# Thread: A series for Zeta^2

1. ## A series for Zeta^2

I have "derived" (I am probably not the 1st one to do so) the following equation for the norm of Zeta^2:

$\displaystyle \| \zeta \left( a, ib \right) \| ^2=\zeta \left( 2a, 0 \right) \left( 1 + \sum^{\infty}_{j=2} \frac{2^{\omega \left(j\right) } \cos(n_1 b \ln(p_1)) \ldots \cos(n_k b \ln(p_k))}{j^a} \right)$

Where each term j has prime factorization $\displaystyle j={p_1}^{n_1} {p_2}^{n_2}...$, and $\displaystyle \omega \left(j\right)$ is the number of different prime factors. For $\displaystyle a=Re\left( s \right) > 1$, the right-hand sum of converges absolutely.

I am trying to figure out how to show if the series diverges/converges or semi-converges for 0.5< a < 1. (Obviously, if the series did converge in that range that would prove the RH so it is a bit unlikely to do so...but I do not know how to approach the proof.)

The only reference to that series I could find is a much simpler form, in the special case of b=0 where it reduces to the known series (e.g. displayed in Wolfram, and in Hardy and Wright 5th edition):

$\displaystyle \frac { \| \zeta \left( a \right) \| ^2 } { \zeta \left( 2a \right) } = \sum^{\infty}_{j=1} \frac{2^{\omega \left(j\right) }} {j^a}$

Has anyone seen the above series and where? Thank you!

2. Originally Posted by jlb
I have "derived" (I am probably not the 1st one to do so) the following equation for the norm of Zeta^2:

$\displaystyle \| \zeta \left( a, ib \right) \| ^2=\zeta \left( 2a, 0 \right) \left( 1 + \sum^{\infty}_{j=2} \frac{2^{\omega \left(j\right) } \cos(n_1 b \ln(p_1)) \ldots \cos(n_k b \ln(p_k))}{j^a} \right)$

Where each term j has prime factorization $\displaystyle j={p_1}^{n_1} {p_2}^{n_2}...$, and $\displaystyle \omega \left(j\right)$ is the number of different prime factors. For $\displaystyle a=Re\left( s \right) > 1$, the right-hand sum of converges absolutely.

I am trying to figure out how to show if the series diverges/converges or semi-converges for 0.5< a < 1. (Obviously, if the series did converge in that range that would prove the RH so it is a bit unlikely to do so...but I do not know how to approach the proof.)

The only reference to that series I could find is a much simpler form, in the special case of b=0 where it reduces to the known series (e.g. displayed in Wolfram, and in Hardy and Wright 5th edition):

$\displaystyle \frac { \| \zeta \left( a \right) \| ^2 } { \zeta \left( 2a \right) } = \sum^{\infty}_{j=1} \frac{2^{\omega \left(j\right) }} {j^a}$

Has anyone seen the above series and where? Thank you!

In an analytic number theory course I've taken, there was a homework problem that asked us to show $\displaystyle \frac { \zeta^2 \left( a \right) } { \zeta \left( 2a \right) } = \sum^{\infty}_{j=1} \frac{2^{\omega \left(j\right) }} {j^a}$, for $\displaystyle \Re(a)>1$.

Proof:

$\displaystyle \sum^{\infty}_{j=1} \frac{2^{\omega \left(j\right) }} {j^a} = \prod_{p} \sum_{k=0}^\infty \frac{2^{\omega(p^k)}}{p^{ak}}$ since $\displaystyle \frac{2^{\omega \left(j\right) }} {j^a}$ is multiplicative.

$\displaystyle = \prod_{p} \left(1+2\sum_{k=1}^\infty \frac{1}{p^{ak}}\right) = \prod_{p} \left(1+\frac{2p^{-a}}{1-p^{-a}}\right)$

$\displaystyle = \prod_{p} \frac{1+p^{-a}}{1-p^{-a}} = \prod_{p} \frac{1-p^{-2a}}{(1-p^{-a})^2}$
$\displaystyle = \frac{\zeta^2(a)}{\zeta(2a)}$, $\displaystyle \Re(a)>1$

With that said I'm wondering how you got $\displaystyle \| \zeta \left( a \right) \| ^2$ instead of $\displaystyle \zeta(a)^2$ like I did...

3. Originally Posted by chiph588@
In an analytic number theory course I've taken, there was a homework problem that asked us to show $\displaystyle \frac { \zeta^2 \left( a \right) } { \zeta \left( 2a \right) } = \sum^{\infty}_{j=1} \frac{2^{\omega \left(j\right) }} {j^a}$, for $\displaystyle \Re{a}>1$.

Proof:

$\displaystyle \sum^{\infty}_{j=1} \frac{2^{\omega \left(j\right) }} {j^a} = \prod_{p} \sum_{k=0}^\infty \frac{2^{\omega(p^k)}}{p^{ak}}$ since $\displaystyle \frac{2^{\omega \left(j\right) }} {j^a}$ is multiplicative.

$\displaystyle = \prod_{p} \left(1+2\sum_{k=1}^\infty \frac{1}{p^{ak}}\right) = \prod_{p} \left(1+\frac{2p^{-a}}{1-p^{-a}}\right)$

$\displaystyle = \prod_{p} \frac{1+p^{-a}}{1-p^{-a}} = \prod_{p} \frac{1-p^{-2a}}{(1-p^{-a})^2}$
$\displaystyle = \frac{\zeta^2(a)}{\zeta(2a)}$, $\displaystyle \Re{a}>1$

With that said I'm wondering how you got $\displaystyle \| \zeta \left( a \right) \| ^2$ instead of $\displaystyle \zeta(a)^2$ like I did...
To get the norm of Zeta(s), s being complex and s= a+ ib,

I derived the following identity:

$\displaystyle \| \frac {1}{1-1/p^{s}} \|^2 = \left( \frac {1}{1-1/p^{2a}} \right) \left(1 + 2 \sum^{\infty}_{j=1} \frac {\cos \left( jb \ln {p} \right)}{p^{ja}} \right) \tag{1}$

where the right-hand sum is absolutely convergent for $\displaystyle Re\left( s \right) > 0$. Note that equation (1) is not the Fourier expansion of the left-hand side, but a real function of 2 independent variables.

I then take equation (1) and use it in an Euler product to get $\displaystyle \| \zeta \left( s \right) \| ^2$.