Help with capital sigma notation please.

[rustynail]

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?

 

[D H]

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]

 

[DonAntonio]

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?

 

[rustynail]

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] ??

 

[CellCoree]

Sure, your approach is correct. You can write the Riemann Zeta function as a sum of infinite sums, where each term in the outer sum corresponds to the coefficient of the inner sum. So, your notation of ζ(s) as a sum of a_k is valid. However, keep in mind that the Riemann Zeta function has a special notation using the capital Greek letter ζ, so you can also write it as:

ζ(s) = 1 + \sum_{n=1}^{\infty} \frac{n!}{(s\;ln\;n)^n}

or in your notation:

ζ(s) = 1 + \sum_{k=1}^{\infty} a_k

Either way, both notations convey the same meaning. Keep in mind that the Riemann Zeta function is a complex-valued function, so s \in \mathbb{C} means that s can take on any complex value. This allows for different values of s to produce different results for the Riemann Zeta function. I hope this helps clarify your notation problems.