
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Cleverbeans
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### Re: Divergent Series

S = 1+(-1)+1+(-1)+...
S = 1 - S
S = 1/2

S = 1+(-1)+1+(-1)+...
S = 0 + 0 + 0 + 0 +...
S = 0

Just to show that moving brackets around on infinite sums can cause trouble.
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dudiobugtron
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### Re: Divergent Series

Voekoevaka wrote:S=1+2S
S=-1

This is the step that doesn't really make sense to me. S is a solution to the equation S = 1 + 2S (in some sense), and so is -1 , but it doesn't mean that S = -1. If S was a finite number then it would be -1 (as well as any other finite number you care to choose), but we know it's not a finite number.

phlip
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### Re: Divergent Series

Cleverbeans wrote:Just to show that moving brackets around on infinite sums can cause trouble.

IIRC you can only arbitrarily move the brackets around in an infinite sum if it's absolutely convergent. If not, you're stuck with what you can get to from finitely-many applications of the associativity of addition... which means that you have finitely many terms grouped however you want, and then the rest of the infinitely-many terms being added left-to-right.

In particular, this means you can't rearrange:
S = 1 + 2 + 4 + 8 + 16 + ...
into:
S = 1 + (2 + 4 + 8 + 16 + ...)
unless S exists. You also can't substitute in to make S = 1 + 2S unless S exists.

So what you end up is not the statement "S = -1" but rather "If S exists, then S = -1". Which is a true statement, as the antecedent is false. But you can't use that as part of an argument that S exists without begging the question.

The difference between this, and doing the same manipulation for things that do converge, like turning
S = 1/2 + 1/4 + 1/8 + 1/16 + ...
into "S either equals 1 or does not exist", is that you find that S does exist, and equals 1.

And these manipulations can be useful when you're trying to figure out if a solution exists or not... you can narrow down "either it doesn't exist or it's some real or complex number" to "either it doesn't exist or it's this particular number", which can make the next step for showing whether or not it exists either easier to find or easier to prove... but it's not the end in itself.

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Voekoevaka
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### Re: Divergent Series

Cleverbeans wrote:S = 1+(-1)+1+(-1)+...
S = 0 + 0 + 0 + 0 +...
S = 0

In my opinion, this second proof if false. After S=0+0+0+0+..., you can factorize this by S=0.∞. The result is indeterminate. The first result (1/2) can be obtained by a lot of methods, like Cesaro summation or by taking the limit of Σn=0(-1)nxn where x→1.

This is the step that doesn't really make sense to me. S is a solution to the equation S = 1 + 2S (in some sense), and so is -1 , but it doesn't mean that S = -1. If S was a finite number then it would be -1 (as well as any other finite number you care to choose), but we know it's not a finite number.

As I previously say, This result is not the real "sum" of the series, but it is a value correspounding to the properties of the series. And, as mathematics are a way to study the properties of mathematical objects, I think such results should be ommited.

phlip wrote:In particular, this means you can't rearrange:
S = 1 + 2 + 4 + 8 + 16 + ...
into:
S = 1 + (2 + 4 + 8 + 16 + ...)
unless S exists. You also can't substitute in to make S = 1 + 2S unless S exists.

Considering that such values for such series doesn't exist is more axiomatic than prooved. Results about divergent series are so opposed to common sense that mathematicians have prefered to avoid such results.
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Xenomortis
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### Re: Divergent Series

Voekoevaka wrote:In my opinion, this second proof if false. After S=0+0+0+0+..., you can factorize this by S=0.∞. The result is indeterminate. The first result (1/2) can be obtained by a lot of methods, like Cesaro summation or by taking the limit of Σn=0(-1)nxn where x→1.

0+0+0+... is stronger than 0*inf
0*inf is indeterminate in the general case, but this is more specific.

Regardless, you do need to be careful when rearranging terms in a series. If your series is absolutely convergent then it's fine, but if it's only conditionally convergent then rearrangement is liable to change the sum. In fact, a conditionally convergent series can be rearranged to converge to any real number (or diverge). Voekoevaka
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### Re: Divergent Series

Xenomortis wrote:Regardless, you do need to be careful when rearranging terms in a series. If your series is absolutely convergent then it's fine, but if it's only conditionally convergent then rearrangement is liable to change the sum. In fact, a conditionally convergent series can be rearranged to converge to any real number (or diverge).

