Proof that the Riemann Zeta Function is holomorphic on $\Re(z)>1$

Question

Ultimately I am trying to prove that that the Riemann Zeta function is holomorphic on $\Re(z)>1$.

So far I have manipulated the series in the following way:

$$\sum_{n=1}^\infty \frac{1}{n^z} = \sum_{n=1}^\infty e^{-z\log n} = \sum_{n=1}^\infty \frac{1}{n^x}\cos(y\log n) - i\sum_{n=1}^\infty\frac{1}{n^x}\sin(y\log n)$$

From this point I have considered using the fact that a function, $f$ say, of a complex variable defined by a power series $\sum_{n=0}^\infty a_n(z-a)^n$ with radius of convergence $R$ is holomorphic on an open disc $D(a,R)$. However, I really don't know how to go about this as I do not know what radius to choose as $\Re(z)>1$ is obviously not a disc. Furthermore, I had no luck trying to find $R$ using the ratio test.

Do I need to continue manipulating the series into a more manageable form? Or am I approaching this problem in the wrong way?

Any help on what to try next would be great!

Not a good approach, I think. Use that a locally uniformly convergent series of holomorphic functions has a holomorphic sum. — Daniel Fischer, Nov 17 '16 at 23:18
Common functions have \cos, \log, and generally \function_name — Simply Beautiful Art, Nov 17 '16 at 23:22
See this answer for showing $\zeta'(z) = -\sum_{n=1}^\infty n^{-z} \ln(n)$ — reuns, Nov 17 '16 at 23:55
I just saw Showing that $\sum_\limits{n=1}^\infty 1/n^z$ defines a holomorphic function pop up on the main list. This looks quite similar. — robjohn, Nov 18 '16 at 00:47

score 7 · Answer 1 · edited Apr 13 '17 at 12:21

For any $z$ so that $\mathrm{Re}(z)\gt1$, the series $$ \sum_{n=1}^\infty\frac1{n^z} $$ converges uniformly. That is, $$ \begin{align} \left|\,\sum_{n=N}^\infty\frac1{n^z}\,\right| &\le\sum_{n=N}^\infty\frac1{n^{\mathrm{Re}(z)}}\\ &\le\frac1{\mathrm{Re}(z)-1}\frac1{N^{\mathrm{Re}(z)-1}} \end{align} $$ Since the uniform limit of holomorphic functions is holomorphic and each partial sum is holomorphic, the limit, $\zeta(z)$, is also holomorphic for $\mathrm{Re}(z)\gt1$.

Proof that the Riemann Zeta Function is holomorphic on $\Re(z)>1$

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