2(5) − cot [4 arctan 0.2 + (i/2) ln i] − 1

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The discussion centers on the evaluation of divergent sums, specifically the expression 2(5) − cot [4 arctan 0.2 + (i/2) ln i] − 1. Participants highlight the complexities of assigning finite values to divergent series, with references to the sum of natural numbers leading to -1/12 and the series Σ(2,∞) 1/n⁴. The conversation emphasizes the necessity of specifying evaluation methods for divergent sums to avoid ambiguity, and it touches on humorous attempts to manipulate mathematical concepts for educational purposes.

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Jenab2
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Don't ever divide anything by the quantity in the title.

Post your favorite "fancy zeros" here.
 
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$$\frac{1}{12} + \sum _{n = 1} ^\infty n$$
 
axmls said:
$$\frac{1}{12} + \sum _{n = 1} ^\infty n$$

That should probably be

−1/12 + Σ(2,∞) 1/n⁴

Edit: whoops, no. That doesn't seem quite right, either. I evaluated ten million terms of the sum and came up with −0.0010100996222299347, so

−1/12 + 1/999 + Σ(2,∞) 1/n⁴

seems to be nearer to zero.
 
Last edited:
Jenab2 said:
That should probably be

−1/12 + Σ(2,∞) 1/n⁴

Edit: whoops, no. That doesn't seem quite right, either. I evaluated ten million terms of the sum and came up with −0.0010100996222299347, so

−1/12 + 1/999 + Σ(2,∞) 1/n⁴

seems to be nearer to zero.

Nope, it's written as I intended.
 
Unless you specify how divergent sums are to be evaluated, the formula is not well-defined.
Yes there is a specific way that leads to -1/12, but this is by far not the only way to assign finite values to divergent sums.
 
mfb said:
Unless you specify how divergent sums are to be evaluated, the formula is not well-defined.
Yes there is a specific way that leads to -1/12, but this is by far not the only way to assign finite values to divergent sums.

It was a somewhat tongue-in-cheek answer, if that wasn't clear.
 
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I tried something like this to mess with my maths teacher in senior year once, replaced pi with some weird sums.
 
1 / Σ(1,∞) n = 0
1 / { 1/a + 1 / Σ(1,∞) n } = a, a≠0.
 
Jenab2 said:
1 / Σ(1,∞) n = 0
1 / { 1/a + 1 / Σ(1,∞) n } = a, a≠0.

See here: https://en.m.wikipedia.org/wiki/1_+_2_+_3_+_4_+_⋯

There are ways to assign a value to that sum using alternate methods. That was the "tongue-in-cheek" aspect of my response.
 
  • #10
## \begin{Vmatrix} \vec\nabla \times \vec\nabla f \end{Vmatrix} ##
 
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  • #11
axmls said:
See here: https://en.m.wikipedia.org/wiki/1_+_2_+_3_+_4_+_⋯
There are ways to assign a value to that sum using alternate methods. That was the "tongue-in-cheek" aspect of my response.
Ah. My difficulty in appreciating the assignment was caused by my thinking of scalars in vector terms. Consider velocities in the same direction, classically being added, tail to head:

v₁ + v₂ + v₃ + v₄ + ...

where each velocity is in the direction of the +x axis and the magnitude of the velocities is proportional to the subscript.

How is it that an object, moving through an infinite succession of changes-of-velocity, all of them forward, might end up moving BACKWARD at a speed of 1/12 velocity units?

I'd figured that this was a case of getting a strange result out of an indeterminate form.
 

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