# Dirichlet Series convergence

• Feb 18th 2010, 02:14 PM
EinStone
Dirichlet Series convergence
Hey guys, while studying dirichlet series I came along a theorem stating that every D.S. has a convergence abscissa (possibly +Infinity). Now there was a corollary stating that any Ordinary D.S. $\displaystyle \sum_{n=1}^{\infty} a_n/n^s$ has a absolute convergence abscissa, which doesn't differ by more than 1 from the normal abscissa, but there is no proof. Can anyone prove this?
• Feb 18th 2010, 02:25 PM
Drexel28
Quote:

Originally Posted by EinStone
Hey guys, while studying dirichlet series I came along a theorem stating that every D.S. has a convergence abscissa (possibly +Infinity). Now there was a corollary stating that any Ordinary D.S. $\displaystyle \sum_{n=1}^{\infty} a_n/n^s$ has a absolute convergence abscissa, which doesn't differ by more than 1 from the normal abscissa, but there is no proof. Can anyone prove this?

For those who aren't familiar with Dirichlet series but might be able to help, could you fill in some of the terminology.
• Feb 18th 2010, 07:34 PM
chisigma
According to...

Dirichlet series - Wikipedia, the free encyclopedia

... a D.S. of the form $\displaystyle \varphi (s)= \sum_{n=1}^{\infty} \frac{a_{n}}{n^{s}}$ converges in the following cases...

a) if $\displaystyle a_{n}$ is a bounded sequence of complex numbers for $\displaystyle Re(s)>1$ ...

b) if $\displaystyle a_{n} = \mathcal{O} (n^{k})$ for $\displaystyle Re(s)> k+1$...

c) if $\displaystyle \sum_{i=n}^{n+k} a_{i}$ is bounded forall $\displaystyle n$ and $\displaystyle k\ge 0$ for $\displaystyle Re(s)>0$...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Feb 19th 2010, 05:36 AM
EinStone
A Dirichlet series is just a series $\displaystyle \sum_{n=1}^{\infty} \dfrac{a_n}{n^s}$, where $\displaystyle a_n$ is any sequence of complex numbers, and $\displaystyle s\in \mathbb{C}$. We consider the convergence of the series depending on s.

I need help with the proof of a corollary, stating that if $\displaystyle \sum_{n=1}^{\infty} \dfrac{a_n}{n^{s_0}}$ converges for some $\displaystyle s_0 \in \mathbb{C}$, then it converges absolutely for any $\displaystyle s_1 \in \mathbb{C}$ with $\displaystyle Re(s_1) > Re(s_0)+1$
• Feb 19th 2010, 07:18 AM
Laurent
Quote:

Originally Posted by EinStone
A Dirichlet series is just a series $\displaystyle \sum_{n=1}^{\infty} \dfrac{a_n}{n^s}$, where $\displaystyle a_n$ is any sequence of complex numbers, and $\displaystyle s\in \mathbb{C}$. We consider the convergence of the series depending on s.

I need help with the proof of a corollary, stating that if $\displaystyle \sum_{n=1}^{\infty} \dfrac{a_n}{n^{s_0}}$ converges for some $\displaystyle s_0 \in \mathbb{C}$, then it converges absolutely for any $\displaystyle s_1 \in \mathbb{C}$ with $\displaystyle Re(s_1) > Re(s_0)+1$

It looks simple: since the series for $\displaystyle s_0$ converges, the general term $\displaystyle \frac{a_n}{n^{s_0}}$ tends to 0.

However, $\displaystyle \left|\frac{a_n}{n^s}\right|=\left|\frac{a_n}{n^{s _0}}\right|\frac{1}{|n^{s-s_0}|}$ and $\displaystyle |n^z|=|e^{z\log n}|=e^{{\rm Re}( z)\log n}=n^{{\rm Re} (z)}$. You have every element you need to conclude!
• Feb 19th 2010, 09:52 AM
EinStone
I see $\displaystyle n^{{\rm Re} (z)}$ converges to 0, for n -> Infinity, but the other term is still in absolute value, so how do you show convergence now, with Cauchy criterion?
• Feb 19th 2010, 01:13 PM
Laurent
Quote:

Originally Posted by EinStone
I see $\displaystyle n^{{\rm Re} (z)}$ converges to 0, for n -> Infinity, but the other term is still in absolute value, so how do you show convergence now, with Cauchy criterion?

Because of my first point, there is $\displaystyle C>0$ such that we have $\displaystyle \left|\frac{a_n}{n^{s_0}}\right|\leq C$ for all $\displaystyle n\geq 1$ (sequence converges to zero hence, in particular, is bounded). Then $\displaystyle \left|\frac{a_n}{n^s}\right|\leq \frac{C}{n^{{\rm Re}(s-s_0)}}$, and we know well that $\displaystyle \sum_n \frac{C}{n^{{\rm Re}(s-s_0)}}$ converges when $\displaystyle {\rm Re}(s-s_0)>1$ (Riemann series). This implies that $\displaystyle \sum_n \left|\frac{a_n}{n^s}\right|$ converges, which by definition is the same as saying that $\displaystyle \sum_n \frac{a_n}{n^s}$ converges absolutely.

I notice that you first mentioned normal convergence, so here it is (almost the same proof): let $\displaystyle s_1$ be such that $\displaystyle {\rm Re}(s_1)>{\rm Re}(s_0)+1$. We have to prove that the series converges normally on the set $\displaystyle \{z\in\mathbb{C}|{\rm Re}(z)>{\rm Re}(s_1)\}$. This is just because, for every $\displaystyle s$ is this set, we have $\displaystyle \left|\frac{a_n}{n^s}\right|\leq \frac{C}{n^{{\rm Re}(s_1-s_0)}}$ and $\displaystyle \sum_n \frac{C}{n^{{\rm Re}(s_1-s_0)}}<\infty$.
• Feb 19th 2010, 04:13 PM
EinStone
Thats it, thanks!