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The sum of natural numbers (1+2+3... )= -1/12

AlbinoBunny
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1/30/2014 6:26:27 PM
Posted: 2 years ago
What do you think?
bladerunner060 | bsh1 , 2014! Presidency campaign!

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Enji
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1/30/2014 7:53:56 PM
Posted: 2 years ago
At 1/30/2014 6:26:27 PM, AlbinoBunny wrote:


What do you think?

Consider the sum S = 1 + 0 - 1 + 1 + 0 - 1...

Using the identity property of addition (x + 0 = x), we can prove that this series is equal to Grandi's series (1 - 1 + 1 - 1...).

The Cesaro sum is defined as the limit of the average of a series partial sums as the number of terms tends towards infinity. For Grandi's series, we can prove that the Cesaro sum is 1/2. The average of the first partial sum is 1, the second is 1/2, the third is 2/3, the fourth is 2/4, etc. For an even number of terms, the Cesaro sum is 1/2 and for an odd number of terms the Cesaro sum is 1/2 + 1/2n -- which tends towards 1/2 as n becomes very large. The series S, however, has a different number of terms. The average of the first partial sum is 1, the second is 1, and the third is 2/3, and so on. Using similar maths as above, we can show that the Cesaro sum of S ends up being 2/3.

S is equal to Grandi's series, yet they have different Cesaro sums. And of course if we apply this new sum to the mathematics in the video, we get a new sum of all the natural numbers. The problem is that the Cesaro sum isn't the sum of the series. Neither Grandi's series nor the infinite series S (nor the natural numbers) are convergent and hence they don't have sums -- the sum of 1 - 1 + 1 - 1... is not equal to one half and the sum of the natural numbers is not equal to -1/12, and if we assert that it is we destroy the consistency of mathematics.

Consequently the claim isn't true. What's interesting, though, is that the sum has meaningful applications.
Enji
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1/30/2014 7:54:41 PM
Posted: 2 years ago
At 1/30/2014 7:53:56 PM, Enji wrote:

Consider the sum S = 1 + 0 - 1 + 1 + 0 - 1...

Using the identity property of addition (x + 0 = x), we can prove that this series is equal to Grandi's series (1 - 1 + 1 - 1...).

The Cesaro sum is defined as the limit of the average of a series partial sums as the number of terms tends towards infinity. For Grandi's series, we can prove that the Cesaro sum is 1/2. The average of the first partial sum is 1, the second is 1/2, the third is 2/3, the fourth is 2/4, etc. For an even number of terms, the Cesaro sum is 1/2 and for an odd number of terms the Cesaro sum is 1/2 + 1/2n -- which tends towards 1/2 as n becomes very large. The series S, however, has a different number of terms. The average of the first partial sum is 1, the second is 1, and the third is 2/3, and so on. Using similar maths as above, we can show that the Cesaro sum of S ends up being 2/3.

S is equal to Grandi's series, yet they have different Cesaro sums. And of course if we apply this new sum to the mathematics in the video, we get a new sum of all the natural numbers. The problem is that the Cesaro sum isn't the sum of the series. Neither Grandi's series nor the infinite series S (nor the natural numbers) are convergent and hence they don't have sums -- the sum of 1 - 1 + 1 - 1... is not equal to one half and the sum of the natural numbers is not equal to -1/12, and if we assert that it is we destroy the consistency of mathematics.

Consequently the claim isn't true. What's interesting, though, is that the sum has meaningful applications.

I bring up Cesaro sums because it's more or less how they got the sum of Grandi's series equals 1/2 elsewhere.
AlbinoBunny
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1/30/2014 8:09:44 PM
Posted: 2 years ago
At 1/30/2014 7:54:41 PM, Enji wrote:
At 1/30/2014 7:53:56 PM, Enji wrote:

Consider the sum S = 1 + 0 - 1 + 1 + 0 - 1...

Using the identity property of addition (x + 0 = x), we can prove that this series is equal to Grandi's series (1 - 1 + 1 - 1...).

The Cesaro sum is defined as the limit of the average of a series partial sums as the number of terms tends towards infinity. For Grandi's series, we can prove that the Cesaro sum is 1/2. The average of the first partial sum is 1, the second is 1/2, the third is 2/3, the fourth is 2/4, etc. For an even number of terms, the Cesaro sum is 1/2 and for an odd number of terms the Cesaro sum is 1/2 + 1/2n -- which tends towards 1/2 as n becomes very large. The series S, however, has a different number of terms. The average of the first partial sum is 1, the second is 1, and the third is 2/3, and so on. Using similar maths as above, we can show that the Cesaro sum of S ends up being 2/3.

Explain the 2/3.


S is equal to Grandi's series, yet they have different Cesaro sums. And of course if we apply this new sum to the mathematics in the video, we get a new sum of all the natural numbers. The problem is that the Cesaro sum isn't the sum of the series. Neither Grandi's series nor the infinite series S (nor the natural numbers) are convergent and hence they don't have sums -- the sum of 1 - 1 + 1 - 1... is not equal to one half and the sum of the natural numbers is not equal to -1/12, and if we assert that it is we destroy the consistency of mathematics.

Consequently the claim isn't true. What's interesting, though, is that the sum has meaningful applications.

I bring up Cesaro sums because it's more or less how they got the sum of Grandi's series equals 1/2 elsewhere.
bladerunner060 | bsh1 , 2014! Presidency campaign!

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Enji
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1/30/2014 8:39:16 PM
Posted: 2 years ago
At 1/30/2014 8:09:44 PM, AlbinoBunny wrote:
At 1/30/2014 7:54:41 PM, Enji wrote:
At 1/30/2014 7:53:56 PM, Enji wrote:

Consider the sum S = 1 + 0 - 1 + 1 + 0 - 1...

Using the identity property of addition (x + 0 = x), we can prove that this series is equal to Grandi's series (1 - 1 + 1 - 1...).

