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Prove for $a\gt0$ that following series is holomorphic

$ \sum_{n=1}^\infty \frac {1}{(a+n)^z} \quad \textrm{for} \quad \operatorname{Re}z \gt 1 $

I'm trying to prove this given that $Re \quad z \gt 1$ implies that there exists an $\varepsilon \gt 0$ such that $ Re \quad z \gt 1+ \varepsilon $.

I've said that if $\frac {1}{(a+n)^z} = f_n(z)$,how do I prove the $|f_n(z)| \le \frac {1}{n^{1+\varepsilon}} $

Can anyone help with this? Thanks

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    Sorry, it is supposed to be the Real part of z, I have now changed it. Thanks for the heads up2012-12-02

1 Answers 1

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The following theorem is relevant: let $\Omega$ a domain in $\mathbb{C}$ and $\{f_n\}$ a sequence of holomorphic functions in $\Omega$. If $\{f_n\}$ converges uniformly to a function $f$ in every compact subset of $\Omega$, then $f$ is holomorphic in $\Omega$ (taken from Stein and Shakarchi).

Does that help?

Edited based on your edits:

You want to show that given $a > 0$ and $\Re(z) \ge 1 + \epsilon$ we have that $|\frac{1}{(n+a)^z}| \le \frac{C}{n^{1 + \epsilon}}$. To do this, first note that if $c$ is a real number larger than $1$, using our branch of the logarithm we have that $|c^z| = |e^{z\log(c)}| = |e^{\Re(z)\log(c)}| = c^{\Re(z)} \ge c^{1 + \epsilon}$. Thus by the sub-triangle inequality, we have that $|\frac{1}{(n + a)^z}| \le \frac{1}{(n - a)^{1 + \epsilon}}$, for large $n$.

Next, observe that since $\frac{n}{n-a} \to 1$ as $n \to \infty$, for large $n$ we have the estimate $n \le \frac{3}{2}(n-a)$. So we may take $C > 0$ to have $n \le C(n-a)$, for large $n$.

Thus: for sufficiently large $n$, we obtain the estimate $|\frac{1}{(n-a)^z}| \le C\frac{1}{n^{1 + \epsilon}}$.

From here, we deduce uniform convergence.

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    Ok I understand what you mean. Thanks for your help2012-12-02