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Gib Z

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Gib Z

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Actually... I have thought of a proof for that.. but without that "R". I hope this helps...

You first consider the "-" summation (from i = 1 to infinity) of i(-r)^i. Let this sum be equal to S. Then,

S = r - 2r^2 + 3r^3 - 4r^4 + ... - ...

Sr = 0 + r^2 - 2r^3 + 3r^4 - ... + ...

S + Sr = r - r^2 + r^3 - r^4 + ... - ...

S(1+ r) = r / [1+r]

S = r/(1+r)^2

When you take the limit of both sides of the equation as r -> 1, we have:

Lim S (r-> 1) = Lim r/(1+r)^2 (r-> 1)

1 - 2(1)^2 + 3(1)^3 - 4(1)^4 + ... - ... = 1/4

(remember how I defined my S awhile ago? )

or simply, 1 - 2 + 3 - 4 + .. - ... = 1/4 (remember this)

Now, If I let T = 1 + 2 + 3 + ... then,

2T = 2(1 + 2 + 3 + ... ).. then,

4T = 4(1 + 2 + 3 + ... ) = 2(2 + 4 + 6 + ... ) (remember this too)

Therefore, 1/4 + 4T = ?.

Look...

1 - 2 + 3 - 4 + .. - ...

+

2(2) + 2(4) + 2(6) + 2(8) +...

As you can see, the odd numbers are left as is, but the even numbers are combined,

say 2(2) - 2 and 2(4) -4.

Thus, 1/4 + 4T = T.

Solving for T, you got -1/12.

So,

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

You first consider the "-" summation (from i = 1 to infinity) of i(-r)^i. Let this sum be equal to S. Then,

S = r - 2r^2 + 3r^3 - 4r^4 + ... - ...

Sr = 0 + r^2 - 2r^3 + 3r^4 - ... + ...

S + Sr = r - r^2 + r^3 - r^4 + ... - ...

S(1+ r) = r / [1+r]

S = r/(1+r)^2

When you take the limit of both sides of the equation as r -> 1, we have:

Lim S (r-> 1) = Lim r/(1+r)^2 (r-> 1)

1 - 2(1)^2 + 3(1)^3 - 4(1)^4 + ... - ... = 1/4

(remember how I defined my S awhile ago? )

or simply, 1 - 2 + 3 - 4 + .. - ... = 1/4 (remember this)

Now, If I let T = 1 + 2 + 3 + ... then,

2T = 2(1 + 2 + 3 + ... ).. then,

4T = 4(1 + 2 + 3 + ... ) = 2(2 + 4 + 6 + ... ) (remember this too)

Therefore, 1/4 + 4T = ?.

Look...

1 - 2 + 3 - 4 + .. - ...

+

2(2) + 2(4) + 2(6) + 2(8) +...

As you can see, the odd numbers are left as is, but the even numbers are combined,

say 2(2) - 2 and 2(4) -4.

Thus, 1/4 + 4T = T.

Solving for T, you got -1/12.

So,

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

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- #3

Gib Z

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1 - 2(1)^2 + 3(1)^3 - 4(1)^4 + ... - ... = 1/4

(remember how I defined my S awhile ago? )

or simply, 1 - 2 + 3 - 4 + .. - ... = 1/4 (remember this)

Now, If I let T = 1 + 2 + 3 + ... then,

2T = 2(1 + 2 + 3 + ... ).. then,

4T = 4(1 + 2 + 3 + ... ) = 2(2 + 4 + 6 + ... ) (remember this too)

Therefore, 1/4 + 4T = ?.

Look...

1 - 2 + 3 - 4 + .. - ...

+

2(2) + 2(4) + 2(6) + 2(8) +...

As you can see, the odd numbers are left as is, but the even numbers are combined,

say 2(2) - 2 and 2(4) -4.

Thus, 1/4 + 4T = T.

Solving for T, you got -1/12.

I dont think this is logically correct. One cannot rearrange the terms of an infinite series unless it is absolutely convergent. The series 1-2+3-4.... is not absolutely convergent.

You shall get fallacy.....what you have derived is 1+2+3+....ad inf =-1/12, sum of all positive numbers is negative!!