The Cesaro sum is defined as the limit of the average of a series partial sums as the number of terms tends towards infinity. For Grandi's series, we can prove that the Cesaro sum is 1/2. The average of the first partial sum is 1, the second is 1/2, the third is 2/3, the fourth is 2/4, etc. For an even number of terms, the Cesaro sum is 1/2 and for an odd number of terms the Cesaro sum is 1/2 + 1/2n -- which tends towards 1/2 as n becomes very large. The series S, however, has a different number of terms. The average of the first partial sum is 1, the second is 1, and the third is 2/3, and so on. Using similar maths as above, we can show that the Cesaro sum of S ends up being 2/3.

Explain the 2/3.

Here's the series of partial sums: 1, 1, 0, 1, 1, 0, 1, 1, 0 ...
Which gets you a series of average partial sums: 1/1, 2/2, 2/3, 3/4, 4/5, 4/6, 5/7, 6/8, 6/9... The average partial sum for every term which is evenly divisible by 3 is 2/3. The average of partial sums for every term with a remainder of 1 is [(1-n)/3 + n]/n, which tends towards 2/3 as n gets big, and the average of partial sums for every term with a remainder of 2 is [(2-n)/3 +n]/n, which again tends towards 2/3 as n tends towards infinity -- getting you your Cesaro sum of 2/3.
AlbinoBunny
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1/31/2014 2:14:53 AM
Posted: 2 years ago
At 1/30/2014 8:39:16 PM, Enji wrote:
At 1/30/2014 8:09:44 PM, AlbinoBunny wrote:
At 1/30/2014 7:54:41 PM, Enji wrote:
At 1/30/2014 7:53:56 PM, Enji wrote:

Consider the sum S = 1 + 0 - 1 + 1 + 0 - 1...

Using the identity property of addition (x + 0 = x), we can prove that this series is equal to Grandi's series (1 - 1 + 1 - 1...).

The Cesaro sum is defined as the limit of the average of a series partial sums as the number of terms tends towards infinity. For Grandi's series, we can prove that the Cesaro sum is 1/2. The average of the first partial sum is 1, the second is 1/2, the third is 2/3, the fourth is 2/4, etc. For an even number of terms, the Cesaro sum is 1/2 and for an odd number of terms the Cesaro sum is 1/2 + 1/2n -- which tends towards 1/2 as n becomes very large. The series S, however, has a different number of terms. The average of the first partial sum is 1, the second is 1, and the third is 2/3, and so on. Using similar maths as above, we can show that the Cesaro sum of S ends up being 2/3.

Explain the 2/3.

Here's the series of partial sums: 1, 1, 0, 1, 1, 0, 1, 1, 0 ...
Which gets you a series of average partial sums: 1/1, 2/2, 2/3, 3/4, 4/5, 4/6, 5/7, 6/8, 6/9... The average partial sum for every term which is evenly divisible by 3 is 2/3. The average of partial sums for every term with a remainder of 1 is [(1-n)/3 + n]/n, which tends towards 2/3 as n gets big, and the average of partial sums for every term with a remainder of 2 is [(2-n)/3 +n]/n, which again tends towards 2/3 as n tends towards infinity -- getting you your Cesaro sum of 2/3.

So the Cesaro sum of all natural numbers is -1/12?
bladerunner060 | bsh1 , 2014! Presidency campaign!

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AlbinoBunny
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1/31/2014 2:23:13 AM
Posted: 2 years ago
"The sum of all natural numbers 1 + 2 + 3 + 4 + " " " is a divergent series. The nth partial sum of the series is the triangular number

\sum_{k=1}^n k = \frac{n(n+1)}{2},

which increases without bound as n goes to infinity. Because the sequence of partial sums fails to converge to a finite limit, the series is divergent, and it does not have a sum in the usual sense of the word.

Although the series seems at first sight not to have any meaningful value at all, it can be manipulated to yield a number of mathematically interesting results, some of which have applications in other fields such as complex analysis, quantum field theory and string theory. There are many different summation methods used in mathematics to assign numerical values even to divergent series. In particular, the methods of zeta function regularization and Ramanujan summation assign the series a value of W22;1/12, which is expressed by a famous formula"

https://en.wikipedia.org...
bladerunner060 | bsh1 , 2014! Presidency campaign!

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Enji
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1/31/2014 11:53:38 AM
Posted: 2 years ago
At 1/31/2014 2:14:53 AM, AlbinoBunny wrote:
At 1/30/2014 8:39:16 PM, Enji wrote:
At 1/30/2014 8:09:44 PM, AlbinoBunny wrote:
At 1/30/2014 7:54:41 PM, Enji wrote:
At 1/30/2014 7:53:56 PM, Enji wrote:

Consider the sum S = 1 + 0 - 1 + 1 + 0 - 1...

Using the identity property of addition (x + 0 = x), we can prove that this series is equal to Grandi's series (1 - 1 + 1 - 1...).

The Cesaro sum is defined as the limit of the average of a series partial sums as the number of terms tends towards infinity. For Grandi's series, we can prove that the Cesaro sum is 1/2. The average of the first partial sum is 1, the second is 1/2, the third is 2/3, the fourth is 2/4, etc. For an even number of terms, the Cesaro sum is 1/2 and for an odd number of terms the Cesaro sum is 1/2 + 1/2n -- which tends towards 1/2 as n becomes very large. The series S, however, has a different number of terms. The average of the first partial sum is 1, the second is 1, and the third is 2/3, and so on. Using similar maths as above, we can show that the Cesaro sum of S ends up being 2/3.

Explain the 2/3.

Here's the series of partial sums: 1, 1, 0, 1, 1, 0, 1, 1, 0 ...
Which gets you a series of average partial sums: 1/1, 2/2, 2/3, 3/4, 4/5, 4/6, 5/7, 6/8, 6/9... The average partial sum for every term which is evenly divisible by 3 is 2/3. The average of partial sums for every term with a remainder of 1 is [(1-n)/3 + n]/n, which tends towards 2/3 as n gets big, and the average of partial sums for every term with a remainder of 2 is [(2-n)/3 +n]/n, which again tends towards 2/3 as n tends towards infinity -- getting you your Cesaro sum of 2/3.

