Help with capital sigma notation please.

rustynail
Messages
53
Reaction score
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} + ... = <br /> 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?
 
Mathematics news on Phys.org
rustynail said:
but since

[tex]\frac{1}{exp(s\;ln\;n)} = 1 + \frac{1!}{(s\;ln\;n)} + \frac{2!}{(s\;ln\;n)^2} + ... = <br /> 1 + \sum_{n=1}^{\infty} \frac{n!}{(s\;ln\;n)^n}[/tex]
No.

What you can say is that
[tex]\exp (s \ln n) = \sum_{r=0}^{\infty} \frac{(s\ln n)^r}{r!}[/tex]
What you did, simply inverting each term on the right hand side to get [itex]1/\exp(s\ln n)[/itex], is invalid.

This is what you need to use for [itex]1/\exp(s\ln n)[/itex]:
[tex]\frac 1{\exp(s\ln n)} = \exp (-s \ln n) =<br /> \sum_{r=0}^{\infty} \frac{(-s\ln n)^r}{r!} =<br /> \sum_{r=0}^{\infty} \frac{(-1)^r(s\ln n)^r}{r!}[/tex]
 
rustynail said:
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]


This only makes sense for [itex]\,Re(s)>1\,[/itex]...careful!

DonAntonio


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} + ... = <br /> 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?
 
Thank you DH for the help, and DonAntonio for your rigor, I need to work on that.

But in this part,

[tex]\sum_{r=0}^{\infty} \frac{(-1)^r(s\ln n)^r}{r!}[/tex]

Don't we need to to take the sum of all terms with both n and r, from 1 to ∞? Could I write it as

[tex]\sum_{n=1}^{\infty} \; \sum_{r=0}^{\infty} \frac{(-1)^r(s\ln n)^r}{r!}[/tex] ??
 
Last edited:

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
Replies
1
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 0 ·
Replies
0
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
4K