# How to write this in terms of ##\zeta (x)##?

• B
• MevsEinstein

#### MevsEinstein

TL;DR Summary
I wrote $$\zeta (x+yi)$$ as ##\zeta(x)\zeta(yi) - \displaystyle\sum_{n=1}^\infty \frac{1}{n^x} *(\displaystyle\sum_{k \in S, \mathbb{Z}\S =n})##. I want to simplify the second term in terms of the zeta function so I can solve for ##\zeta (x)##.
How do I re-write $$\displaystyle\sum_{n=1}^\infty \frac{1}{n^x} *(\displaystyle\sum_{k \in S, \mathbb{Z}\S =n})$$ in terms of ##\zeta (x)## ? I want to solve for ##\zeta (x)## and simplifying the above expression in terms of ##\zeta (x)## would avoid repetition.

Last edited:
You have to fix the latex in your post. You can use dollar signs instead of hashes to make the whole line display style which might make it look better.

You have to fix the latex in your post.
It took me forever to find the error. Turns out it was just a parentheses >D.

Oh wait instead of \ it gives me ##\S=##

How do I re-write $$\displaystyle\sum_{n=1}^\infty \frac{1}{n^x} *(\displaystyle\sum_{k \in S, \mathbb{Z}\S =n})$$
Before you can simplify the 2nd term, you need to say what is being summed.

Before you can simplify the 2nd term, you need to say what is being summed.
OOPS! $$\displaystyle\sum_{n=1}^\infty \frac{1}{n^x}*(\displaystyle\sum_{k \in S, \mathbb{Z} \S =n} \frac{1}{k^{yi}})$$ There you go. Note that ##\mathbb{Z} \S =n## is actually ##\mathbb{Z}##\S = n, I couldn't fix the bug.