So the Cesaro sum of all natural numbers is -1/12?

No, the natural numbers don't have a Cesaro sum because the series of average partial sums is divergent. Cesaro sums allow you to assign a value to divergent series if the series of average partial sums converges. The video roughly uses the Cesaro sum as the sum of Grandi's series and does some arithmetic with infinite series which isn't explicitly allowed to get -1/12 as the sum of the natural numbers, but you could just as easily end up with a different final number -- the Cesaro sum is easily changed by adding terms of 0 or rearranging terms, all of which is permitted based on the axioms of arithmetic; the Cesaro sum is not equal to the sum of a divergent series.

At 1/31/2014 2:23:13 AM, AlbinoBunny wrote:
"The sum of all natural numbers 1 + 2 + 3 + 4 + " " " is a divergent series. The nth partial sum of the series is the triangular number

\sum_{k=1}^n k = \frac{n(n+1)}{2},

which increases without bound as n goes to infinity. Because the sequence of partial sums fails to converge to a finite limit, the series is divergent, and it does not have a sum in the usual sense of the word.

Although the series seems at first sight not to have any meaningful value at all, it can be manipulated to yield a number of mathematically interesting results, some of which have applications in other fields such as complex analysis, quantum field theory and string theory. There are many different summation methods used in mathematics to assign numerical values even to divergent series. In particular, the methods of zeta function regularization and Ramanujan summation assign the series a value of W22;1/12, which is expressed by a famous formula"

Note the language here, it is important. The partial sums of a convergent series converge to a finite limit. For example, we can prove that sum of the series 1, 1/2, 1/4... converges to 2, and it would be accurate to say that the infinite sum of that series equals 2. With a divergent series, the partial sums don't converge to any real value. For example, the series 1, 2, 4, 8... is divergent. There are different methods with which we can assign values to divergent series. One which you learn in calculus is using limits -- and using limits we would say the sum 1 + 2 + 4... tends towards infinity. In this example it's somewhat more clear that this is distinct from saying the sum 1 + 2 + 4 equals infinity because infinity is not a number. And similarly, we would say that the sum of the natural numbers tends towards infinity. There are other methods of assigning values to divergent series, each with their own strengths and weaknesses (for example, we can't use limits to assign a value to Grandi's series because the partial sums don't diverge to infinity - instead they alternate between 0 and 1). Cesaro sums are one such method, and some other methods include the ones mentioned in the wiki link. However it is never accurate to say that the sum of a divergent series equals the value given by any of these methods -- divergent series don't converge to any such value -- these methods only provide rules which allow us to consistently assign values to divergent series, and attempting to equate the assigned value with the sum of the series destroys the consistency of mathematics.
Enji
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1/31/2014 12:00:05 PM
Posted: 2 years ago
At 1/31/2014 11:45:22 AM, Such wrote:
Waaaiiiiit... isn't the average of -1 and +1, 0?

Yes, but that's different from the average of the partial sums. The partial sums of the series 1 - 1 + 1 - 1... are 0 for an even number of terms and 1 for an odd number of terms which gets you the average partial sum being 1/2.
Such
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1/31/2014 12:00:47 PM
Posted: 2 years ago
At 1/31/2014 12:00:05 PM, Enji wrote:
At 1/31/2014 11:45:22 AM, Such wrote:
Waaaiiiiit... isn't the average of -1 and +1, 0?

Yes, but that's different from the average of the partial sums. The partial sums of the series 1 - 1 + 1 - 1... are 0 for an even number of terms and 1 for an odd number of terms which gets you the average partial sum being 1/2.

Ahhh, right, well done, thanks. ^_^
AlbinoBunny
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1/31/2014 3:52:50 PM
Posted: 2 years ago
At 1/31/2014 11:53:38 AM, Enji wrote:
At 1/31/2014 2:14:53 AM, AlbinoBunny wrote:
At 1/30/2014 8:39:16 PM, Enji wrote:
At 1/30/2014 8:09:44 PM, AlbinoBunny wrote:
At 1/30/2014 7:54:41 PM, Enji wrote:
At 1/30/2014 7:53:56 PM, Enji wrote:

Consider the sum S = 1 + 0 - 1 + 1 + 0 - 1...

Using the identity property of addition (x + 0 = x), we can prove that this series is equal to Grandi's series (1 - 1 + 1 - 1...).

The Cesaro sum is defined as the limit of the average of a series partial sums as the number of terms tends towards infinity. For Grandi's series, we can prove that the Cesaro sum is 1/2. The average of the first partial sum is 1, the second is 1/2, the third is 2/3, the fourth is 2/4, etc. For an even number of terms, the Cesaro sum is 1/2 and for an odd number of terms the Cesaro sum is 1/2 + 1/2n -- which tends towards 1/2 as n becomes very large. The series S, however, has a different number of terms. The average of the first partial sum is 1, the second is 1, and the third is 2/3, and so on. Using similar maths as above, we can show that the Cesaro sum of S ends up being 2/3.

Explain the 2/3.

Here's the series of partial sums: 1, 1, 0, 1, 1, 0, 1, 1, 0 ...
Which gets you a series of average partial sums: 1/1, 2/2, 2/3, 3/4, 4/5, 4/6, 5/7, 6/8, 6/9... The average partial sum for every term which is evenly divisible by 3 is 2/3. The average of partial sums for every term with a remainder of 1 is [(1-n)/3 + n]/n, which tends towards 2/3 as n gets big, and the average of partial sums for every term with a remainder of 2 is [(2-n)/3 +n]/n, which again tends towards 2/3 as n tends towards infinity -- getting you your Cesaro sum of 2/3.

So the Cesaro sum of all natural numbers is -1/12?

No, the natural numbers don't have a Cesaro sum because the series of average partial sums is divergent. Cesaro sums allow you to assign a value to divergent series if the series of average partial sums converges. The video roughly uses the Cesaro sum as the sum of Grandi's series and does some arithmetic with infinite series which isn't explicitly allowed to get -1/12 as the sum of the natural numbers, but you could just as easily end up with a different final number -- the Cesaro sum is easily changed by adding terms of 0 or rearranging terms, all of which is permitted based on the axioms of arithmetic; the Cesaro sum is not equal to the sum of a divergent series.

