$\sum\limits_{n}\frac{1}{z+n}$ diverges 
How to bound the sequence $\sum\limits_{n\ge0}\frac{1}{z+n}$ from below ?

Actually the problem is from complex analysis, and it is given that $z\in U=\mathbb C\setminus\mathbb R_{\le0}$, but I think it can be treated as in the real case. 
So for $z>0, n\in \mathbb N$, I have to show the divergence of the sequence 
Is it OK, if I write $\displaystyle\frac{1}{z+n}=\frac{1}{n}\cdot\frac{1}{1+\frac zn}\ge\frac{1}{n}\cdot\frac{1}{1+z},\quad\forall n$
then $\displaystyle \sum\limits_{n\ge0}\frac{1}{z+n}\ge\left(\frac{1}{1+z}\cdot \sum\limits_{n\ge1}\frac{1}{n}\right)+\frac1z\longrightarrow\infty$
$\bf{EDIT:}$ Original task is to show that $\sum_{n\ge0}f_n$ with $f_n=\frac{(-1)^n}{z+n}$ and $z\in\mathbb C\setminus\mathbb R_{\le0}=:U$ does not converge locally normally on $U$.

You've only shown that $\sum_{n\ge 0}^\infty \frac{1}{z + n}$ diverges for $z > 0$. However, you missed the case when $z = a + bi$ with $b\neq 0$. Write
$$\frac{1}{z + n} = \frac{1}{(a + n) + bi} = \frac{a + n}{(a + n)^2 + b^2} - \frac{b}{(a + n)^2 + b^2}i.\tag{*} $$
The first term on the right-most side of $(*)$ dominates $\frac{c}{a + n}$ (with $c = \frac{a^2}{a^2 + 4b^2}$) for $n \ge \frac{|a|}{2}$. Indeed, if $n \ge \frac{|a|}{2}$, then $n + a \ge n - |a| \ge \frac{|a|}{2}$. So 
$$\frac{a + n}{(a + n)^2 + b^2} = \frac{1}{a + n}\cdot \frac{1}{1 + \frac{b^2}{(a + n)^2}} \ge \frac{1}{a + n}\cdot \frac{1}{1 + \frac{4b^2}{a^2}} = \frac{c}{a + n}.$$
Since $\sum_{n\ge 1} \frac{c}{a + n}$ diverges, it follows that $\sum_{n\ge 0} \frac{a + n}{(a + n)^2 + b^2}$ diverges. The second term is dominated by $\frac{1}{b(a + n)^2}$, and $\sum_{n \ge 0} \frac{1}{b(a + n)^2}$ converges. So $\sum_{n\ge 1} \frac{b}{(a + n)^2 + b^2}$ converges. Therefore, $\sum_{n \ge 0} \frac{1}{z + n}$ diverges.
