Divergent series sum, versus integral from -1 to 0

  • #1
Some popular math videos point out that, for example, the value of -1/12 for the divergent sum 1 + 2 + 3 + 4 .... can be found by integrating n/2(n+1) from -1 to 0. We can easily verify a similar result for the sum of k^2, k^3 and so on.

Is there an elementary way to connect this with the more formal definitions of divergent sums, e.g. analytic continuation? By elementary, I mean something that the popular math channels (e.g. Mathologer or Numberphile) might come up with, to show that the two procedures are somehow equivalent.

Also, it seems that this can be generalized to all sorts of arbitrary sums via Taylor's theorem -- is that the case?

Edit: Are there cases where this fails?
 
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Answers and Replies

  • #3
By the way, it works for arbitrary power terms, including complex powers:

Code:
In[85]:= fnZ[x_, n_] := Evaluate[Sum[k^n, {k, 1, x}]]

In[86]:= fnF[k_] := Integrate[fnZ[x, k], {x, -1, 0}]

In[87]:= fnF[1]

Out[87]= -(1/12)

In[89]:= fnF[3]

Out[89]= 1/120

In[90]:= fnF[2.345 + I*4.321]

Out[90]= 0.0992824 - 0.363493 I

In[91]:= Zeta[-(2.345 + I*4.321)]

Out[91]= 0.0992824 - 0.363493 I
which seems to be an amazingly simple way to find zeta(x). Is this utterly trivial, or is it a historic find? :cool:

On reflection, the integral evaluation is probably non-trivial in most cases? Hmmm...
 
  • #8
This is one of the videos I had in mind in my OP.

He does state the fact that
## \int_{-1}^0 \frac n 2(n+1) \, dn ##
gives you the same answer as ##\zeta(-1)## , but he doesn't explain how it comes about.

And what's more, it works for other arguments z in ##\zeta(z)##, including complex z. (See my Mathematica code above). I'm hoping there might be an elementary way to explain these things at the level of a Mathologer or Numberphile video.
 
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  • #9
PeroK
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Some popular math videos point out that, for example, the value of -1/12 for the divergent sum 1 + 2 + 3 + 4 .... can be found by integrating n/2(n+1) from -1 to 0. We can easily verify a similar result for the sum of k^2, k^3 and so on.

Is there an elementary way to connect this with the more formal definitions of divergent sums, e.g. analytic continuation? By elementary, I mean something that the popular math channels (e.g. Mathologer or Numberphile) might come up with, to show that the two procedures are somehow equivalent.

Also, it seems that this can be generalized to all sorts of arbitrary sums via Taylor's theorem -- is that the case?

Edit: Are there cases where this fails?
You can see something similar without looking at complex analytic continuation. Consider: $$f(x) = \frac{1}{1-x}$$ and using the Taylor expansion we have $$f(x) = g(x) = 1 + x + x^2 \dots \ \ (|x| < 1)$$ These functions agree on the interval ##(-1, 1)##, but ##g(x)## is not defined outside this range. Let's compare, however, the function values for ##f(x)## with the series for ##g(x)## outside this range:
$$f(2) = -1 \ \ \text{and} \ \ g(2) = 1 + 2 + 4 + 8 \dots$$ $$f(-1) = \frac 1 2 \ \ \text{and} \ \ g(-1) = 1 -1 + 1 -1 \dots$$
Is there some sense in which the information for the function value at ##f(x)## is somehow encoded in the divergent sum? So that in some sense:
$$1 + 2 + 4 + 8 \dots = -1 \ \ \text{and} \ \ 1 -1 + 1 -1 \dots = \frac 1 2$$
Let's define a sum function that acts on sequences and has two properties:
$$S(a_0 + a_1 + a_2 + \dots) = a_0 + S(a_2 + a_2 + a_3 \dots)$$ $$S(ka_0 + ka_1 + ka_2 \dots) = kS(a_0 + a_1 + a_2 \dots)$$ If we apply that sum function to our divergent series we get: $$s = S(1 -1 + 1- 1 \dots) = 1 + S(-1 + 1 -1\dots) = 1 -S(1 -1 + 1 \dots) = 1 - s$$ and hence ##s = \frac 1 2##.

$$s = S(1 + 2 + 4 + 8 \dots ) = 1 + S(2 + 4 + 8 \dots ) = 1 + 2s$$ And, in this case, ##s = -1##.

And, in general, using the properties of the sum function we have:
$$s(x) = S(1 + x + x^2 + \dots) = 1 + xs(x) \ \ \text{hence} \ \ s(x) = \frac 1 {1-x}$$

You can see, therefore, that in some sense the function values for ##f(x)## are encoded in the divergent series for ##g(x)##.
 
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  • #10
You can see something similar without looking at complex analytic continuation.

.... etc ...

You can see, therefore, that in some sense the function values for ##f(x)## are encoded in the divergent series for ##g(x)##.
That's a very nice introduction to how a value can be assigned to the sum of a divergent series.

But it doesn't address my question as to how, if we take the formula for a finite sum over ##n^z## and integrate it from -1 to 0, we get the same answer as a more advanced formalism like analytic continuation.
 
  • #11
PeroK
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That's a very nice introduction to how a value can be assigned to the sum of a divergent series.

But it doesn't address my question as to how, if we take the formula for a finite sum over ##x^n## and integrate it from -1 to 0, we get the same answer as a more advanced formalism like analytic continuation.
It was intended to address this part of your question:

Is there an elementary way to connect this with the more formal definitions of divergent sums ... By elementary, I mean something that the popular math channels (e.g. Mathologer or Numberphile) might come up with, to show that the two procedures are somehow equivalent.

Also, it seems that this can be generalized to all sorts of arbitrary sums via Taylor's theorem -- is that the case?
 
  • #12
It was intended to address this part of your question:
"Is there an elementary way to connect this with the more formal definitions of divergent sums ... By elementary, I mean something that the popular math channels (e.g. Mathologer or Numberphile) might come up with, to show that the two procedures are somehow equivalent."
Maybe that part of my question could have been worded more clearly, because when I said "show that the two procedures are somehow equivalent" I meant the following two procedures:

(a) Integrating the finite-sum formula from -1 to 0 (which I had referred to in the first paragraph of the OP)

and

(b) Analytic continuation and other formal definitions

Sorry for the ambiguity.
 
  • #13
anuttarasammyak
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According to your observation,
[tex]\int_{-1}^0 dx \sum_{k=1}^x k^{n}[/tex]
where n=0,1,2,.. and summation is carried out as if x is integer, equals to
[tex]\zeta(-n)=\frac{(-1)^n}{n+1}(\frac{d}{dx})^n\frac{x}{e^x-1}|_{x=0}[/tex]
As a detailed prescription of summation first write Faulhaber's formula (ref. https://en.wikipedia.org/wiki/Faulhaber's_formula ) and then replace n, the maximum integer, with real number x[-1,0].

I hope careful calculation explains equality though I have not done it by myself.
 
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  • #14
As a detailed prescription of summation first write Faulhaber's formula (ref. https://en.wikipedia.org/wiki/Faulhaber's_formula ) and then replace n, the maximum integer, with real number x[-1,0]. I hope careful calculation explains equality...
The actual task is probably beyond my capacity, but it's still nice to see the roadmap that a real mathematician would follow.
 

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