At 1/31/2014 2:23:13 AM, AlbinoBunny wrote:
"The sum of all natural numbers 1 + 2 + 3 + 4 + " " " is a divergent series. The nth partial sum of the series is the triangular number

\sum_{k=1}^n k = \frac{n(n+1)}{2},

which increases without bound as n goes to infinity. Because the sequence of partial sums fails to converge to a finite limit, the series is divergent, and it does not have a sum in the usual sense of the word.

Although the series seems at first sight not to have any meaningful value at all, it can be manipulated to yield a number of mathematically interesting results, some of which have applications in other fields such as complex analysis, quantum field theory and string theory. There are many different summation methods used in mathematics to assign numerical values even to divergent series. In particular, the methods of zeta function regularization and Ramanujan summation assign the series a value of W22;1/12, which is expressed by a famous formula"

Note the language here, it is important. The partial sums of a convergent series converge to a finite limit. For example, we can prove that sum of the series 1, 1/2, 1/4... converges to 2, and it would be accurate to say that the infinite sum of that series equals 2. With a divergent series, the partial sums don't converge to any real value. For example, the series 1, 2, 4, 8... is divergent. There are different methods with which we can assign values to divergent series. One which you learn in calculus is using limits -- and using limits we would say the sum 1 + 2 + 4... tends towards infinity. In this example it's somewhat more clear that this is distinct from saying the sum 1 + 2 + 4 equals infinity because infinity is not a number. And similarly, we would say that the sum of the natural numbers tends towards infinity. There are other methods of assigning values to divergent series, each with their own strengths and weaknesses (for example, we can't use limits to assign a value to Grandi's series because the partial sums don't diverge to infinity - instead they alternate between 0 and 1). Cesaro sums are one such method, and some other methods include the ones mentioned in the wiki link. However it is never accurate to say that the sum of a divergent series equals the value given by any of these methods -- divergent series don't converge to any such value -- these methods only provide rules which allow us to consistently assign values to divergent series, and attempting to equate the assigned value with the sum of the series destroys the consistency of mathematics.

"There are many different summation methods used in mathematics to assign numerical values even to divergent series. In particular, the methods of zeta function regularization and Ramanujan summation assign the series a value of -1/12, which is expressed by a famous formula:[2]

1+2+3+4+...= -1/12 "
https://en.wikipedia.org...

Not that I actually understand it.
bladerunner060 | bsh1 , 2014! Presidency campaign!

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sadolite
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1/31/2014 4:06:13 PM
Posted: 2 years ago
http://www.siteadvisor.com...
It's not your views that divide us, it's what you think my views should be that divides us.

If you think I will give up my rights and forsake social etiquette to make you "FEEL" better you are sadly mistaken

If liberal democrats would just stop shooting people gun violence would drop by 90%
sadolite
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1/31/2014 4:06:59 PM
Posted: 2 years ago
At 1/31/2014 4:06:13 PM, sadolite wrote:
http://www.siteadvisor.com...

Please delete this link says it is bad.
It's not your views that divide us, it's what you think my views should be that divides us.

If you think I will give up my rights and forsake social etiquette to make you "FEEL" better you are sadly mistaken

If liberal democrats would just stop shooting people gun violence would drop by 90%
sadolite
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1/31/2014 4:09:54 PM
Posted: 2 years ago
Oh look, it's another one of these number threads that try to defy reality.

http://www.integritystaffing.com...
It's not your views that divide us, it's what you think my views should be that divides us.

If you think I will give up my rights and forsake social etiquette to make you "FEEL" better you are sadly mistaken

If liberal democrats would just stop shooting people gun violence would drop by 90%
Enji
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1/31/2014 4:15:44 PM
Posted: 2 years ago
At 1/31/2014 3:52:50 PM, AlbinoBunny wrote:

"There are many different summation methods used in mathematics to assign numerical values even to divergent series. In particular, the methods of zeta function regularization and Ramanujan summation assign the series a value of -1/12, which is expressed by a famous formula:[2]

1+2+3+4+...= -1/12 "
https://en.wikipedia.org...

Not that I actually understand it.

Intuitively, I think you realise that if you add all infinitely many positive integers numbers together, you'll never get a fraction, much less a negative fraction, and the value that first comes to mind for the value of the sum would be infinity. And this intuitive sum can be defended rigorously -- with limits we can show that the sum tends towards infinity. And similarly, intuitively you can realise that the sum of Grandi's series is only ever going to be 0 or 1 no matter how many terms you add together -- never 1/2 -- although we can't use limits to assign the sum of Grandi's series a value as we can with the natural numbers.

With different mathematics, we can assign divergent series values even though they don't converge to any such value. The sum from 1 to infinity of k does not converge to or equal -1/12, and asserting this equality breaks stuff which is bad. Assigning a value to a divergent infinite series does not mean that the sum of that series is the assigned value.
AlbinoBunny
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1/31/2014 4:22:56 PM
Posted: 2 years ago
At 1/31/2014 4:15:44 PM, Enji wrote:
At 1/31/2014 3:52:50 PM, AlbinoBunny wrote:

"There are many different summation methods used in mathematics to assign numerical values even to divergent series. In particular, the methods of zeta function regularization and Ramanujan summation assign the series a value of -1/12, which is expressed by a famous formula:[2]

1+2+3+4+...= -1/12 "
https://en.wikipedia.org...

Not that I actually understand it.

Intuitively, I think you realise that if you add all infinitely many positive integers numbers together, you'll never get a fraction, much less a negative fraction, and the value that first comes to mind for the value of the sum would be infinity. And this intuitive sum can be defended rigorously -- with limits we can show that the sum tends towards infinity. And similarly, intuitively you can realise that the sum of Grandi's series is only ever going to be 0 or 1 no matter how many terms you add together -- never 1/2 -- although we can't use limits to assign the sum of Grandi's series a value as we can with the natural numbers.

