All Questions
Tagged with divergent-series regularization
20 questions
2
votes
1
answer
254
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New (?) Regularization Method for Divergent Series [closed]
Playing with identities ($1$) and ($2$) from this blog post and infinite geometric series, I've noticed the following.
For $x > 1$, the following series is convergent:
$$\sum_{n=0}^{\infty} e^{(2n ...
37
votes
2
answers
5k
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1+2+3+4+… and −⅛
Is there some deeper meaning to the following derivation (or rather one-parameter family of derivations) associating the divergent series $1+2+3+4+…$ with the value $-\frac 1 8$ (as opposed to the ...
- 19.7k
3
votes
0
answers
171
views
The divergent sum $\sum_{n=1}^\infty (-1)^n (n^2)! x^n$
Question
I'm interested in assigning a value to the divergent series $F(x)=\sum_{n=1}^\infty (-1)^n (n^2)! x^n$. I'm hoping that (1) the definition for $F(x)$ has (one-sided) derivatives of $(-1)^n (n^...
- 1,730
2
votes
2
answers
249
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Assigning values to divergent oscillating integrals
I have recently run into a number of divergent oscillating integrals in various contexts. Thus, I have been led to desire general methods for assigning values to divergent oscillating integrals. All ...
- 1,730
6
votes
0
answers
312
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Is there any intuition of why the both, regularized logarithm of zero is $-\gamma$ and the regularized logarithm of Bernoulli umbra is $-\gamma$?
If we take the MacLaurin series for $\ln(x+1)$ and evaluate it at $x=-1$, we will get the Harmonic series with the opposite sign: $-\sum_{k=1}^\infty \frac1x$. Since the regularized sum of the ...
- 10.1k
1
vote
2
answers
319
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What's the true regularized value of product of all natural numbers?
Muñoz Garcia and Pérez-Marco - The product over all primes is $4\pi^2$ claims that the regularized value of product $\prod_{k=1}^\infty k$ is $\sqrt{2\pi}$ and of $\prod_{k=1}^\infty p_k$ over primes $...
- 10.1k
3
votes
0
answers
276
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Evaluating $\sum_{n=0}^\infty n^k n!$ in p-adics, and its connection to the summation of divergent series
Often, in the discussion of the regularization of the geometric series it is mentioned that $\sum_{n=0}^\infty p^n$ converges in the p-adics, and indeed, that it converges to $\frac{1}{1-p}$. I had ...
- 1,730
3
votes
2
answers
462
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A proposition for summing divergent series, but how should partial summation be defined at non-natural values?
Introduction
I have been in search of methods of assinging values to divergent series that have a nice intuitive or geometric interpretation. One fairly straightforward method I've considered for ...
- 1,730
12
votes
1
answer
1k
views
Divergent series summation beyond natural boundaries
I'm hoping to investigate the effects of divergent summation methods on series which cannot be analytically continued due to a dense set of singularities. At least a priori, it doesn't seem that a ...
- 1,730
10
votes
2
answers
2k
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Value of divergent sum $\sum_{n=0}^\infty (-1)^n n^n$
I'm hoping to find a reasonable value to assign to the divergent series $\sum_{n=0}^\infty (-1)^n n^n$ and $\sum_{n=0}^\infty (-1)^n (xn)^n$. For the first one, I have obtained something around 0.71, ...
- 1,730
1
vote
0
answers
106
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What intuitive meaning "determinant" of a divergency (divergent integral, series, germ, pole or a singularity) can have?
I am working on the algebra of "divergencies", that is, infinite integrals, series, and germs.
So, I decided to construct something similar to the modulus or determinant of a matrix of these ...
- 10.1k
2
votes
0
answers
232
views
Did anyone ever propose the distinction between "divergent to infinity" as opposed to "divergent but with finite average"?
There are different regularization methods that allow us to ascribe finite values to divergent integrals, series or sequences.
Still, in my view there is fundamental difference between divergent ...
- 10.1k
3
votes
2
answers
388
views
Theta-function in the lower half-plane
Standard theta function
$$\vartheta(q)=\sum_{n=-\infty}^\infty q^{n^2} \qquad\qquad(1)$$
has a natural boundary of analyticity at $|q|=1$. This means that it can not be used to regularize expressions ...
- 319
7
votes
3
answers
679
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Is regularization of infinite sums by analytic continuation unique?
There are ill-posed summations that we can assign values to, take for concreteness,
$$ S = \sum_{k=0}^\infty k $$
to which we can assign $-1/12$ by several methods. Is there a fundamental and rigorous ...
- 1,324
7
votes
2
answers
976
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Regularizing the sum of all primes
In the spirit of a similar question for the harmonic series, is there a way to regularize the (divergent) sum of all primes?
$$ \sum_{p \text{ prime}} p $$
Neither of these questions obtained a ...
- 2,203
3
votes
0
answers
358
views
New/useful method for summation of divergent series?
Questions
$$ S(n,x) = x+e^x + e^{e^x} + e^{e^{e^x}} + \dots \text{$n$ times}$$
Also obeys (see background for argument):
$$ \frac{1}{2 \pi i} \oint e^{S(k,x)} \frac{\partial \ln(\frac{\int_0^\...
- 237
1
vote
1
answer
218
views
Why we cannot speak about the main or natural regularization?
Often when asking about a regularized value of an integral or series, I encounter a negative reaction of the sorts that "regularization is what you define it".
But in practice if we consider some ...
- 10.1k
2
votes
0
answers
109
views
What is the generalized sum of the following series? $\sum _{x=1}^{\infty } \sqrt{s^2 x^2-1}$ [closed]
I tried Mathematica, various regularization methods, including Borel, with no result.
On Math.SE the question was attacked with claims that divergent series cannot have a sum, so I decided to ask at ...
- 10.1k
1
vote
0
answers
280
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Regularization on divergent series [closed]
I want to compute $$ \sum ^{\infty }_{n=1}\dfrac {\left( 2n+k-2\right) !\zeta \left( 2n\right) \left( -1\right) ^{n-1}}{\left( 2\pi \right) ^{2n}}. $$
This series is surely not convergent for any ...
- 11
9
votes
1
answer
1k
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What are some geometric / physical / probabilistic interpretations of the Riemann zeta function at integer arguments n ≤ 1?
Introduction: This is slightly edited and generalised version of a question I asked on the Physics Stack Exchange website. This question has a twin brother asked here on MO, only now we consider ...
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