# Riemann Zeta Function showing converges uniformly for s>1

1. Oct 23, 2016

### binbagsss

1. The problem statement, all variables and given/known data

$g(s) = \sum\limits^{\infty}_{n=1} 1/n^{-s},$

Show that $g(s)$ converges uniformly for $Re(s>1)$

2. Relevant equations

Okay, so I think the right thing to look at is the Weistrass M test. This tells me that if I can find a $M_{n}$, a real number, such that for each $n$ , $| f_{n} | \leq M_{n}$, and $\sum\limits^{\infty}_{n=1} M_{n}$ converges, then $\sum\limits^{\infty}_{n=1} f_{n}(s)$ converges, where $f_{n}(s)= 1/n^{-s}$ here.

3. The attempt at a solution

Okay, so if I consider the real part of $s$ only, it's pretty obvious that such a $M_{n}$ can be found for $s>1$, i.e. $M_{n} = 1/n$.

However I'm pretty stuck on how to incorporate $Im(s)$ into this, which has no bounds specified right?

So say I assume $Re (s) =1$, and we know that the series is then less than :

$\frac{1}{1^{1+iy}} + \frac{1}{2^{1+iy}} + \frac{1}{1^{3+iy}} + ...$
= $\frac{1}{1 . iy} + \frac{1}{2 . iy} + \frac{1}{3 . iy} +...$

where $s = 1 + iy$,

but surely as $Im(s) -> 0$, the imaginary part of each term in the series blows up, so I'm having a hard time understanding how it is bounded within any contraints on $Im(s)$ and only $Re(s)$.

2. Oct 25, 2016

### stevendaryl

Staff Emeritus
First of all, I think you mean $\frac{1}{n^{+s}}$ not $\frac{1}{n^{-s}}$.

Second, you've got the right idea, that if you can find some $M_n$ such that $|\frac{1}{n^s}| \leq M_n$, and $\sum_n M_n$ converges, then $\sum_n \frac{1}{n^s}$ converges. The easiest choice is to just let $M_n = |\frac{1}{n^s}|$

Third, you need to figure out what $|\frac{1}{n^s}|$ is. What you wrote is wrong: $2^{1+i} \neq 2 \cdot i$. Try this: Write $n = e^{log(n)}$, where $log$ means the natural log. And then write $s = Re(s) + i Im(s)$.