Determining A Finite Value of Infinity

AI Thread Summary
The discussion centers on the controversial assertion that the series 1+2+3+4... equals -1/12, a concept derived from zeta regularization. Participants argue that while this value is accepted in certain mathematical contexts, it does not imply that the series itself converges to -1/12, as the series is divergent. The original poster attempts to derive a finite value for infinity using a quadratic equation, but others point out that infinity cannot be treated as a real number and that the reasoning is flawed. The conversation highlights the distinction between regularization methods and actual sums, emphasizing that the manipulation of infinite values differs fundamentally from finite ones. Ultimately, the thread concludes with a consensus that the original claim lacks validity.
willr12
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okay...if you accept that the sequence
1+2+3+4...=-1/12,
I think I have determined a finite value of infinity.
To find the value of the sums of all natural numbers up to a number, you can use the equation
((x^2)+x)/2.
An example would be 4.
4+3+2+1=10.
((4^2)+4)/2 also equals 10.
following this logic,
((x^2)+x)/2=-1/12
is true for the above sequence. this can be rearranged to
(x^2)+x+1/6
This is the resulting quadratic equation. Using the quadratic formula, one obtains the x intercepts as
(-3+-(sqrt3))/6.
and since x is infinity in this situation (since the highest value is infinity), the x intercepts are the values of infinity in the equation. Therefore, by assigning a value of -1/12 to riemann zeta(-1), you also assign finite values to infinity, approximately
-0.211324...
and
-0.788675...
If there is any faulty reasoning, please remember I'm 15 and I most likely have no clue what I'm talking about.
 
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willr12 said:
...
I think I have determined a finite value of infinity...
Since this is a contraction in terms, I stopped reading right there.
 
willr12 said:
okay...if you accept that the sequence
1+2+3+4...=-1/12,
I think I have determined a finite value of infinity.
http://en.wikipedia.org/wiki/Divergent_series

There is no sum of the integers from one to infinity. The series is divergent. The assignment of the value -1/12 as the "sum" of this series does not correspond to anything which you should regard as being an actual sum.
 
phinds said:
Since this is a contraction in terms, I stopped reading right there.
Most people would stop reading when someone says that 1+2+3...=-1/12, that's counterintuitive but generally accepted...so thanks for the ignorance
 
jbriggs444 said:
http://en.wikipedia.org/wiki/Divergent_series

There is no sum of the integers from one to infinity. The series is divergent. The assignment of the value -1/12 as the "sum" of this series does not correspond to anything which you should regard as being an actual sum.
Makes sense
 
willr12 said:
okay...if you accept that the sequence
1+2+3+4...=-1/12,

No. The series 1+2+3+... is divergent. So it is infinite.
It's Zeta regularization has value -1/12.
That is not the same thing as claiming that the series is "equal" to -1/12

To find the value of the sums of all natural numbers up to a number, you can use the equation
((x^2)+x)/2.
An example would be 4.
4+3+2+1=10.
((4^2)+4)/2 also equals 10.
following this logic,
((x^2)+x)/2=-1/12
is true for the above sequence.

I fail to see how. Just because a statement is true for finite values of x does not mean it is valid for infinite values of x. You need to prove that "((x^2)+x)/2=-1/12" is a valid statement for some x. You have not.

and since x is infinity in this situation (since the highest value is infinity), the x intercepts are the values of infinity in the equation.

This is not possible because (1) infinity is not a real number, (2) infinite values cannot be manipulated in the same way as finite values.

If there is any faulty reasoning

Unfortunately, everything.
 
pwsnafu said:
No. The series 1+2+3+... is divergent. So it is infinite.
It's Zeta regularization has value -1/12.
That is not the same thing as claiming that the series is "equal" to -1/12
I fail to see how. Just because a statement is true for finite values of x does not mean it is valid for infinite values of x. You need to prove that "((x^2)+x)/2=-1/12" is a valid statement for some x. You have not.
This is not possible because (1) infinity is not a real number, (2) infinite values cannot be manipulated in the same way as finite values.
Unfortunately, everything.
Good to know. Thanks for the reply
 
willr12 said:
Most people would stop reading when someone says that 1+2+3...=-1/12, that's counterintuitive but generally accepted ...
No, as has been pointed out, it is NOT generally accepted.
 
phinds said:
No, as has been pointed out, it is NOT generally accepted.
It is generally accepted as a representation of Riemann zeta (-1)
 
  • #10
willr12 said:
It is generally accepted as a representation of Riemann zeta (-1)

No you have this backwards.
##\zeta(-1)## is a possible regularization of ##\sum_{n=1}^{\infty} n##.
##\sum_{i=1}^{\infty} n## is not a representation of ##\zeta(-1)##.
Regularization and representation are separate concepts.
 
  • #11
pwsnafu said:
No. The series 1+2+3+... is divergent. So it is infinite.
It's Zeta regularization has value -1/12.
That is not the same thing as claiming that the series is "equal" to -1/12
I fail to see how. Just because a statement is true for finite values of x does not mean it is valid for infinite values of x. You need to prove that "((x^2)+x)/2=-1/12" is a valid statement for some x. You have not.
This is not possible because (1) infinity is not a real number, (2) infinite values cannot be manipulated in the same way as finite values.
Unfortunately, everything.
Y-value of the vertex in the resulting quadratic equation is -1/12. Any significance do you think?
pwsnafu said:
No you have this backwards.
##\zeta(-1)## is a possible regularization of ##\sum_{n=1}^{\infty} n##.
##\sum_{i=1}^{\infty} n## is not a representation of ##\zeta(-1)##.
Regularization and representation are separate concepts.
good to know thanks
 
  • #12
willr12 said:
okay...if you accept that the sequence
1+2+3+4...=-1/12

I don't.
 
  • #13
This thread has run its course, so I'm closing it.
 
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