Example:

ln(2)= 1- 1/2 + 1/3 - 1/4+.......ad inf

=(1 + 1/2 + 1/3 + 1/4 +....) - 2(1/2 + 1/4 + 1/6 +...)

=(1 + 1/2 + 1/3 + 1/4 +....)-(1 + 1/2 + 1/3 + 1/4 +....)

=0

=ln(1)

Therefore ln(2)=ln(1) , that is 1=2.

Since 1- 1/2 + 1/3 - 1/4+.......ad inf is not absolutely convergent, its terms cannot be rearranged.

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Gib Z

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LOL, I am not a man of Physics, unable to understand the physical interpretation of String theory facts.

- #8

Gib Z

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I know enough though to understand the link. Theres 2 proofs for the question, look there if you like.

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S = r - 2r^2 + 3r^3 - 4r^4 + ... - ...

Sr = 0 + r^2 - 2r^3 + 3r^4 - ... + ...

S + Sr = r - r^2 + r^3 - r^4 + ... - ...

S(1+ r) = r / [1+r]

S = r/(1+r)^2

When you take the limit of both sides of the equation as r -> 1, we have:

Lim S (r-> 1) = Lim r/(1+r)^2 (r-> 1)

1 - 2(1)^2 + 3(1)^3 - 4(1)^4 + ... - ... = 1/4

(remember how I defined my S awhile ago? )

[/tex]

'Gib Z' I dont know what to say.....

My konwledge goes in this way,

r - r^2 + r^3 - r^4....ad inf

= r/(1+r) if and only if "-1< r <1" holds. The inequalities being strict.

When one uses the limit r->1, that means he calculates limit r->1- and limit r ->1+ and found them to be equal. But by taking limit r ->1+ one violets the necessary and sufficient condiditon |r|<1 for this summation.

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Gib Z

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Forgive the lack of LaTeX but the notation is self-explanatory! The following combined with the analytic extension theorems of complex analysis provide suitable Euler-Abel sums for negative values of the Riemann Zeta function.

Note that <B*_k> stands for the old Bernoulli numbers(the k th one here) and <B_k> the new ones(I explain in the end)

1 + e^iy + e^2iy + e^3iy + ... = (1 - e^iy)^-1 = 1/2 + (1/2)i.cot y/2

Equating real and imaginary parts we have,

cos y + cos 2y + cos 3y + ... = -1/2 --------(1)

and,

sin y + sin 2y + sin 3y + ... = (1/2)cot y/2 ------------(2)

Replace y by (y + pi) and we have,

cos y - cos 2y + cos 3y - ... = 1/2

differentiate 2k times and we have

Sum [n=1 to infinity] (-1)^(n-1) n^2k cos ny = 0

differentiate 2k-1 times and we have,

Sum [n=1 to infinity] (-1)^(n-1) n^(2k-1) sin ny = 0

for (2) we get,

sin y - sin 2y + sin 3y - .... = (1/2)tan y/2

since the Taylor expansion of (1/2)tan y/2 is,

Sum [k=1 to infinity] {(2^2k - 1) <B*_k> y^(2k-1)} / (2k)!

Differentiate 2k-1 times, put y=1 and equate to get,

Sum [n=1 to infinity] (-1)^(n-1) n^(2k-1) = [(-1)^(k-1) (2^2k - 1) <B*_k>] / 2k

Now recall Z(1- 2k) = 1^(2k-1) + 2^(2k-1) + 3^(2k-1) + ....

where Z is Riemann's Zeta function, we see that the function we handled,

N(1-2k) = Sum [n=1 to infinity] (-1)^(n-1) n^(2k-1) = 1^(2k-1) - 2^(2k-1) + 3^(2k-1) - ...

satisfies,

(1 - 2^2k) Z(1-2k) = N(1-2k)

Then,

Z(1-2k) = (-1)^k <B*_k> / 2k

But with the modern notation for the Bernoulli numbers,

<B_2n> = (-1)^(n+1) <B*_n>

therefore,

Z(1-2k) = (-1)^(2k+1) <B_2k> / 2k = - <B_2k> / 2k

Quite generally,

Z(1-n) = (-1)^(n-1) <B_n> / n

for all positive integer k(even zero in a sense, more below), defining the Zeta function for odd negative integers (the case we considered first obviously shows that even negative integers make the function zero, these are its trivial zeros, the other zeros constitute the Riemann Hypothesis, the greatest intellectual challenge faced by man)