With different mathematics, we can assign divergent series values even though they don't converge to any such value. The sum from 1 to infinity of k does not converge to or equal -1/12, and asserting this equality breaks stuff which is bad. Assigning a value to a divergent infinite series does not mean that the sum of that series is the assigned value.

What does assigning a value mean? Also, I don't really trust intuition much any more. lol
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Enji
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1/31/2014 7:17:25 PM
Posted: 2 years ago
At 1/31/2014 4:22:56 PM, AlbinoBunny wrote:
At 1/31/2014 4:15:44 PM, Enji wrote:
At 1/31/2014 3:52:50 PM, AlbinoBunny wrote:

"There are many different summation methods used in mathematics to assign numerical values even to divergent series. In particular, the methods of zeta function regularization and Ramanujan summation assign the series a value of -1/12, which is expressed by a famous formula:[2]

1+2+3+4+...= -1/12 "
https://en.wikipedia.org...

Not that I actually understand it.

Intuitively, I think you realise that if you add all infinitely many positive integers numbers together, you'll never get a fraction, much less a negative fraction, and the value that first comes to mind for the value of the sum would be infinity. And this intuitive sum can be defended rigorously -- with limits we can show that the sum tends towards infinity. And similarly, intuitively you can realise that the sum of Grandi's series is only ever going to be 0 or 1 no matter how many terms you add together -- never 1/2 -- although we can't use limits to assign the sum of Grandi's series a value as we can with the natural numbers.

With different mathematics, we can assign divergent series values even though they don't converge to any such value. The sum from 1 to infinity of k does not converge to or equal -1/12, and asserting this equality breaks stuff which is bad. Assigning a value to a divergent infinite series does not mean that the sum of that series is the assigned value.

What does assigning a value mean? Also, I don't really trust intuition much any more. lol

It doesn't mean anything -- it's just a method of consistently dealing with divergent series.
AlbinoBunny
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2/1/2014 5:08:05 AM
Posted: 2 years ago
At 1/31/2014 7:17:25 PM, Enji wrote:
At 1/31/2014 4:22:56 PM, AlbinoBunny wrote:
At 1/31/2014 4:15:44 PM, Enji wrote:
At 1/31/2014 3:52:50 PM, AlbinoBunny wrote:

"There are many different summation methods used in mathematics to assign numerical values even to divergent series. In particular, the methods of zeta function regularization and Ramanujan summation assign the series a value of -1/12, which is expressed by a famous formula:[2]

1+2+3+4+...= -1/12 "
https://en.wikipedia.org...

Not that I actually understand it.

Intuitively, I think you realise that if you add all infinitely many positive integers numbers together, you'll never get a fraction, much less a negative fraction, and the value that first comes to mind for the value of the sum would be infinity. And this intuitive sum can be defended rigorously -- with limits we can show that the sum tends towards infinity. And similarly, intuitively you can realise that the sum of Grandi's series is only ever going to be 0 or 1 no matter how many terms you add together -- never 1/2 -- although we can't use limits to assign the sum of Grandi's series a value as we can with the natural numbers.

With different mathematics, we can assign divergent series values even though they don't converge to any such value. The sum from 1 to infinity of k does not converge to or equal -1/12, and asserting this equality breaks stuff which is bad. Assigning a value to a divergent infinite series does not mean that the sum of that series is the assigned value.

What does assigning a value mean? Also, I don't really trust intuition much any more. lol

It doesn't mean anything -- it's just a method of consistently dealing with divergent series.

So they just randomly assign a meaningless and arbitrary number to a divergent series?
bladerunner060 | bsh1 , 2014! Presidency campaign!

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http://www.debate.org... - Running for president.
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SeventhProfessor
Posts: 5,086
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2/1/2014 4:44:08 PM
Posted: 2 years ago
At 2/1/2014 5:08:05 AM, AlbinoBunny wrote:
At 1/31/2014 7:17:25 PM, Enji wrote:
At 1/31/2014 4:22:56 PM, AlbinoBunny wrote:
At 1/31/2014 4:15:44 PM, Enji wrote:
At 1/31/2014 3:52:50 PM, AlbinoBunny wrote:

"There are many different summation methods used in mathematics to assign numerical values even to divergent series. In particular, the methods of zeta function regularization and Ramanujan summation assign the series a value of -1/12, which is expressed by a famous formula:[2]

1+2+3+4+...= -1/12 "
https://en.wikipedia.org...

Not that I actually understand it.

Intuitively, I think you realise that if you add all infinitely many positive integers numbers together, you'll never get a fraction, much less a negative fraction, and the value that first comes to mind for the value of the sum would be infinity. And this intuitive sum can be defended rigorously -- with limits we can show that the sum tends towards infinity. And similarly, intuitively you can realise that the sum of Grandi's series is only ever going to be 0 or 1 no matter how many terms you add together -- never 1/2 -- although we can't use limits to assign the sum of Grandi's series a value as we can with the natural numbers.

With different mathematics, we can assign divergent series values even though they don't converge to any such value. The sum from 1 to infinity of k does not converge to or equal -1/12, and asserting this equality breaks stuff which is bad. Assigning a value to a divergent infinite series does not mean that the sum of that series is the assigned value.

What does assigning a value mean? Also, I don't really trust intuition much any more. lol

It doesn't mean anything -- it's just a method of consistently dealing with divergent series.

So they just randomly assign a meaningless and arbitrary number to a divergent series?

