Simplest proof that $\sum 1/n^s$ is analytic. [duplicate]

Show that the series $$\sum_{n=1}^\infty \frac{1}{n^z}$$ defines a holomorphic function on the set $\{z \in \Bbb{C}\ :\ \operatorname{Re}(z) > 1\}$.

Does it suffice to show that the function is analytic - for instance, by using the Cauchy-Riemann equations? If so, what is the geometric representation of this series?

I presume you mean the Riemann zeta function. You might benefit from reading the linked article and then going on from there. You might also benefit from reading about Dirichlet series and their abscissae of convergence.

To your specific question, this proof of the result uses elementary estimates and the $p$-test for convergence of a series.

This is a duplicate, but let me write the most elementary proof I know :

If $g(z)$ is holomorphic and $|g''(z)| \le C$ then $$ \left|\frac{g(z+h)-g(z)}{h}-g'(z)\right| = \left|\frac{1}{h}\int_z^{z+h} (g'(u)-g'(z))du\right|=\left|\frac{1}{h}\int_z^{z+h}\int_z^u g''(v)dvdu\right|$$ $$\le \frac{1}{|h|}\int_z^{z+h}\int_z^u Cdvdu< |h| C$$ Now let $f_n(z) = n^{-z} = e^{-z \ln n}$, it is holomorphic with derivatives $f_n'(z)= -n^{-z} \ln n,f_n''(z)= n^{-z} \ln^2 n$ and for $Re(z) > 1+\epsilon$ : $|f_n''(z)| < n^{-1-\epsilon}\ln^2 n$.

Let $F(z) = \sum_{n=1}^\infty f_n'(z) = -\sum_{n=1}^\infty n^{-z} \ln n$, note it converges absolutely on $Re(z) > 1$, and for $Re(z) > 1+2\epsilon, |h| < \epsilon$ : $$\left|\frac{\zeta(z+h)-\zeta(z)}{h}-F(z)\right| \le \sum_{n=1}^\infty \left|\frac{f_n(z+h)-f_n(z)}{h}-f_n'(z)\right|\le|h|\sum_{n=1}^\infty n^{-1-\epsilon}\ln^2 n$$ i.e. for $Re(z) > 1+2\epsilon$ : $$\zeta'(z) = \lim_{h \to 0} \frac{\zeta(z+h)-\zeta(z)}{h} = F(z) = -\sum_{n=1}^\infty n^{-z} \ln n$$ so that $\zeta(z)$ is holomorphic on $Re(z) > 1+2\epsilon$ for every $\epsilon > 0$,

and hence on $Re(z) > 1$.