A problem with sums 1+2+3+4+5+6

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In summary, the conversation discusses using an exponential regulator to calculate the sum of infinite series, such as 1+2+3+4+5+6+7+... and 1+8+27+64+125+... It is suggested that by removing certain terms and imposing renormalization conditions, the calculations can be simplified. The use of a power regulator is also mentioned, with the suggestion that it can lead to a functional determinant without a divergent term proportional to log(\epsilon). The conversation also touches on a related problem proposed by Conway and Baez, which involves finding integer values for U when solving for 1+2^{m}+3^{m}+...+ N^m= U
  • #1
zetafunction
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using an exponential regulator [tex] exp(-\epsilon n) [/tex] the sum

[tex] 1+2+3+4+5+6+7+...= -1/12+ 1/\epsilon^{2} [/tex]

and for Casimir effect [tex] 1+8+27+64+125+...= -1/120+ 1/\epsilon^{4} [/tex]

can i simply remove in the calculations of divergent series 1+2+3+4+5.. and similar the epsilon terms imposing renormalization conditions ??

how about for the rest of sums [tex] 1+2^{m}+3^{m}+...= \zeta (-m) + 1/\epsilon ^{m+1} [/tex]

if i introducte a power regulator [tex] n^{-s} [/tex] in the limit s-->0+ i would get

[tex] \zeta(s-m)=\zeta(-m) [/tex] but i am not sure, why this work

for example in the definition of a functional determinant (in differential geommetry )

[tex] \prod_{i} \lambda_{i} [/tex] apparently there is no divergent term proportional to [tex] log(\epsilon) [/tex] as one would expect since the product is divergent
 
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  • #2
Now I see the series, a related problem comes from Conway via Baez:
Let [tex]
1+2^{m}+3^{m}+...+ N^m= U^m
[/tex]
For which values of N, m do we get an integer value for U?

In some strange way, string theory relates the solution N=24 (or 26?), m=2 with the regulation of N=infinity, m=1
 
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1. What is the problem with sums 1+2+3+4+5+6?

The problem with this sum is that it is an arithmetic progression with a finite number of terms, which makes it easy to calculate.

2. Can this sum be simplified?

Yes, this sum can be simplified by using the formula for the sum of an arithmetic progression: Sn = (n/2)(a1 + an), where n is the number of terms, a1 is the first term, and an is the last term. In this case, n = 6, a1 = 1, and an = 6, so the simplified sum would be (6/2)(1 + 6) = 21.

3. How can this sum be solved using a loop?

This sum can be solved using a for loop, where the number of iterations is set to the number of terms (6 in this case) and the loop adds each term to a running total.

4. Is there a pattern in this sum?

Yes, this sum follows a specific pattern known as an arithmetic progression, where each term is the sum of the previous term and a constant value (in this case, the constant value is 1).

5. Can this sum be represented visually?

Yes, this sum can be represented visually using a bar model or a number line. On a bar model, each term can be represented by a bar with a length equal to its value, and the sum can be found by adding all the lengths together. On a number line, each term can be represented by a point on the line, and the sum can be found by counting the number of points from the starting point to the end point.

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