Actually, you can use mathematical proofs other than Cesaro sums. For example:

x=1-1+1-...
1-x=1-(1-1+1-...)
Now distribute the "-"
1-x=1-1+1-...
1-x=x
1=2x
1/2=x

In a number theory class I took, every textbook that mentions it, and within the mathematical community in general, it is agreed that the proof is valid. Same with 1+2+3+...=-1/12.
#UnbanTheMadman

#StandWithBossy

#BetOnThett

"bossy r u like 85 years old and have lost ur mind"
~mysteriouscrystals

"I've honestly never seen seventh post anything that wasn't completely idiotic in a trying-to-be-funny way."
~F-16

https://docs.google.com...
AlbinoBunny
Posts: 3,781
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2/1/2014 6:13:54 PM
Posted: 2 years ago
At 2/1/2014 4:44:08 PM, SeventhProfessor wrote:
At 2/1/2014 5:08:05 AM, AlbinoBunny wrote:
At 1/31/2014 7:17:25 PM, Enji wrote:
At 1/31/2014 4:22:56 PM, AlbinoBunny wrote:
At 1/31/2014 4:15:44 PM, Enji wrote:
At 1/31/2014 3:52:50 PM, AlbinoBunny wrote:

"There are many different summation methods used in mathematics to assign numerical values even to divergent series. In particular, the methods of zeta function regularization and Ramanujan summation assign the series a value of -1/12, which is expressed by a famous formula:[2]

1+2+3+4+...= -1/12 "
https://en.wikipedia.org...

Not that I actually understand it.

Intuitively, I think you realise that if you add all infinitely many positive integers numbers together, you'll never get a fraction, much less a negative fraction, and the value that first comes to mind for the value of the sum would be infinity. And this intuitive sum can be defended rigorously -- with limits we can show that the sum tends towards infinity. And similarly, intuitively you can realise that the sum of Grandi's series is only ever going to be 0 or 1 no matter how many terms you add together -- never 1/2 -- although we can't use limits to assign the sum of Grandi's series a value as we can with the natural numbers.

With different mathematics, we can assign divergent series values even though they don't converge to any such value. The sum from 1 to infinity of k does not converge to or equal -1/12, and asserting this equality breaks stuff which is bad. Assigning a value to a divergent infinite series does not mean that the sum of that series is the assigned value.

What does assigning a value mean? Also, I don't really trust intuition much any more. lol

It doesn't mean anything -- it's just a method of consistently dealing with divergent series.

So they just randomly assign a meaningless and arbitrary number to a divergent series?

Actually, you can use mathematical proofs other than Cesaro sums. For example:

x=1-1+1-...
1-x=1-(1-1+1-...)
Now distribute the "-"
1-x=1-1+1-...
1-x=x
1=2x
1/2=x

In a number theory class I took, every textbook that mentions it, and within the mathematical community in general, it is agreed that the proof is valid. Same with 1+2+3+...=-1/12.

I wasn't suggesting that people do that, the previous poster and I said;

"With different mathematics, we can assign divergent series values even though they don't converge to any such value."

"What does assigning a value mean? "

"It doesn't mean anything"
bladerunner060 | bsh1 , 2014! Presidency campaign!

http://www.debate.org...
http://www.debate.org... - Running for president.
http://www.debate.org... - Running as his vice president.

May the best man win!
SeventhProfessor
Posts: 5,086
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2/1/2014 6:16:02 PM
Posted: 2 years ago
At 2/1/2014 6:13:54 PM, AlbinoBunny wrote:
At 2/1/2014 4:44:08 PM, SeventhProfessor wrote:
At 2/1/2014 5:08:05 AM, AlbinoBunny wrote:
At 1/31/2014 7:17:25 PM, Enji wrote:
At 1/31/2014 4:22:56 PM, AlbinoBunny wrote:
At 1/31/2014 4:15:44 PM, Enji wrote:
At 1/31/2014 3:52:50 PM, AlbinoBunny wrote:

"There are many different summation methods used in mathematics to assign numerical values even to divergent series. In particular, the methods of zeta function regularization and Ramanujan summation assign the series a value of -1/12, which is expressed by a famous formula:[2]

1+2+3+4+...= -1/12 "
https://en.wikipedia.org...

Not that I actually understand it.

Intuitively, I think you realise that if you add all infinitely many positive integers numbers together, you'll never get a fraction, much less a negative fraction, and the value that first comes to mind for the value of the sum would be infinity. And this intuitive sum can be defended rigorously -- with limits we can show that the sum tends towards infinity. And similarly, intuitively you can realise that the sum of Grandi's series is only ever going to be 0 or 1 no matter how many terms you add together -- never 1/2 -- although we can't use limits to assign the sum of Grandi's series a value as we can with the natural numbers.

With different mathematics, we can assign divergent series values even though they don't converge to any such value. The sum from 1 to infinity of k does not converge to or equal -1/12, and asserting this equality breaks stuff which is bad. Assigning a value to a divergent infinite series does not mean that the sum of that series is the assigned value.

What does assigning a value mean? Also, I don't really trust intuition much any more. lol

It doesn't mean anything -- it's just a method of consistently dealing with divergent series.

So they just randomly assign a meaningless and arbitrary number to a divergent series?

Actually, you can use mathematical proofs other than Cesaro sums. For example:

x=1-1+1-...
1-x=1-(1-1+1-...)
Now distribute the "-"
1-x=1-1+1-...
1-x=x
1=2x
1/2=x

In a number theory class I took, every textbook that mentions it, and within the mathematical community in general, it is agreed that the proof is valid. Same with 1+2+3+...=-1/12.

I wasn't suggesting that people do that, the previous poster and I said;

"With different mathematics, we can assign divergent series values even though they don't converge to any such value."

"What does assigning a value mean? "

"It doesn't mean anything"

Sorry, it looked like you were saying random numbers were assigned. I was showing that there are multiple ways, so the numbers decided aren't random.
#UnbanTheMadman

#StandWithBossy

#BetOnThett

"bossy r u like 85 years old and have lost ur mind"
~mysteriouscrystals

"I've honestly never seen seventh post anything that wasn't completely idiotic in a trying-to-be-funny way."
~F-16

https://docs.google.com...
AlbinoBunny
Posts: 3,781
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2/1/2014 6:18:47 PM
Posted: 2 years ago
At 2/1/2014 6:16:02 PM, SeventhProfessor wrote:
At 2/1/2014 6:13:54 PM, AlbinoBunny wrote:
At 2/1/2014 4:44:08 PM, SeventhProfessor wrote:
At 2/1/2014 5:08:05 AM, AlbinoBunny wrote:
At 1/31/2014 7:17:25 PM, Enji wrote:
At 1/31/2014 4:22:56 PM, AlbinoBunny wrote:
At 1/31/2014 4:15:44 PM, Enji wrote:
At 1/31/2014 3:52:50 PM, AlbinoBunny wrote:

"There are many different summation methods used in mathematics to assign numerical values even to divergent series. In particular, the methods of zeta function regularization and Ramanujan summation assign the series a value of -1/12, which is expressed by a famous formula:[2]

1+2+3+4+...= -1/12 "
https://en.wikipedia.org...