Now the fun bit, put k=1 and we have the magical,

Z(-1 ) = 1 + 2 + 3+ 4 + .... = -1 / 12 , since B<2> = 1/6

This value actually holds a meaning(though we are hard put to explain in any connotation of the word brief with regard to use in physics, but it has to do with what may be termed 'measure' of infinities, thankfully in math definition is all that matters!) and is extensively used with the case k=2 in theoretical physics(namely the Casimir Effect and Perturbation theory on quantum fields), but applications are always of secondary importance!

Rigorous enough eh? Yes, but only if we consider it in a special summation definition (the extended A definition due to Euler, though the Ramanujan sum coincides too)and note the analyic extention of the zeta function to the whole plane.

Actually this is most awe-inspiring since it fits with the (seemingly independant) analytic extention perfectly. As in how k=0 gives the simple pole!

The 'trickery' per se is that le^iyl = 1 for all y, but since we avoid the actual pole there is no damage done!!

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Then we have,

$\sum_{n = 0}^{\infty} f(n) = \sum_{n = 0}^{\infty} {\exp}{n{\delta}} f(0)$

When t is limited to 0.

We can sum the exponential series as a geometric progression(this works even for the operator)and,

$\sum_{n = 0}^{\infty} f(n) = (1 - {\exp}{{\delta}})^{ - 1} f(0)$

since,

$\frac {{\delta}}{{\exp}{{\delta}} - 1} = \sum_{n = 0}^{\infty} \frac {B_{n}}{n!} {\delta}^{n}$

we substitute and we have,

$\sum_{n = 0}^{\infty} f(n) = - \sum_{n = 0}^{\infty} \frac {B_{n}}{n!} {\delta}^{n - 1}f(0)$

(the operations on the operator are valid)

put f(n) = n and we have 1+2+3+4+...= -1/12

etc.

- #13

Gib Z

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- #14

Hurkyl

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And [ itex ] for stuff inlined with ordinary text.

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The beauty of mathematics is that logic can at times have very little bearing and all that matters is definition!(if you can follow it up that is)

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Hurkyl

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Of course, that's where you need logic. :tongue:(if you can follow it up that is)

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please can someone clarify whether or not this is a FACT?!

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Hurkyl

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The common meaning of the symbols you wrote is equivalent to "the infinite summation operation you learned in calculus". And in that meaning, there does not exist a proof of that statement.If there do exist rigorous proofs of the fact that 1 + 2 + 3 + .... = -1/12

Only when you consider different kinds of 'summation' operations can statements like that be true.

- #20

Gib Z

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please can someone clarify whether or not this is a FACT?!

I didn't put an R in that equation for no reason you know =]

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Note the connexion between the Bernoulli/differential operator derivation I gave secondly, the Euler MacLaurin sum formula and Ramanujan definition of a sum (which is perhaps most unorthodox, being, very bluntly put- the difference between the actual sum and integral!

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https://www.physicsforums.com/showthread.php?p=1474865#post1474865

- #23

Hurkyl

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Of course it does. The sumBut you must understand(forgive my apparent reduction to metaphysics, but that's not what it is) that whether it be the R, A, E, B, C, etc. definitions it does not make the relations any less true.

1 + 2 + 3 + .... = -1/12

is an explicit example of this -- this relation is patently - #24

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NOT!

It IS valid for many sorts of generalized functions as is apparent from my 2nd post (though that's not explicitly Ramanujan summation, for more consider the Euler MacLaurin formula)

These new summation operators are most all legitimate, can be made rigorous, consistent and are most inconveniently for those who curl up in their comfy math beds(or dare I say Couches) - true.

It is all in the definition as Hurkyl apparently concedes and does not invade upon the rigours of pure mathematics(as that is indeed my main area of interest)

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which is all very well and appreciated, but we mustn't overlook other possibilities. I'd imagine you'd be more open to this sort of thing on 'physicsforums' since almost textbook results from this neck of the woods are required in perturbation theory sometimes as I understand!!

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