## Saturday, November 24, 2012

### Dirichlet Series of Divisor Function

DF.ID.1: Dirichlet series of first and second order  divisor function:

$\sum _{n=1}^{\infty } \frac{d(n)}{n^{s}} = \zeta ^{2}(s)$

$\sum _{n=1}^{\infty } \frac{d^{2}(n)}{n^{s}} = \frac{\zeta ^{4}(s)}{\zeta (2s)}$

proof
Euler product representation of Dirichlet series of divisor function:

$\sum _{n=1}^{\infty } \frac{d(n)}{n^{s}} = \prod _{p}(1+\frac{d(p)}{p^{s}} + \frac{d(p^{2})}{p^{2s}}+\cdot \cdot \cdot +\frac{d(p^{k})}{p^{ks}}+\cdot \cdot \cdot )$
(1.a)

$\sum _{n=1}^{\infty } \frac{d^{2}(n)}{n^{s}} = \prod _{p}(1+\frac{d^{2}(p)}{p^{s}} + \frac{d^{2}(p^{2})}{p^{2s}}+\cdot \cdot \cdot +\frac{d^{2}(p^{k})}{p^{ks}}+\cdot \cdot \cdot )$
(1.b)
(2.a)
$d^{2}(p^{k})= (k+1)^{2}$

(2.b)

according to
$n=p_{1}^{k_{1}}p_{2}^{k_{2}}p_{3}^{k_{3}}\cdot \cdot \cdot p_{r}^{k_{r}}$
from (1.a) and (2.a):

$\sum _{n=1}^{\infty } \frac{d(n)}{n^{s}} = \prod _{p}(1+\frac{2}{p^{s}} + \frac{3}{p^{2s}}+\cdot \cdot \cdot +\frac{k+1}{p^{ks}}+\cdot \cdot \cdot )$
(3.a)
from (1.b) and (2.b):

$\sum _{n=1}^{\infty } \frac{d^{2}(n)}{n^{s}} = \prod _{p}(1+\frac{4}{p^{s}} + \frac{9}{p^{2s}}+\cdot \cdot \cdot +\frac{(k+1)^{2}}{p^{ks}}+\cdot \cdot \cdot )$

(3.b)
The derivative of geometric series (GS.ID.1):

$1+2x+3x^{2}+4x^{3}+\cdot \cdot \cdot = \frac{1}{(1-x)^{2}}$
(4)
$1+4x+9x^{2}+16x^{3}+\cdot \cdot \cdot = \frac{(1-x)^{2}+ 2x(1-x))}{(1-x)^{4}} = \frac{1+x}{(1-x)^{3}}$
(5.a)
$\frac{1+x}{(1-x)^{3}} = \frac{1-x^{2}}{(1-x)^{4}}$

$1+4x+9x^{2}+16x^{3}+\cdot \cdot \cdot = \frac{1-x^{2}}{(1-x)^{4}}$
(5.b)
from ( 3.a) and (4) , the Dirichlet series of first order divisor function:

$\sum _{n=1}^{\infty } \frac{d(n)}{n^{s}} = \prod _{p}\frac{1}{(1-p^{-s})^{2}} =\left ( \prod _{p}\frac{1}{(1-p^{-s})} \right ) ^{2} =\zeta (s)^{2}$
$\sum _{n=1}^{\infty } \frac{d(n)}{n^{s}} = \zeta ^{2}(s)$

(6)
from (3.b) and (5.b) , the Dirichlet series of second order divisor function:

$\sum _{n=1}^{\infty } \frac{d^{2}(n)}{n^{s}} = \prod _{p}\frac{(1-p^{-2s})}{(1-p^{-s})^{4}} =\frac{\zeta (s)^{4}}{\zeta (2s)}$
$\sum _{n=1}^{\infty } \frac{d^{2}(n)}{n^{s}} = \frac{\zeta ^{4}(s)}{\zeta (2s)}$
(7)

DF.ID.2: Dirichlet series of first  order  divisor function: of squared number

$\sum _{n=1}^{\infty } \frac{d(n^{2})}{n^{s}} = \frac{\zeta ^{3}(s)}{\zeta (2s)}$

proof
$\sum _{n=1}^{\infty } \frac{d(n^{2})}{n^{s}} = \prod _{p}(1+\frac{d(p^{2})}{p^{s}} + \frac{d(p^{4})}{p^{2s}}+\cdot \cdot \cdot +\frac{d(p^{2k})}{p^{ks}}+\cdot \cdot \cdot )$
(8)
$d(p^{2k})= 2k+1$
(8.b)
$\sum _{n=1}^{\infty } \frac{d(n^{2})}{n^{s}} = \prod _{p}(1+\frac{3}{p^{s}} + \frac{5}{p^{2s}}+\cdot \cdot \cdot +\frac{2k+1}{p^{ks}}+\cdot \cdot \cdot )$
(8.c)
given that :
$x+x^{3}+x^{5}+x^{7}+\cdot \cdot \cdot+ x^{2k+1} = \frac{x}{1-x^{2}}$
(9)
The derivative of previous series:
$x+3x^{2}+5x^{4}+7x^{6}+\cdot \cdot \cdot+ 2k+1x^{2k} = \frac{1-x^{4}}{(1-x^{2})^{3}}$
(10)
$x^{2} = p^{-s}$
(11)
from (8.c)(10) and (11),
$\sum _{n=1}^{\infty } \frac{d(n^{2})}{n^{s}} = \prod _{p}\frac{1-p^{-2s}}{(1-p^{-s})^{3}}$
(12)
$\sum _{n=1}^{\infty } \frac{d(n^{2})}{n^{s}} = \frac{\zeta ^{3}(s)}{\zeta (2s)}$
(13)

