Showing that the series for $\zeta(z)$ defines a holomorphic function

Question

I was told that is the Riemann Zeta function, but I'm very stuck on where to begin a mathematical proof of this question and I would greatly appreciate any help to understand the question and where to begin.

Answer 1

When $Re(z)>1$, we have that $$\zeta(z) = \lim_{x\to\infty} \sum_{n\leq x} \frac{1}{n^z} = \lim_{x\to\infty} (1+\int_1^x u^{-z}du -z\int_1^x \{u\}u^{-z-1}du -\frac{\{x\}}{x^z}).$$ This follows from an application of partial summation, where $\{u\}$ denotes the fractional part of $u$. We then have that this equals

$$\lim_{x\to\infty} (1+\frac{1}{1-z}(x^{-z+1}-1)-z\int_1^x \{u\} u^{-z-1}du -\frac{\{x\}}{x^z}) =\frac{1}{z-1} +1-z\int_1^\infty \{u\}u^{-z-1} du$$

The integral on the right is absolutely convergent when $Re(z)>0$, and we have that $\zeta(z)$ has a simple pole at $z=1$, so we have that $\zeta(z)$ is holomorphic whenever $Re(z)>1$. We have in fact shown more than that; we have shown that $\zeta(z)$ is meromorphic where $Re(z)>0$.

Answer 2

What is asked is to say that $f_n(z) = n^{-z} = e^{-z \ln n}$ is holomorphic with derivative $f_n'(z) = -e^{-z \ln n} \ln n$.

Also $$\left|f_n(z+h)-f_n(z) - h f_n'(z) \right|= \left|\int_z^{z+h} (f_n'(u)-f_n'(z))du\right|=\left|\int_z^{z+h} \int_z^u f_n''(v)dv du\right|$$ $$ \le \int_z^{z+h} \int_z^u \sup_{|w| < |h|}|f_n''(w)|dv du<|h|^2\sup_{|v| < |h|} |f_n''(z+v)| =|h|^2 n^{|Re(h)|-Re(z)}\ln^2 n $$

Now since for any $\epsilon > 0$ : $\lim_{n \to \infty} n^{-\epsilon}\ln^2 n = 0$

you have that $ \sum_{n=1}^\infty n^{|Re(h)|-Re(z)}\ln^2 n$ is bounded when say $Re(z) > 1+\epsilon$ and $|h|<\frac{\epsilon}{2}$,

and with $g(z) = \sum_{n=1}^\infty f_n'(z) $ : $$\left|\frac{ \zeta(z+h)-\zeta(z)}{h} - g(z)\right| \le \sum_{n=1}^\infty \left|\frac{ f_n(z+h)-f_n(z)}{h} - f_n'(z)\right| < |h| \sum_{n=1}^\infty n^{-1-\alpha}\ln^2n $$ i.e $$Re(z) > 1 , \qquad\qquad \lim_{h \to 0}\frac{\zeta(z+h)-\zeta(z)}{h} = g(z)$$ so that $\zeta'(z)= g(z) = -\sum_{n=1}^\infty n^{-z} \ln n$ and $\zeta(z)$ is holomorphic on $Re(z) > 1$.