Help with capital sigma notation please.

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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?
 

Answers and Replies

  • #2
D H
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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]
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) =
\sum_{r=0}^{\infty} \frac{(-s\ln n)^r}{r!} =
\sum_{r=0}^{\infty} \frac{(-1)^r(s\ln n)^r}{r!}[/tex]
 
  • #3
606
1
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} + ... =
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?
 
  • #4
53
0
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:

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