Not that I actually understand it.

Intuitively, I think you realise that if you add all infinitely many positive integers numbers together, you'll never get a fraction, much less a negative fraction, and the value that first comes to mind for the value of the sum would be infinity. And this intuitive sum can be defended rigorously -- with limits we can show that the sum tends towards infinity. And similarly, intuitively you can realise that the sum of Grandi's series is only ever going to be 0 or 1 no matter how many terms you add together -- never 1/2 -- although we can't use limits to assign the sum of Grandi's series a value as we can with the natural numbers.

With different mathematics, we can assign divergent series values even though they don't converge to any such value. The sum from 1 to infinity of k does not converge to or equal -1/12, and asserting this equality breaks stuff which is bad. Assigning a value to a divergent infinite series does not mean that the sum of that series is the assigned value.

What does assigning a value mean? Also, I don't really trust intuition much any more. lol

It doesn't mean anything -- it's just a method of consistently dealing with divergent series.

So they just randomly assign a meaningless and arbitrary number to a divergent series?

Actually, you can use mathematical proofs other than Cesaro sums. For example:

x=1-1+1-...
1-x=1-(1-1+1-...)
Now distribute the "-"
1-x=1-1+1-...
1-x=x
1=2x
1/2=x

In a number theory class I took, every textbook that mentions it, and within the mathematical community in general, it is agreed that the proof is valid. Same with 1+2+3+...=-1/12.

I wasn't suggesting that people do that, the previous poster and I said;

"With different mathematics, we can assign divergent series values even though they don't converge to any such value."

"What does assigning a value mean? "

"It doesn't mean anything"

Sorry, it looked like you were saying random numbers were assigned. I was showing that there are multiple ways, so the numbers decided aren't random.

Enji seemed to be saying they're picked out of a hat. Although I think I'm just misunderstanding what they're saying.

So do you know what the summation methods being used mean when they say that 1 + 2 + 3... = -1/12?
bladerunner060 | bsh1 , 2014! Presidency campaign!

http://www.debate.org...
http://www.debate.org... - Running for president.
http://www.debate.org... - Running as his vice president.

May the best man win!
SeventhProfessor
Posts: 5,086
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2/1/2014 6:20:03 PM
Posted: 2 years ago
At 2/1/2014 6:18:47 PM, AlbinoBunny wrote:
At 2/1/2014 6:16:02 PM, SeventhProfessor wrote:
At 2/1/2014 6:13:54 PM, AlbinoBunny wrote:
At 2/1/2014 4:44:08 PM, SeventhProfessor wrote:
At 2/1/2014 5:08:05 AM, AlbinoBunny wrote:
At 1/31/2014 7:17:25 PM, Enji wrote:
At 1/31/2014 4:22:56 PM, AlbinoBunny wrote:
At 1/31/2014 4:15:44 PM, Enji wrote:
At 1/31/2014 3:52:50 PM, AlbinoBunny wrote:

"There are many different summation methods used in mathematics to assign numerical values even to divergent series. In particular, the methods of zeta function regularization and Ramanujan summation assign the series a value of -1/12, which is expressed by a famous formula:[2]

1+2+3+4+...= -1/12 "
https://en.wikipedia.org...

Not that I actually understand it.

Intuitively, I think you realise that if you add all infinitely many positive integers numbers together, you'll never get a fraction, much less a negative fraction, and the value that first comes to mind for the value of the sum would be infinity. And this intuitive sum can be defended rigorously -- with limits we can show that the sum tends towards infinity. And similarly, intuitively you can realise that the sum of Grandi's series is only ever going to be 0 or 1 no matter how many terms you add together -- never 1/2 -- although we can't use limits to assign the sum of Grandi's series a value as we can with the natural numbers.

With different mathematics, we can assign divergent series values even though they don't converge to any such value. The sum from 1 to infinity of k does not converge to or equal -1/12, and asserting this equality breaks stuff which is bad. Assigning a value to a divergent infinite series does not mean that the sum of that series is the assigned value.

What does assigning a value mean? Also, I don't really trust intuition much any more. lol

It doesn't mean anything -- it's just a method of consistently dealing with divergent series.

So they just randomly assign a meaningless and arbitrary number to a divergent series?

Actually, you can use mathematical proofs other than Cesaro sums. For example:

x=1-1+1-...
1-x=1-(1-1+1-...)
Now distribute the "-"
1-x=1-1+1-...
1-x=x
1=2x
1/2=x

In a number theory class I took, every textbook that mentions it, and within the mathematical community in general, it is agreed that the proof is valid. Same with 1+2+3+...=-1/12.

I wasn't suggesting that people do that, the previous poster and I said;

"With different mathematics, we can assign divergent series values even though they don't converge to any such value."

"What does assigning a value mean? "

"It doesn't mean anything"

Sorry, it looked like you were saying random numbers were assigned. I was showing that there are multiple ways, so the numbers decided aren't random.

Enji seemed to be saying they're picked out of a hat. Although I think I'm just misunderstanding what they're saying.

So do you know what the summation methods being used mean when they say that 1 + 2 + 3... = -1/12?

Yes.
#UnbanTheMadman

#StandWithBossy

#BetOnThett

"bossy r u like 85 years old and have lost ur mind"
~mysteriouscrystals

"I've honestly never seen seventh post anything that wasn't completely idiotic in a trying-to-be-funny way."
~F-16

https://docs.google.com...
Enji
Posts: 1,022
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2/1/2014 7:36:59 PM
Posted: 2 years ago
At 2/1/2014 6:18:47 PM, AlbinoBunny wrote:

Enji seemed to be saying they're picked out of a hat. Although I think I'm just misunderstanding what they're saying.

