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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?

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  • $\begingroup$ This is the riemann zeta function . $\endgroup$ – vidyarthi Nov 14 '16 at 17:11
  • $\begingroup$ And yes, the Cauchy-Riemann equations are enough, but you have to prove the partial derivatives first (which is what I did, without using any theorem) $\endgroup$ – reuns Nov 14 '16 at 17:45
  • $\begingroup$ Possible duplicate of Euler Product formula for Riemann zeta function proof $\endgroup$ – user3.14259 Nov 21 '16 at 10:13
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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.

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  • $\begingroup$ Showing convergence is easy. It's analyticity that's harder. $\endgroup$ – Vik78 Nov 14 '16 at 17:58
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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 $Re(z) > 1$.

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