Riemann zeta function is analytic for $\operatorname{Re}(z)>1$

Question

I want to show that, Riemann zeta function is analytic on the domain $\operatorname{Re}(z)>1$. I know that it is absolutely and uniformly convergent on the right half of the line $Re(z)=1$. Now, I am trying to invoke Morera's theorem to prove its analyticity, but I am not quite sure how to do that. Can anyone help?

In general, if anyone can point me to some other direction then that is also welcome.

Answer 1

Otherwise it is immediate that for $\Re(c)> 1$ and $|s-c|<\Re(c)-1$, by absolute convergence $$\sum_{n\ge 1}n^{-s}=\sum_{n\ge 1}n^{-c}\sum_{k\ge 0}\frac{((c-s)\log n)^k}{k!}=\sum_{k\ge 0} (c-s)^k \sum_{n\ge 1}n^{-c} \frac{(\log n)^k}{k!} $$

We need the whole analytic continuation of $\zeta(s)$ and the Cauchy integral formula to prove that the latter series in fact converges for $|s-c|<|c-1|$.

Answer 2

From analyticity of $\sum_{n=1}^k \frac{1}{n^z}$, we get that $ \int_{\mathfrak{D}}\sum_{n=1}^k \frac{1}{n^z}=0 $, from Goursat's Theorem. Now, from uniform convergence of the partial sums, $\sum_{n=1}^k \frac{1}{n^z}$, to $\zeta(z)$ in $Re(z)>1$, we get

$$ lim_{k\rightarrow\infty}\int_{\mathfrak{D}}\sum_{n=1}^k \frac{1}{n^z}=\int_{\mathfrak{D}}\zeta(z)=0 $$ Therefore, analytic.