That's why I think a lot of rearrangements are not possible with divergent series. It il also the case with semi-convergent series. Take for example S=Σn=1(-1)nn-1. This series is not absolutly convergent, but its sum is equal to ln(2). But, by rearranging the terms, you can make this series converge to every value you want.

For example :

1-1/2+1/3-1/4+...=ln(2)
1+1/3-1/2-1/4-1/6+1/5+1/7-1/8-1/10-1/12+1/9+...=ln(2)/2 (the same terms, but 2 positive, then 3 negative, then you repeat this pattern)

That's why I think there should be some rules about rearranging terms, for semi-convergent series, but also for divergent series.
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### Re: Divergent Series

Voekoevaka wrote:Most of the mathematicians claim that divergent series don't have a result, or tait their result is ∞.

But Euler himself (happy 306th birthday) has published himself an article entitled De Seriebus Divergentibus, where he explains that we can find results for divergent series. In this essay, he described some methods to sum divergent series, then he tried to compute one of them (1-1+2-6+24-120+...=Σn=0n!), using four differents methods that leads him to the same result. Studies about divergent series were made by Ramanujan too.

Euler was a mathematical pioneer, with broad interests (and prolific output). Sure, he played around with ways of assigning sums to divergent series, but he knew he was exploring the fringes of mathematics, and he wasn't sure what such sums really meant. Euler was certainly a great & gifted mathematician, but mathematics was far less rigorous back in his day.

Newton had introduced some techniques for dealing with infinite sums that appeared legitimate, but it wasn't until well after Euler that people like Weierstrass & Cauchy were able to provide solid foundations for dealing with such things. The study of real numbers was still in its infancy in Euler's time, for that matter, it wasn't even clear back then that negative integers were "proper" numbers and not just a mere bookkeeping convention.

Ramanujan was also a gifted mathematician, but his mathematical training was rather informal, and he was a lot less rigorous than his contemporaries. Hardy bemoaned the fact that he couldn't make Ramanujan understand why formal proof was so important; Ramanujan was far more interested in finding interesting results and much less interested in rigorously proving them. OTOH, Ramanujan's patchy mathematical training gave him a kind of advantage - he was happy to investigate things like divergent series that other better-trained mathematicians would dismiss out of hand.

Voekoevaka wrote:As I previously say, This result is not the real "sum" of the series, but it is a value correspounding to the properties of the series.

Indeed. As the Wikipedia article on Ramanujan summation says:
Although the Ramanujan summation of a divergent series is not a sum in the traditional sense, it has properties which make it mathematically useful in the study of divergent infinite series, for which conventional summation is undefined.

Ramanujan summation essentially is a property of the partial sums, rather than a property of the entire sum, as that doesn't exist.

Still, Ramanujan sums aren't just a weird result of pure mathematics - they have a very practical use in the renormalization of Quantum Field theories, as the Wikipedia article mentions.

Voekoevaka
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### Re: Divergent Series

The series 1+2+3+4+5+6+7+...=-1/12 is one of my favorites. I know a simple proof for it :

We call :

O=1+3+5+7+9+... the sum of odd numbers
E=2+4+6+8+10+... the sum of even numbers
so O-E=1-2+3-4+5-6+7-...

Let's assume f(x)=1-2x+3x2-4x3+5x4-...
=Σ( (-1)n(n+1)xn , n∈⟦0,∞⟦)

Let's g(x) be a primitive of f(x)
g(x)=x-x2+x3-x4+x5-...
=Σ( (-1)n+1xn , n∈⟦1,∞⟦)

Let's assume h(x)=1-g(x), we can found an analytic form of h(x) :
h(x)=1-x+x2-x3+x4-x5+...
=1-x.1+x.x-x.x2+x.x3-x.x4+...
=1-x.(1-x+x2-x3+x4-...
=1-x.h(x)
⇒ h(x).(1+x)=1
⇒ h(x)=1/(1+x)

⇒ g(x)=1-1/(1+x)

⇒ h(x)=(1+x)-2 by differenciating.

By assuming x=1, we get :
f(1)=O-E=1-2+3-4+5-6+7-...=(1+1)-2

Now take :
E=2+4+6+8+10+...
=2.(1+2+3+4+5+...
=2.(1+3+5+...+2+4+6+...
=2(O+E)

So let's solve the system :
O-E=¼
E=2(O+E)

⇒-E=2O
⇒3O=¼

⇒O=⅟12
⇒E=-⅙

O+E=1+2+3+4+5+6+7+...=⅟12-⅙=-⅟12
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