# Riemann Zeta function

• Feb 23rd 2010, 03:31 AM
EinStone
Riemann Zeta function
Hey, so following problem:

I want to express $\displaystyle (1-2^{1-s})*\sum_{n=1}^{\infty}\frac{1}{n^s}$

as a Dirichlet series, i.e. as a series of the form

$\displaystyle \sum_{n=1}^{\infty}\frac{a_n}{n^s}$ where $\displaystyle a_n$ is any complex sequence.

Who can help?
• Feb 23rd 2010, 04:02 AM
tonio
Quote:

Originally Posted by EinStone
Hey, so following problem:

I want to express $\displaystyle (1-2^{1-s})*\sum_{n=1}^{\infty}\frac{1}{n^s}$

as a Dirichlet series, i.e. as a series of the form

$\displaystyle \sum_{n=1}^{\infty}\frac{a_n}{n^s}$ where $\displaystyle a_n$ is any complex sequence.

Who can help?

I'm not sure I follow you, but what if you define $\displaystyle a_n=1-2^{1-s}\,\,\forall n\in\mathbb{N}$ ? This certainly will do the trick...(Cool)

Now perhaps you're on to something else, since $\displaystyle (1-2^{1-s})\zeta(s)$ is the first step to express the zeta function as an Euler product, so...well, so who knows what you're on to actually.

Tonio
• Feb 23rd 2010, 04:04 AM
NonCommAlg
Quote:

Originally Posted by EinStone
Hey, so following problem:

I want to express $\displaystyle (1-2^{1-s})*\sum_{n=1}^{\infty}\frac{1}{n^s}$

as a Dirichlet series, i.e. as a series of the form

$\displaystyle \sum_{n=1}^{\infty}\frac{a_n}{n^s}$ where $\displaystyle a_n$ is any complex sequence.

Who can help?

i have a good reason to believe that $\displaystyle a_n=(-1)^{n+1}$ also works but i need some time to prove it. (Nerd) EDIT: it's actually very easy to prove! (Rofl)
• Feb 23rd 2010, 05:39 AM
EinStone
Quote:

Originally Posted by tonio
I'm not sure I follow you, but what if you define $\displaystyle a_n=1-2^{1-s}\,\,\forall n\in\mathbb{N}$ ? This certainly will do the trick...(Cool)

Now perhaps you're on to something else, since $\displaystyle (1-2^{1-s})\zeta(s)$ is the first step to express the zeta function as an Euler product, so...well, so who knows what you're on to actually.

Tonio

I'm sorry I didnt clarify this, $\displaystyle a_n$ should not depend on s or anything else, otherwise its trivial ^^.

Quote:

Originally Posted by NonCommAlg
i have a good reason to believe that $\displaystyle a_n=(-1)^{n+1}$ also works but i need some time to prove it. (Nerd) EDIT: it's actually very easy to prove! (Rofl)

I also found that result online, but I don't see how to prove this :(, can you help?
• Feb 23rd 2010, 05:50 AM
NonCommAlg
$\displaystyle \sum_{n \geq 1} \frac{(-1)^{n+1}}{n^s}=\sum_{n \geq 1} \frac{1}{(2n-1)^s} - \sum_{n \geq 1} \frac{1}{(2n)^s}=\sum_{n \geq 1} \frac{1}{n^s} - \sum_{n \geq 1} \frac{1}{(2n)^s} - \sum_{n \geq 1} \frac{1}{(2n)^s}$

$\displaystyle =\sum_{n \geq 1} \frac{1}{n^s} -2\sum_{n \geq 1} \frac{1}{(2n)^s}= \sum_{n \geq 1} \frac{1}{n^s} -2^{1-s} \sum_{n \geq 1} \frac{1}{n^s}=(1-2^{1-s}) \sum_{n \geq 1} \frac{1}{n^s}.$
• Feb 23rd 2010, 06:03 AM
EinStone
Quote:

Originally Posted by NonCommAlg
$\displaystyle \sum_{n \geq 1} \frac{1}{(2n-1)^s} - \sum_{n \geq 1} \frac{1}{(2n)^s}=\sum_{n \geq 1} \frac{1}{n^s} - \sum_{n \geq 1} \frac{1}{(2n)^s} - \sum_{n \geq 1} \frac{1}{(2n)^s}$

I understand everything except the quoted one above, so why is

$\displaystyle \sum_{n \geq 1} \frac{1}{(2n-1)^s} = \sum_{n \geq 1} \frac{1}{n^s} - \sum_{n \geq 1} \frac{1}{(2n)^s}$ ?
• Feb 23rd 2010, 06:44 AM
tonio
Quote:

Originally Posted by EinStone
I understand everything except the quoted one above, so why is

$\displaystyle \sum_{n \geq 1} \frac{1}{(2n-1)^s} = \sum_{n \geq 1} \frac{1}{n^s} - \sum_{n \geq 1} \frac{1}{(2n)^s}$ ?

$\displaystyle \sum_{n \geq 1} \frac{1}{(2n-1)^s}=\frac{1}{1^s}+\frac{1}{3^s}+\frac{1}{5^s}+\l dots=\frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+\ld ots -\left(\frac{1}{2^s}+\frac{1}{4^s}+\ldots\right)$

Tonio
• Feb 23rd 2010, 02:09 PM
EinStone
Ah too easy xD.

So I have another really similar problem. Namely expressing
$\displaystyle (1-3^{1-s})*\sum_{n=1}^{\infty}\frac{1}{n^s}$ as a Dirichlet series.

I tried again to split up the n on three series for the modulo 3 residues, but it somehow didnt work. Do you have an idea?

• Feb 24th 2010, 08:30 AM
NonCommAlg
Quote:

Originally Posted by EinStone
Ah too easy xD.

So I have another really similar problem. Namely expressing
$\displaystyle (1-3^{1-s})*\sum_{n=1}^{\infty}\frac{1}{n^s}$ as a Dirichlet series.

I tried again to split up the n on three series for the modulo 3 residues, but it somehow didnt work. Do you have an idea?

$\displaystyle a_n=-2 \cos \left(\frac{2 \pi n}{3} \right)$ works.

in general, let $\displaystyle d \in \mathbb{N}$ and define $\displaystyle f : \mathbb{N} \longrightarrow \{0,1\}$ by $\displaystyle f(n)=\begin{cases}1 & \text{if} \ d \mid n \\ 0 & \text{otherwise} \end{cases}.$ let $\displaystyle a_n=1-df(n).$ then it's easy to prove that $\displaystyle \sum_{n \geq 1} \frac{a_n}{n^s}=(1-d^{1-s})\sum_{n \geq 1} \frac{1}{n^s}.$