All Questions
6 questions
2
votes
0
answers
131
views
Limit of scaled infinite sum with Dirichlet characters modulo 4: is it zero?
I am trying to get an asymptotic formula such as
$$ L_4(s, n) \sim L_4(s) + \rho_n(s)\Lambda_n + \frac{\alpha(s)}{\sqrt{n}} + \frac{\beta(s)}{\sqrt{n\log n}}+\cdots$$
where $L_4(s, n)$ is the first $n$...
- 3,098
11
votes
3
answers
866
views
Conditions under which $\lim_{s\to1^+}\sum_{n=1}^{\infty}\frac{a_n}{n^s}=\sum_{n=1}^{\infty}\frac{a_n}{n}$
I was working with some Dirichlet series and I realized that I have never seen any general conditions under which
\begin{equation}
\sum_{n=1}^{\infty}\frac{a_n}{n}=\lim_{s\to1^+}\sum_{n=1}^{\infty}\...
- 2,902
1
vote
0
answers
129
views
Asymptotic of $\sum_{1\leq n\leq x}a_n$ where $\exp(\sum_{n=1}^\infty\alpha\operatorname{rad}(n)n^{-s})=\sum_{n=1}^\infty\frac{a_n}{n^s}$
Yesterday I tried to study the article [1] in wich were showed incredible expressions related to Dirichlet series. In the same way I wondered about next question.
We denote for integers $m>1$ the ...
4
votes
1
answer
244
views
The function $\sum_{n=0}^\infty\frac{(-1)^n\mu(2n+1)}{(2n+1)^s}$: reference request or particular values at integers and abscissa of convergence
We denote for integers $m\geq 1$ the Möbius function as $\mu(m)$. With the help of a CAS, Wolfram Alpha online calculator, I was calculating certain values of $$\sum_{n=0}^\infty\frac{(-1)^n\mu(2n+1)}{...
12
votes
2
answers
382
views
Vanishing of certain periodic series: A question related to $L(1 , \chi) \neq 0$.
Fix $q$ to be a positive integer. Let $$f : \mathbb{N} \to \{-1 ,0, 1\}$$ be a $q$-periodic arithmetic function such that $$\sum_{n = 1}^q f(n) = 0.$$ If $f$ is not identically zero, is it true that $...
- 2,283
3
votes
1
answer
963
views
Abscissa of absolute convergence of the product of two Dirichlet series
I first asked the following question on Mathematics StackExchange (a few weeks ago), since the content of MathOverflow is mostly above my academic level. I didn't want to bother people on this forum ...
