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#1 2016-02-14 05:21:05

evene
Member
Registered: 2015-10-18
Posts: 272

Summing

Need help with this: Prove that 1+2+3+4+5...=-1/12

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#2 2016-02-14 21:43:12

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Summing

Hi;

Ramanujan summation is a technique invented by the mathematician Srinivasa Ramanujan for assigning a sum to infinite divergent series. 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.

I do not know if you should call it proving because he is using a whole different concept of convergence.

https://www.youtube.com/watch?v=w-I6XTVZXww


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If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#3 2016-02-14 23:17:05

Nehushtan
Member
Registered: 2013-03-09
Posts: 957

Re: Summing

How about the old division-by-zero trick?

Last edited by Nehushtan (2016-02-15 02:08:31)


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#4 2016-02-15 00:57:17

zetafunc
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Registered: 2014-05-21
Posts: 2,432
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Re: Summing

evene wrote:

Need help with this: Prove that 1+2+3+4+5...=-1/12

You need to be careful what you mean by this. It is not true that the sum on the LHS converges to -1/12. In fact, most elementary courses in analysis will show you that the series

converges only if p > 1. However, you can instead define the Riemann zeta function:

. Note that s is complex! In other words,
, where
.

The sum on the RHS converges for Re(s) > 1. However, using the principle of analytic continuation from complex analysis, it is possible to analytically continue
over the whole complex plane, so that it satisfies Riemann's functional equation:

.

In particular,
is analytic everywhere, except for a simple pole at s = 1 which has residue 1. Setting s = -1 yields the result that you're referring to. There are two ways I know of to derive this functional equation: one uses the Poisson summation formula, whilst the idea of the other method uses an interesting toy contour, but is somewhat long-winded. Unfortunately, this does not imply that:

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#5 2016-02-15 01:52:26

Relentless
Member
Registered: 2015-12-15
Posts: 631

Re: Summing

I have seen a proof before. It goes like this...

Grandi's series:

S1:

2S1 (shift the second series one column):

Last edited by Relentless (2016-02-15 01:54:30)

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#6 2016-02-15 02:30:56

evene
Member
Registered: 2015-10-18
Posts: 272

Re: Summing

Thanks Relentless

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#7 2016-02-15 19:47:45

zetafunc
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Registered: 2014-05-21
Posts: 2,432
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Re: Summing

evene wrote:

Thanks Relentless

You should be aware that proofs of this type (in particular, rearranging the terms of a divergent series) are strictly not valid -- unless you were searching for that particular manipulation as an exercise in things that can go wrong if you do treat divergent series in that way.

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#8 2016-02-15 23:38:16

Relentless
Member
Registered: 2015-12-15
Posts: 631

Re: Summing

Hi zetafunc,

I basically just parroted a chain of reasoning I have seen before. What goes wrong with this argument? smile

Last edited by Relentless (2016-02-15 23:38:39)

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#9 2016-02-16 01:44:48

zetafunc
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Registered: 2014-05-21
Posts: 2,432
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Re: Summing

Relentless wrote:

Grandi's series:

The problem with Grandi's series is that it does not converge under the standard definition of convergence of a series: that is, that the sequence of partial sums converge (which means if you take the sum of the first 2 terms, the first 3 terms, the first N terms -- they form a sequence which converges). Suppose S(n) denotes the partial sums of Grandi's series. Then one has

S(1) = 1
S(2) = 1 - 1 = 0
s(3) = 1 - 1 + 1 = 1
etc.

Then, this sequence of partial sums clearly does not converge (it oscillates between 0 and 1, depending on the parity of n). Thus, if you want to assign a value to Grandi's series, then you need an alternative way of defining the "sum" of a divergent series. Typically, one uses something like Cesaro summation or analytic continuation (which I talked a bit about above). What Cesaro summation does is it assigns the divergent series a value by taking the limit of the arithmetic mean of the partial sums, rather than the partial sums themselves. Here, Grandi's series is Cesaro-summable, and has value 1/2. The nice thing about Cesaro's definition is that any convergent series is also Cesaro-summable.

The rest of your proof is somewhat more problematic, in that it involves re-arranging and adding terms to divergent series. In general, this is not valid: in fact, one loses associativity of addition with arbitrarily many terms that don't have a constant sign. An example of how problematic this can be is the Riemann rearrangement theorem, which says that you can rearrange the terms of a conditionally convergent series so that it converges to any value you like (or diverges).

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#10 2016-02-16 20:13:15

Nehushtan
Member
Registered: 2013-03-09
Posts: 957

Re: Summing

How about this?

Also

Substituting from [2]:

[1] + [3] ⇒

But we also have

Substituting [5] into [4]:

Finally

[6] & [7] ⇒

Last edited by Nehushtan (2016-02-16 20:15:14)


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#11 2016-02-16 20:44:31

Relentless
Member
Registered: 2015-12-15
Posts: 631

Re: Summing

Hi,

If you were to take:

And divide it by two, you would get:

This is false for any number.
I think there is something arbitrary about adding zeroes as placeholders, something relating to the point that zetafunc has been making

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#12 2016-02-16 22:39:13

Nehushtan
Member
Registered: 2013-03-09
Posts: 957

Re: Summing

The fact is that the series 1+2+3+4+5+… diverges to +∞. Most of the statements in the mock proof don't make sense unless you take S =  ∞ and treat infinity as a number (so you would have

[list=*]
[*]

[/*]
[/list]

etc). Of course infinity isn't a number and people who don't know what they're doing should never treat it as a number. The above was only a mock proof of a mock statement – just an exercise in meaningless symbol manipulation.


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#13 2016-02-20 21:59:12

Anuj Naru
Member
Registered: 2016-02-20
Posts: 2

Re: Summing

Hey Bobbym,
I am Anuj Naru. I have heard about you from Amartyanil. I have heard that you are very intelligent.
Then you say that why 0.9999.......is greater than 1. I can't understand this. What is the method? Can you tell me Bobbym.

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#14 2016-02-20 22:44:45

Agnishom
Real Member
From: Riemann Sphere
Registered: 2011-01-29
Posts: 24,974
Website

Re: Summing

Hi Anuj;

This is Agnishom. I am not very intelligent.

0.99... = 1

We spell bobbym with a small 'b'

Welcome to the forum!

Last edited by Agnishom (2016-02-20 22:45:36)


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'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
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#15 2016-02-21 02:05:47

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Summing

Hi Anuj Naru;

I have never said that .999... is greater than 1. In my view, it is equal to 1. I am also not very intelligent, I just work very hard. Welcome to the forum.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#16 2016-02-21 03:40:46

evene
Member
Registered: 2015-10-18
Posts: 272

Re: Summing

Oh my... you guys are so funny...

How about this:

Let
. So,
. Subtracting x from 10x gives us
which is 9.



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#17 2016-02-21 04:50:43

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Summing

That is the standard proof.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#18 2016-02-21 07:13:36

evene
Member
Registered: 2015-10-18
Posts: 272

Re: Summing

I know! Which is why I put it on here.

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