- #1
rustynail
- 53
- 0
I was playing a bit with the Riemann Zeta function, and have been struggling with some notation problems.
The function is defined as follows
[tex] \zeta (s) = \sum_{n=1}^{\infty} \frac{1}{n^s} [/tex]
where [tex] s \in \mathbb{C} [/tex]
we know that
[tex] n^s = exp(s\;ln\;n)[/tex]
so I can write
[tex] \zeta (s) = \sum_{n=1}^{\infty} \frac{1}{exp(s\;ln\;n)} [/tex]
but since
[tex] \frac{1}{exp(s\;ln\;n)} = 1 + \frac{1!}{(s\;ln\;n)} + \frac{2!}{(s\;ln\;n)^2} + ... =
1 + \sum_{n=1}^{\infty} \frac{n!}{(s\;ln\;n)^n} [/tex]
how can I write this ''sum within a sum''? ζ(s) here, if I am correct, would be an infinite sum of terms which are infinite sums.
Thank you for taking the time to help!edit :
Could I say
[tex]1 + \sum_{n=1}^{\infty} \frac{n!}{(s\;ln\;n)^n} = a_k[/tex]
then
[tex] \zeta (s) = \sum_{k=1}^{\infty} a_k[/tex]
Does that make any sense?
The function is defined as follows
[tex] \zeta (s) = \sum_{n=1}^{\infty} \frac{1}{n^s} [/tex]
where [tex] s \in \mathbb{C} [/tex]
we know that
[tex] n^s = exp(s\;ln\;n)[/tex]
so I can write
[tex] \zeta (s) = \sum_{n=1}^{\infty} \frac{1}{exp(s\;ln\;n)} [/tex]
but since
[tex] \frac{1}{exp(s\;ln\;n)} = 1 + \frac{1!}{(s\;ln\;n)} + \frac{2!}{(s\;ln\;n)^2} + ... =
1 + \sum_{n=1}^{\infty} \frac{n!}{(s\;ln\;n)^n} [/tex]
how can I write this ''sum within a sum''? ζ(s) here, if I am correct, would be an infinite sum of terms which are infinite sums.
Thank you for taking the time to help!edit :
Could I say
[tex]1 + \sum_{n=1}^{\infty} \frac{n!}{(s\;ln\;n)^n} = a_k[/tex]
then
[tex] \zeta (s) = \sum_{k=1}^{\infty} a_k[/tex]
Does that make any sense?