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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:

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

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!

1 Answers1

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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 $\mathrm{Re}(z)\gt1$.

robjohn
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