DF.ID.3: Dirichlet series of  sum of higher order  divisor function:

$\frac{\zeta (s)\zeta (s-a)\zeta (s-b)\zeta (s-a-b)}{\zeta (2s-a-b)}= \sum _{n=1}^{\infty } \frac{\sigma _{a}(s)\sigma _{b}(s)}{n^{s}}$

Proof
$\zeta (s)= \prod _{p \: \: \: prime} \frac{1}{1-p^{-s}}= \prod _{p \: \: \: prime}\left ( 1+\frac{1}{p^{s}} +\frac{1}{p^{2s}}+... \right )$

(1)

$\zeta (s-a)= \prod _{p \: \: \: prime} \frac{1}{1-p^{a-s}}= \prod _{p \: \: \: prime}\left ( 1+\frac{p^{a}}{p^{s}} +\frac{p^{2a}}{p^{2s}}+... \right )$

(2)

$\zeta (s)\zeta (s-a)= \prod _{p \: \: \: prime}\left ( 1+\frac{1}{p^{s}} +\frac{1}{p^{2s}}+... \right )\left ( 1+\frac{p^{a}}{p^{s}} +\frac{p^{2a}}{p^{2s}}+... \right )$
(3)

$\zeta (s)\zeta (s-a)= \prod _{p \: \: \: prime}\left ( 1+\frac{1+p^{a}}{p^{s}} +\frac{1+p^{a}+p^{2a}}{p^{2s}}+... \right )$

(4)

$\zeta (s)\zeta (s-a)= \prod _{p \: \: \: prime}\left ( 1+\frac{1-p^{2a}}{1-p^{a}}\frac{1}{p^{s}} +... \right )$

(5)

$\frac{\zeta (s)\zeta (s-a)\zeta (s-b)\zeta (s-a-b)}{\zeta (2s-a-b)}= \prod _{p \: \: \: prime} \frac{1-p^{-2s+a+b}}{(1-p^{-s})(1-p^{-s+a})(1-p^{-s+b})(1-p^{-s+a+b})}$

(6)

$p^{-s}=z \: \: ;\: \: p^{-2s}=z^{2}$

(7)
replacing (7) in (6) gives:
$\frac{1-p^{-2s+a+b}}{(1-p^{-s})(1-p^{-s+a})(1-p^{-s+b})(1-p^{-s+a+b})} = \frac{1-p^{a+b}z^{2} }{(1-z)(1-p^{a}z)(1-p^{b}z)(1-p^{a+b}z)}$
(8)

$\frac{1-p^{a+b}z^{2}}{(1-z)(1-p^{a}z)(1-p^{b}z)(1-p^{a+b}z)}= \frac{1}{(1-p^{a})(1-p^{b})}\left \{ \frac{1}{1-z}-\frac{p^{a}}{1-p^{a}z} -\frac{p^{b}}{1-p^{b}z} +\frac{p^{a+b}}{1-p^{a+b}z}\right \}$
(9)

$\frac{1}{(1-p^{a})(1-p^{b})}\left \{ \frac{1}{1-z}-\frac{p^{a}}{1-p^{a}z} -\frac{p^{b}}{1-p^{b}z} +\frac{p^{a+b}}{1-p^{a+b}z}\right \} = \frac{1}{(1-p^{a})(1-p^{b})} \sum _{m=0}^{\infty }\left \{ 1- p^{(m+1)a} - p^{(m+1)b}+ p^{(m+1)(a+b)}\right \} z^{m}$
(10)

$\frac{1}{(1-p^{a})(1-p^{b})} \sum _{m=0}^{\infty }\left \{ 1- p^{(m+1)a} - p^{(m+1)b}+ p^{(m+1)(a+b)}\right \} z^{m} = \frac{1}{(1-p^{a})(1-p^{b})} \sum _{m=0}^{\infty }\left \{ 1- p^{(m+1)a} \right \}\left \{ 1- p^{(m+1)b} \right \} z^{m}$
(11)

$\frac{\zeta (s)\zeta (s-a)\zeta (s-b)\zeta (s-a-b)}{\zeta (2s-a-b)}=\prod _{p \: \: \: prime} \sum _{m=0}^{\infty }\left \{ \frac{1- p^{(m+1)a}}{(1-p^{a})}\cdot \frac{1- p^{(m+1)b}}{(1-p^{b})}. \frac{1}{p^{ms}} \right \}$

(12)

$\frac{\zeta (s)\zeta (s-a)\zeta (s-b)\zeta (s-a-b)}{\zeta (2s-a-b)}= \sum _{n=1}^{\infty } \frac{\sigma _{a}(s)\sigma _{b}(s)}{n^{s}}$
(13)