# Thread: Riemann Zeta function

1. ## Riemann Zeta function

Hey, so following problem:

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

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

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

Who can help?

2. Originally Posted by EinStone
Hey, so following problem:

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

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

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

Who can help?

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

Now perhaps you're on to something else, since $(1-2^{1-s})\zeta(s)$ 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

3. Originally Posted by EinStone
Hey, so following problem:

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

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

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

Who can help?
i have a good reason to believe that $a_n=(-1)^{n+1}$ also works but i need some time to prove it. EDIT: it's actually very easy to prove!

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

Now perhaps you're on to something else, since $(1-2^{1-s})\zeta(s)$ 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, $a_n$ should not depend on s or anything else, otherwise its trivial ^^.

Originally Posted by NonCommAlg
i have a good reason to believe that $a_n=(-1)^{n+1}$ also works but i need some time to prove it. EDIT: it's actually very easy to prove!
I also found that result online, but I don't see how to prove this , can you help?

5. $\sum_{n \geq 1} \frac{(-1)^{n+1}}{n^s}=\sum_{n \geq 1} \frac{1}{(2n-1)^s} - \sum_{n \geq 1} \frac{1}{(2n)^s}=\sum_{n \geq 1} \frac{1}{n^s} - \sum_{n \geq 1} \frac{1}{(2n)^s} - \sum_{n \geq 1} \frac{1}{(2n)^s}$

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

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

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

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

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

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

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

Tonio

8. Ah too easy xD.

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

9. Originally Posted by EinStone
Ah too easy xD.

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

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

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