Prove uniform convergence of the series $\sum_{n=1}^{\infty}\frac{1}{(n-x)^2}$ 
Prove uniform convergence of the series.
  $$
\sum_{n=1}^{\infty}\frac{1}{(n-x)^2},\ \ x\in\mathbb{R}
$$

Well, I tried to apply Weierstrass M-test. However, it did not help since I could not properly bound $1/(n-x)^2$ (Especially because $x\in\mathbb{R}$).
So, I would be grateful if anyone could help.
 A: The series converges uniformly for $x \in (-\infty,0]$ by the Weierstrass M-test since for $x \leqslant 0$,
$$|f_n(x)| =\frac{1}{(n-x)^2} \leqslant \frac{1}{n^2}$$
However convergence is not uniform for all $x \in \mathbb{R}$.
Technically, the series as written is not well-defined for any $x=n   \in \mathbb{N}$ since the $n-$th term is not defined. However, this can be rectified by assigning a value to $f_n(n)$. Regardless of how this is done the convergence cannot be uniform for $x \in \mathbb{R}$, since we can show it is not uniform for $x \in [0,\infty)\setminus \mathbb{N}$.
A necessary condition for uniform convergence of a series $\sum f_n(x)$ for $x \in D$ is that $f_n(x) \to 0$ uniformly as $n \to \infty$.  This follows from the Cauchy criterion, and  is equivalent to $\lim_{n \to \infty} \sup_{x \in D} |f_n(x)| = 0.$
In this case, we have with $x_n = n - 1/n \in [0,\infty)\setminus \mathbb{N},$
$$\sup_{x \in [0,\infty) \setminus \mathbb{N}}|f_n(x)| =\sup_{x \in [0,\infty) \setminus \mathbb{N}} \frac{1}{(n-x)^2} \geqslant \frac{1}{(n- x_n)^2} = n^2 $$
Since the RHS diverges to $+\infty$ the convergence cannot be uniform.
A: Assume $x\not\in\mathbb{Z}$. Then all but a finite number of values (until when $n\gt x$) can be bounded by a corresponding $\frac1{(n-1)^2}\gt\frac1{(n-\{x\})^2}$, where $[x]=x-\{x\}$ as usual.
