Prove that $\sum_{n=1}^{\infty} \frac{a_n}{n^z}$ converges absolutely and uniformly 
Let $(a_n)$ be a sequence in $\mathbb{C}$. Assume that
  $\sum_{n=1}^{\infty} \frac{a_n}{n^z}$ converges absolutely for some 
  $z= z_0 \in \mathbb{C}$. Prove that the series converges absolutely
  and uniformly on $\{z \in \mathbb{C}: \text{Re} z \ge \text{Re} z_0 \}$

This is another revision question I am working on. Here is how I have approached this question, would greatly appreciate any feedback if I have done anything wrong.
We are given that $$\sum_{n=1}^{\infty} \frac{a_n}{n^{z_0}}$$ converges absolutely. Then letting $z_0 = a+bi$, I have that
$$\sum_{n=1}^{\infty} ||\frac{a_n}{n^a\cdot n^{bi}}||$$ converges. So it follows that 
$$||\frac{a_n}{n^a\cdot n^{bi}}|| \le ||\frac{a_n}{n^a}||$$ Then if $z_0 = a'+b'i$ with $ a' \ge a$, it follows that 
$$||\frac{a_n}{n^{a'}\cdot n^{b'i}}|| \le||\frac{a_n}{n^{a'}}||\le ||\frac{a_n}{n^a}||$$
Then since  $\frac{a_n}{n^a} \rightarrow 0 $ as $n \rightarrow \infty$, by the Weierstrass-M Test, we have that $\sum_{n=1}^{\infty} \frac{a_n}{n^z}$ converges absolutely and uniformly for $\{z \in \mathbb{C}: \text{Re} z \ge \text{Re} z_0\}$
Would this be correct? Many thanks in advance!

Things you need to change
First
$$|n^{ib}|=|\exp{i\cdot b\cdot ln(n)}|=|cos(b\cdot ln(n))+i\sin(b\cdot ln(n))|=\sqrt{cos(b\cdot ln(n))^2+sin(b\cdot ln(n))^2}=1$$
Second
You need to say $|a_n|\geq a_n$ and $|n^z|= n^a$ so $\sum_{n=1}^{\infty} \frac{a_n}{n^{\alpha}}$ converge uniformly on $]z_o,\infty[$
Third
Just forget "since $\frac{a_n}{n^a} \rightarrow 0$" and write :
$\sum_{n=1}^{\infty} \frac{a_n}{n^{\alpha}}$ converge so by the Weierstrass-M Test, we have that $\sum_{n=1}^{\infty} \frac{a_n}{n^{\alpha '}}$ converges   
and you can conclude !
