Skip to main content

All Questions

Filter by
Sorted by
Tagged with
3 votes
2 answers
625 views

Continuation up to zero of a Dirichlet series with bounded coefficients

Let $a_n$ be a bounded sequence of positive real numbers. Is it the case that the Dirichlet series $\sum \frac{a_n}{n^s}$ can be meromorphically continued up to the right of zero, with at the most a ...
Anweshi's user avatar
  • 7,442
2 votes
1 answer
275 views

Binomial transform of Dirichlet series

Let $\Theta(s)$ be a Dirichlet series , and let $\beta$ be its abscissa of convergence: $$\Theta(s)=\sum_{n=1}^{\infty}\frac{\theta(n)}{n^{s}}\;\;\;\;\;\;\Re(s)>\beta$$ And let $\left\{a_{n}\right\}...
mohammad-83's user avatar
0 votes
1 answer
290 views

Analytic continuation of Euler product $\prod_{p} (1 - e^{-2 \pi i p \alpha}p^{-s})^{-1}$

Is anything useful known about the function defined by \[ f(s, \alpha) = \prod_{p} (1 - e^{-2 \pi i p \alpha}p^{-s})^{-1} \quad ? \] Here, $\alpha$ is real. When $\alpha = 1$, this is certainly the ...
Mr. SnowRemover's user avatar
0 votes
2 answers
392 views

What is the relationship between the abscissa of holomorphy and abscissa of convergence of a Dirichlet series

Given a Dirichlet series $$\phi(s)=\sum_{n\ge1}\frac{a_n}{n^s}$$ let $\sigma_{\text{conv}}\in\bar{\mathbb{R}}$ its abscissa of convergence, then we know that $\phi(s)$ is holomorphic on the half-plan $...
Adam's user avatar
  • 93