Summing

evene · 2016-02-14 05:21:05

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

bobbym · 2016-02-14 21:43:12

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

Nehushtan · 2016-02-14 23:17:05

How about the old division-by-zero trick?

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

zetafunc · 2016-02-15 00:57:17

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:

Relentless · 2016-02-15 01:52:26

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)

evene · 2016-02-15 02:30:56

Thanks Relentless

zetafunc · 2016-02-15 19:47:45

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.

Relentless · 2016-02-15 23:38:16

Hi zetafunc,

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

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

zetafunc · 2016-02-16 01:44:48

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).

Nehushtan · 2016-02-16 20:13:15

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)

Relentless · 2016-02-16 20:44:31

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

Nehushtan · 2016-02-16 22:39:13

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.

Anuj Naru · 2016-02-20 21:59:12

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.

Agnishom · 2016-02-20 22:44:45

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)

bobbym · 2016-02-21 02:05:47

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.

evene · 2016-02-21 03:40:46

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

How about this:

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

bobbym · 2016-02-21 04:50:43

That is the standard proof.

evene · 2016-02-21 07:13:36

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

Math Is Fun Forum

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

Summing

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

Re: Summing

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

Re: Summing

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

Re: Summing

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

Re: Summing

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

Re: Summing

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

Re: Summing

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

Re: Summing

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

Re: Summing

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

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

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

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

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

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#15 2016-02-21 02:05:47

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

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

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