What is the total sum of all numbers?

zero?

I'd say 8 - and then rotate it 90degrees.

By analytic continuation of the Riemann zeta function,

 

[math]\sum_{n=1}^{\infty} n = -\,\frac 1 {12}[/math]

 

Ask any physicist.

forty-two.

 

But seriously, I guess it would be zero. [math]\infty - \infty = 0[/math]?

Isn't the set of all numbers unbounded (continuous), such that asking the sum of them is undefined?

But if the sum of all positive integers is infinity, and the sum of all positive numbers which are multiples of tenths is infinity, and so on and so forth, the limit as the precision of the numbers included in the set goes to infinity should be the same, shouldn't it?

Kyrisch: Infinity is not a number. Don't treat it as one.

 

iNow: Correct. The sum of all integers, [math]\sum_{n=-\infty}^{\infty} n[/math] is not an absolutely convergent series. The terms can be rearranged to give whatever answer one desires. The correct answer is of course 42.

So I was right after all xP

What is the total sum of all numbers?

zero?

 

Zero? Does it not seem common sense that the sum of all numbers in any sequence would be bigger than the biggest number in the sequence you're summing up?

Zero? Does it not seem common sense that the sum of all numbers in any sequence would be bigger than the biggest number in the sequence you're summing up?

 

Not to me.

For every positive number there is a negative number and they form a pair whose sum is zero.

Then you sum all the zeros and get zero.

That's exactly what I was going to say. We know that [math]|\Re^+ |= |\Re^- |[/math]. Pair these up, and you'll get 0. Sum the zeros, and you get zero. The only thing I'm not sure about here that since you're adding an infinite amount of zeros, wouldn't this be like multiplying by infinity? I'm sort of loosening the leash on my imagination here, so feel free to step in a kick me back in place

I don't know but the virtual photon- Anitphoton pairs that form near Black holes have a preference based on gravity...

that the negative photons tend to fall in and the positive ones tend to radiate.

Thus the Entropy of a black hole as per Hawking...

As such because of gravity I am quite sure that the number cannot be exactly zero so I vote for 42 also.

You are after all adding up all numbers in the presence of gravity.

~minus

I don't know but the virtual photon- Anitphoton pairs that form near Black holes have a preference based on gravity...

that the negative photons tend to fall in and the positive ones tend to radiate.

Thus the Entropy of a black hole as per Hawking...

As such because of gravity I am quite sure that the number cannot be exactly zero so I vote for 42 also.

You are after all adding up all numbers in the presence of gravity.

~minus

 

Um I am not sure on the physics of your answer but I am pretty sure that the answer is a mathematical one that has nothing to do with gravity, black holes, or Hawking Radiation. That said I would have to agree with Shadow that if you add all real numbers for every positive real number [math]n[/math]there must be a negative real number [math]-n[/math] and since [math](n)+(-n)=0[/math] I would assume that the series of all real numbers would be 0.

Um I am not sure on the physics of your answer but I am pretty sure that the answer is a mathematical one that has nothing to do with gravity, black holes, or Hawking Radiation. That said I would have to agree with Shadow that if you add all real numbers for every positive real number [math]n[/math]there must be a negative real number [math]-n[/math] and since [math](n)+(-n)=0[/math] I would assume that the series of all real numbers would be 0.

 

Sorry about the bad humor...

I think the correct answer is the limit of N as N approaches zero.

~minus

Not to me.

For every positive number there is a negative number and they form a pair whose sum is zero.

Then you sum all the zeros and get zero.

You can't do that. The reason is that you can reorganize a conditionally convergent series to give any answer you want according to the Riemann series theorem.

You can't do that. The reason is that you can reorganize a conditionally convergent series to give any answer you want according to the Riemann series theorem.

Cool then we are all right *GROUP HUG*

This kind of math certainly explains global warming and coming ice age at the same time.

Those climate guys just need to keep adding those numbers up and they can give us any answer we are willing to pay for...

It is still 42 as is global warming.