So do you know what the summation methods being used mean when they say that 1 + 2 + 3... = -1/12?

I'm not saying numbers are randomly selected. For example, you can use limits to determine the value of the sum of a series and the answer you get is non-random and can be rigorously defended given the methodology of limits. But saying that the sum of the natural numbers of infinity is of limited use -- we can't do much with infinity. Alternative methods (analytic continuation) allow you to get the sum -1/12, and like the approach using limits, the rules which allow you to get this sum are self-consistent and mathematically rigorous. Neither set of rules is wrong, and the answers they give are non-random, but -1/12 is more useful in some circumstances (like physics). But -1/12 is not the value of the sum of all the natural numbers.
AlbinoBunny
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2/1/2014 7:41:57 PM
Posted: 2 years ago
At 2/1/2014 7:36:59 PM, Enji wrote:
At 2/1/2014 6:18:47 PM, AlbinoBunny wrote:

Enji seemed to be saying they're picked out of a hat. Although I think I'm just misunderstanding what they're saying.

So do you know what the summation methods being used mean when they say that 1 + 2 + 3... = -1/12?

I'm not saying numbers are randomly selected. For example, you can use limits to determine the value of the sum of a series and the answer you get is non-random and can be rigorously defended given the methodology of limits. But saying that the sum of the natural numbers of infinity is of limited use -- we can't do much with infinity. Alternative methods (analytic continuation) allow you to get the sum -1/12, and like the approach using limits, the rules which allow you to get this sum are self-consistent and mathematically rigorous. Neither set of rules is wrong, and the answers they give are non-random, but -1/12 is more useful in some circumstances (like physics). But -1/12 is not the value of the sum of all the natural numbers.

What is it?
bladerunner060 | bsh1 , 2014! Presidency campaign!

http://www.debate.org...
http://www.debate.org... - Running for president.
http://www.debate.org... - Running as his vice president.

May the best man win!
Enji
Posts: 1,022
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2/1/2014 8:36:36 PM
Posted: 2 years ago
At 2/1/2014 7:41:57 PM, AlbinoBunny wrote:
At 2/1/2014 7:36:59 PM, Enji wrote:
At 2/1/2014 6:18:47 PM, AlbinoBunny wrote:

Enji seemed to be saying they're picked out of a hat. Although I think I'm just misunderstanding what they're saying.

So do you know what the summation methods being used mean when they say that 1 + 2 + 3... = -1/12?

I'm not saying numbers are randomly selected. For example, you can use limits to determine the value of the sum of a series and the answer you get is non-random and can be rigorously defended given the methodology of limits. But saying that the sum of the natural numbers of infinity is of limited use -- we can't do much with infinity. Alternative methods (analytic continuation) allow you to get the sum -1/12, and like the approach using limits, the rules which allow you to get this sum are self-consistent and mathematically rigorous. Neither set of rules is wrong, and the answers they give are non-random, but -1/12 is more useful in some circumstances (like physics). But -1/12 is not the value of the sum of all the natural numbers.

What is it?

There isn't one.
AlbinoBunny
Posts: 3,781
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2/1/2014 9:50:50 PM
Posted: 2 years ago
At 2/1/2014 8:36:36 PM, Enji wrote:
At 2/1/2014 7:41:57 PM, AlbinoBunny wrote:
At 2/1/2014 7:36:59 PM, Enji wrote:
At 2/1/2014 6:18:47 PM, AlbinoBunny wrote:

Enji seemed to be saying they're picked out of a hat. Although I think I'm just misunderstanding what they're saying.

So do you know what the summation methods being used mean when they say that 1 + 2 + 3... = -1/12?

I'm not saying numbers are randomly selected. For example, you can use limits to determine the value of the sum of a series and the answer you get is non-random and can be rigorously defended given the methodology of limits. But saying that the sum of the natural numbers of infinity is of limited use -- we can't do much with infinity. Alternative methods (analytic continuation) allow you to get the sum -1/12, and like the approach using limits, the rules which allow you to get this sum are self-consistent and mathematically rigorous. Neither set of rules is wrong, and the answers they give are non-random, but -1/12 is more useful in some circumstances (like physics). But -1/12 is not the value of the sum of all the natural numbers.

What is it?

There isn't one.

I mean what is -1/12.
bladerunner060 | bsh1 , 2014! Presidency campaign!

http://www.debate.org...
http://www.debate.org... - Running for president.
http://www.debate.org... - Running as his vice president.

May the best man win!
Enji
Posts: 1,022
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2/1/2014 11:12:14 PM
Posted: 2 years ago
At 2/1/2014 9:50:50 PM, AlbinoBunny wrote:
At 2/1/2014 8:36:36 PM, Enji wrote:
At 2/1/2014 7:41:57 PM, AlbinoBunny wrote:
At 2/1/2014 7:36:59 PM, Enji wrote:
At 2/1/2014 6:18:47 PM, AlbinoBunny wrote:

Enji seemed to be saying they're picked out of a hat. Although I think I'm just misunderstanding what they're saying.

So do you know what the summation methods being used mean when they say that 1 + 2 + 3... = -1/12?

I'm not saying numbers are randomly selected. For example, you can use limits to determine the value of the sum of a series and the answer you get is non-random and can be rigorously defended given the methodology of limits. But saying that the sum of the natural numbers of infinity is of limited use -- we can't do much with infinity. Alternative methods (analytic continuation) allow you to get the sum -1/12, and like the approach using limits, the rules which allow you to get this sum are self-consistent and mathematically rigorous. Neither set of rules is wrong, and the answers they give are non-random, but -1/12 is more useful in some circumstances (like physics). But -1/12 is not the value of the sum of all the natural numbers.

What is it?

There isn't one.

I mean what is -1/12.

It's more or less information about how the sum diverges which can be used to do some other useful maths like define determinants.