Proving $\sum_\limits{n=1}^{\infty}\frac{1}{(n+z)^2}$ converges uniformly 
Using the Weierstrass test show that the series $\sum_\limits{n=1}^{\infty}\frac{1}{(n+z)^2}$ converge uniformly on $E=Re(z)\geqslant 1$.

This solution was given to me but I am not understanding certain steps:
$|\frac{1}{(n+z)^2}|\leqslant\frac{1}{|n+z|^2}\leqslant \frac{1}{|n-|z||^2}\leqslant\frac{1}{|n-1|^2} $
Questions:
Which inequality was used here $\frac{1}{|n+z|^2}\leqslant \frac{1}{|n-|z||^2}$? How can the modulus of $|z|$ be subtracted? What is backing up that step?
Thanks in advance!
 A: HINT:
Rather than using the proof based on the triangle inequality, $|x+y|\ge ||x|-|y||$, simply write
$$|n+z|^2=n^2+|z|^2+2n\text{Re}(z)\ge n^2+1+2n=(n+1)^2$$
A: What was used is the reverse triangle inequality $|z-w|\geq ||z|-|w||$.
A: The triangle inequality gives
$$|n+z|\ge n-|-z|=n-|z|$$
and
$$|n+z|\ge |z|-|-n|=-(n-|z|).$$
In both cases
$$|n+z|\ge|n-|z||$$ and
so
$$|n+z|^2\ge|n-|z||^2.$$
Then
$$\frac1{|n+z|^2}\le\frac1{|n-|z||^2}$$
provided that $|z|\ne n$.
A: If $\mathrm{Re}\,z\ge 1$, then
$$
|z+n|\ge |\mathrm{Re}\,z+n|=\mathrm{Re}\,z+n\ge n+1
$$
and hence
$$
\left|\frac{1}{(z+n)^2}\right|\le \frac{1}{(n+1)^2}.
$$
Comparison Test implies that $\sum \frac{1}{(z+n)^2}$ converges ABSOLUTELY, and since that right hand side does not depend on $z$, then it converges uniformly on $z$, for $\mathrm{Re}\,z\ge 1$.
To understand this, observe that if
$$
s_n(z)=\sum_{k=1}^n\frac{1}{(z+k)^2}
$$
then
$$
|s_m(z)-s_n(z)|\le \sum_{k=n+1}^m\frac{1}{|z+k|^2}\le \sum_{k=n+1}^m\frac{1}{(k+1)^2}
$$
and the right hand side becomes less that $\varepsilon$, for $m,n\ge N$, when $N$ is sufficiently large, and this $N$ clearly depends only on $\varepsilon$ and not on $z$.
