All Questions
Tagged with divergent-series cv.complex-variables
14 questions
21
votes
6
answers
1k
views
What is the relationship between $\sum_{n=0}^\infty f(n) x^n$ and $-\sum_{n=1}^\infty f(-n) x^{-n}$?
Background
Taking a relatively arbitrary combination of exponential and polynomial terms, for instance
$$\sum_{n=0}^\infty \left(n^{2}\sin\left(n\right)+n\cos\left(3n-2\right)\right)\cos\left(5n+1\...
6
votes
1
answer
241
views
Fractional integrals and $\sum f(n) n^x$
Preamble
The following is a rather unrigorous way to obtain the Euler-Maclaurin formula. Consider some $\sum_{n=1}^\infty f(n)$. We may rewrite this as
$$\sum_{n=1}^\infty f(n)=\sum_{n=1}^\infty \sum_{...
25
votes
1
answer
2k
views
Can we just use the linear term of exponential sums to sum divergent series
Suppose you want to compute the sum $\sum_{n=0}^{\infty} a_n $
You could consider the expression $f(x) = \sum_{n=0}^{\infty} e^{a_n x}$ and try to compute the coefficient of an $x^1$ term in the ...
2
votes
2
answers
260
views
Laurent Series $\sum_{n=-1}^\infty a_n x^n$ when $a_{-1} = \infty$
When dealing with complex functions, if $f(x)$ has a simple pole, then we can find the coefficient $a_{-1}$ in the Laurent expansion $f(x) = \sum_{n=-1}^\infty a_n x^n$ by evaluating the limit $\lim_{...
7
votes
1
answer
337
views
If $f(x) = \sum_{n=0}^\infty a_n x^n$, then $\int_{-\infty}^\infty f(x^2) dx = \pi i a_{-\frac{1}{2}}$
I've noticed a curious relationship between the coefficient $a_n$ for a power series and the integral of the real line. For instance, take $f(x) = e^{-x} = \sum_{n=0}^\infty \frac{(-1)^n}{n!} x^n$. ...
6
votes
0
answers
171
views
Computing residues at $\infty$
As an initial note, let me show by example what I mean by the terminology 'residue at $\infty$' I use in the title. I assume there is some standard terminology for this stuff, so I'd appreciate it if ...
4
votes
0
answers
160
views
Correct way to extend a sequence defined on the naturals into the complex plane
Preamble
Sequences $a_n$ defined on the natural numbers are clearly not uniquely interpolated by only one function. In particular, given an interpolation $f(n) = a_n$, then $f(n) + \sin(2\pi n)$ is ...
7
votes
0
answers
306
views
Gottfried Helms' tetra-eta series
Here Gottfried Helms introduces the following fascinating divergent series
$$ T_2(x)=- \sum_{n=1}^\infty (-1)^n n^{n^x}$$
The terms don't go to zero, so technically the series does not converge ...
9
votes
0
answers
313
views
Switching the order of a summation and replacing a series by its analytical continuation
Background
A useful trick when trying to analyze a series $\sum_{n=0}^\infty f(n)$ is to expand $f(n)$ as some kind of series, swap the order of summation, and then evaluate the inner infinite sum. ...
0
votes
1
answer
200
views
A holomorphic function in the open unit disk satisfying certain properties
Does there exist a function which is holomorphic in $|z|<1,$ continuous in $|z|\leq1$ and such that the series $\sum |a_n|$ is divergent, where $a_n$'s coefficients in the Taylor series expansion ...
4
votes
1
answer
401
views
How to correctly renormalize this function at the pole $x=1$? Evaluating: $\sum_{n=1}^{\infty} e^{e^n}$
So I was considering the divergent everywhere but 0 power series
$$ f(x) = \sum_{n=0}^{\infty} e^{e^n} x^n $$
Now one can do the following "questionable" manipulation
$$ f(x) = \sum_{n=0}^{\...
3
votes
2
answers
388
views
Theta-function in the lower half-plane
Standard theta function
$$\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 ...
5
votes
3
answers
343
views
Evaluating the series $\sum_{n=0}^{\infty} n! x^n$ and inverse variable-fractional-derivatives
So I was interested in formally assigning values to the completely divergent series $G(x) = \sum_{n=0}^{\infty} n!x^n $. I guess the question COULD end here if you already have an idea of how to ...
4
votes
2
answers
420
views
Borel summation and the Abel function of $e^z-1$
This is a question that has bothered myself and Gottfried Helms a fair amount of late. He has made his case for the following result, but a proof escapes both of us. The question is deceptively simple,...