Divergent series sum, versus integral from -1 to 0

In summary, the conversation discusses the value of -1/12 for the divergent sum 1 + 2 + 3 + 4 ... and how it can be found by integrating n/2(n+1) from -1 to 0. The conversation also explores the connection between this method and more formal definitions of divergent sums, such as analytic continuation. It is noted that this method can be generalized to other arbitrary sums via Taylor's theorem. However, it is also questioned whether there are cases where this method fails. A solution is proposed by defining a sum function that can assign values to divergent series, showing that the function values for f(x) can be encoded in the divergent series for g(x). While this
  • #1
Swamp Thing
Insights Author
914
587
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?
 
Last edited:
Mathematics news on Phys.org
  • #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...
 
  • #4
@anuttarasammyak I have read that page, but I'm not sure if it will help make a simple connection between the two things I mentioned in my OP..
 
  • #5
Swamp Thing said:
can be found by integrating n/2(n+1) from -1 to 0
Integrate with respect to what?
 
  • #6
Svein said:
Integrate with respect to what?
With respect to n, treating n as a real variable.
 
  • #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.
 
Last edited:
  • #9
Swamp Thing said:
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)##.
 
  • Like
Likes nasu and fresh_42
  • #10
PeroK said:
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
Swamp Thing said:
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:

Swamp Thing said:
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
PeroK said:
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
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.
 
Last edited:
  • Informative
Likes Swamp Thing
  • #14
anuttarasammyak said:
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.
 
  • #15
I posted a comment on a recent video of Mathologer's, suggesting that he might do a video about this topic. The comment got a like from Mathologer, with this reply:

"Yes, that would be nice. Pretty challenging to get this right. Have you see the video on the Euler-Maclaurin series. It all starts with that :) "
 

Related to Divergent series sum, versus integral from -1 to 0

1. What is a divergent series sum?

A divergent series sum is a mathematical concept that refers to the sum of an infinite series where the terms do not approach a finite limit. This means that the sum of the series does not have a specific value and can continue to increase without bound.

2. What is an integral?

An integral is a mathematical concept that represents the area under a curve in a graph. It is used to calculate the total value of a function over a given interval. In the context of this question, the integral from -1 to 0 refers to the area under the curve between the x-axis and the function between the x-values of -1 and 0.

3. How are divergent series sums and integrals related?

Divergent series sums and integrals are related through the concept of convergence. If a series is convergent, then its corresponding integral will also be convergent. However, if a series is divergent, then its corresponding integral will also be divergent.

4. Why is it important to understand the difference between divergent series sums and integrals?

Understanding the difference between divergent series sums and integrals is important in mathematics and science because it helps us determine whether a series or function is convergent or divergent. This information is crucial in many applications, such as calculating probabilities and predicting outcomes in various fields.

5. How can we determine whether a series is convergent or divergent?

There are several tests that can be used to determine the convergence or divergence of a series, such as the comparison test, ratio test, and integral test. These tests involve analyzing the behavior of the series' terms and comparing them to a known convergent or divergent series. Additionally, computer algorithms and software can also be used to calculate the sum or integral of a series and determine its convergence or divergence.

Similar threads

Replies
15
Views
2K
Replies
1
Views
2K
Replies
7
Views
1K
  • Topology and Analysis
Replies
0
Views
245
Replies
3
Views
821
Replies
4
Views
509
Replies
1
Views
1K
Replies
6
Views
2K
Replies
15
Views
2K
Back
Top