Help with capital sigma notation please.

I was playing a bit with the Riemann Zeta function, and have been struggling with some notation problems.

The function is defined as follows

$$\zeta (s) = \sum_{n=1}^{\infty} \frac{1}{n^s}$$

where $$s \in \mathbb{C}$$

we know that

$$n^s = exp(s\;ln\;n)$$

so I can write

$$\zeta (s) = \sum_{n=1}^{\infty} \frac{1}{exp(s\;ln\;n)}$$

but since

$$\frac{1}{exp(s\;ln\;n)} = 1 + \frac{1!}{(s\;ln\;n)} + \frac{2!}{(s\;ln\;n)^2} + ... = 1 + \sum_{n=1}^{\infty} \frac{n!}{(s\;ln\;n)^n}$$

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

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

then

$$\zeta (s) = \sum_{k=1}^{\infty} a_k$$

Does that make any sense?

D H
Staff Emeritus
but since

$$\frac{1}{exp(s\;ln\;n)} = 1 + \frac{1!}{(s\;ln\;n)} + \frac{2!}{(s\;ln\;n)^2} + ... = 1 + \sum_{n=1}^{\infty} \frac{n!}{(s\;ln\;n)^n}$$
No.

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

This is what you need to use for $1/\exp(s\ln n)$:
$$\frac 1{\exp(s\ln n)} = \exp (-s \ln n) = \sum_{r=0}^{\infty} \frac{(-s\ln n)^r}{r!} = \sum_{r=0}^{\infty} \frac{(-1)^r(s\ln n)^r}{r!}$$

I was playing a bit with the Riemann Zeta function, and have been struggling with some notation problems.

The function is defined as follows

$$\zeta (s) = \sum_{n=1}^{\infty} \frac{1}{n^s}$$

where $$s \in \mathbb{C}$$

This only makes sense for $\,Re(s)>1\,$...careful!

DonAntonio

we know that

$$n^s = exp(s\;ln\;n)$$

so I can write

$$\zeta (s) = \sum_{n=1}^{\infty} \frac{1}{exp(s\;ln\;n)}$$

but since

$$\frac{1}{exp(s\;ln\;n)} = 1 + \frac{1!}{(s\;ln\;n)} + \frac{2!}{(s\;ln\;n)^2} + ... = 1 + \sum_{n=1}^{\infty} \frac{n!}{(s\;ln\;n)^n}$$

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

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

then

$$\zeta (s) = \sum_{k=1}^{\infty} a_k$$

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,

$$\sum_{r=0}^{\infty} \frac{(-1)^r(s\ln n)^r}{r!}$$

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

$$\sum_{n=1}^{\infty} \; \sum_{r=0}^{\infty} \frac{(-1)^r(s\ln n)^r}{r!}$$ ??

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