# zeta function

• Feb 27th 2010, 12:50 PM
bigdoggy
zeta function
why is this correct?
If $\zeta(s)= \Sigma ^\infty _{n=1} {1 \over {n^s}}$ then $\zeta(s)= \Sigma ^\infty _{N=1} {\tau(N) \over {N^s}}$ where $\tau(N)$ the number of divisors of N...

Also
$\zeta(s)\zeta(s-1) = \Sigma {\sigma(N) \over N^s}$ where $\sigma(N)$ is the sum of the divisors of N....
• Feb 27th 2010, 02:56 PM
NonCommAlg
Quote:

Originally Posted by bigdoggy
why is this correct?
If $\zeta(s)= \Sigma ^\infty _{n=1} {1 \over {n^s}}$ then $\zeta(s)= \Sigma ^\infty _{N=1} {\tau(N) \over {N^s}}$ where $\tau(N)$ the number of divisors of N...

Also
$\zeta(s)\zeta(s-1) = \Sigma {\sigma(N) \over N^s}$ where $\sigma(N)$ is the sum of the divisors of N....

the LHS of the first identity is wrong. it should be $\zeta^2(s).$ (Nod)
• Feb 27th 2010, 03:11 PM
NonCommAlg
ok, for the integers $k \geq 0, \ n \geq 1,$ let $f_k(n)=\sum_{d \mid n} d^k.$ then, assuming that $\zeta(s-k)$ is defined we have: $\sum_{n=1}^{\infty} \frac{f_k(n)}{n^s}=\sum_{n=1}^{\infty} \sum_{d \mid n} \frac{d^k}{n^s}=\sum_{d=1}^{\infty} d^k \sum_{m=1}^{\infty} \frac{1}{(md)^s}=\sum_{d=1}^{\infty} \frac{1}{d^{s-k}} \sum_{m=1}^{\infty} \frac{1}{m^s}=\zeta(s-k) \zeta(s).$
now use the fact that $f_0(n)=\tau(n)$ and $f_1(n)=\sigma(n).$