~minus

Cool then we are all right *GROUP HUG*

This kind of math certainly explains global warming and coming ice age at the same time.

Those climate guys just need to keep adding those numbers up and they can give us any answer we are willing to pay for...

It is still 42 as is global warming.

~minus

 

Smooth segue, guy.

Hmm. That's an interesting point. For each number n, there is a negative -n, whose sum is zero. But there are an infinite number of these pairs, so the answer would be [math]\infty * 0[/math] which is an indeterminate form.

Hmm. That's an interesting point. For each number n, there is a negative -n, whose sum is zero. But there are an infinite number of these pairs, so the answer would be [math]\infty * 0[/math] which is an indeterminate form.

 

It's more indeterminate than that, actually. You can add them that way, or an infinity of other ways. For example, you can say "for every positive number n, there is a negative number -n/2." It's still a one to one correspondence, but the positive increases twice as fast as the negative, and the "sum" is infinitely large.

 

In other words, it's not just undefined but actually meaningless to talk about the sum of all numbers.

With infinity multiplied by zero, which takes precedence, the infinity or the zero?

 

Or does it equal 1?

With infinity multiplied by zero, which takes precedence, the infinity or the zero?

 

Or does it equal 1?

 

You can't multiply infinity by anything, inasmuch as it is not a number.

In other words, it's not just undefined but actually meaningless to talk about the sum of all numbers.

The sum of all counting numbers,

 

[math]1+2+3+4+\cdots = \sum_{n=1}^{\infty}n[/math]

 

does have meaning. By zeta function regularization (see http://motls.blogspot.com/2007/09/zeta-function-regularization.html, http://en.wikipedia.org/wiki/Zeta_function_regularization, http://math.ucr.edu/home/baez/week126.html, and (PDF) http://math.ucr.edu/home/baez/qg-winter2004/zeta.pdf),

 

[math]\sum_{n=1}^{\infty}n = -\frac 1 {12}[/math]

 

Zeta function regularization similarly says

 

[math]1+1+1+1+1+\cdots = \sum_{n=1}^{\infty} 1 = -\frac 1 2[/math]

 

Abel summation yields things like

 

[math]1-1+1-1+1-1+\cdots = \frac 1 2[/math]

 

 

Some other interesting answers for divergent series:

 

[math]1+2+4+8+\cdots = \sum_{n=0}^{\infty}2^n = -1[/math]

 

[math]1-2+4-8+\cdots = \sum_{n=0}^{\infty}(-2)^n = \frac 1 3[/math]

 

 

Various tricks for evaluating such sums go all the way back to Euler. That [math]1+2+3+4+\cdots=-1/12[/math] has also been verified experimentally in measurements of the Casimir force.

Edited January 26, 200916 yr by D H

Could you post a source for the [math] 1+2+4+8+\cdots = \sum_{n=0}^{\infty}2^n = -1 [/math] one? I'd be interested to know more about this...Thanks

Suppose [math]1+2+4+8+\cdots[/math] exists. Call this sum S:

 

[math]S\equiv\sum_{n=0}^{\infty}2^n[/math]

 

Multiply S by two:

 

[math]2S = 2\sum_{n=0}^{\infty}2^n = \sum_{n=0}^{\infty}2\cdot2^n = \sum_{n=0}^{\infty}2^{n+1} = \sum_{n=1}^{\infty}2^n[/math]

 

The last expression is just the original sum sans the zeroth element: [math]2S=S-1[/math] and thus [math]S=-1[/math].

 

See http://en.wikipedia.org/wiki/1_%2B_2_%2B_4_%2B_8_%2B_·_·_·

Haven't you mathematicians figured out yet that Euler was just messing with you guys? Anyway, as enraging as an answer of -1/12 is, that's just that specific series, right? If you calculate it differently, can you get different results? Also, there wouldn't be anything analogous if you were trying to, say, sum all rational numbers? It seems more clear that you could get any answer you wanted in that case, right? (These are non-rhetorical questions.)