Examine if $f(x) = \sum\limits_{n=1}^\infty \frac{(-1)^n}{n-x}$ is term by term differentiable

Question

Series $\sum\limits_{n=1}^\infty \frac{(-1)^n}{n-x},$ $x\not = n$ is convergent as:
1) $\big|\sum\limits_{n=1}^k (-1)^n \big|\le 1$
2) $\frac{1}{n-x}$ is decreasing to $0$ as $n\rightarrow +\infty$
Hence by Dirichlet test series is convergent
$u_n(x)= \frac{(-1)^n}{n-x}, \; u_n'(x) = \frac{(-1)^n}{(n-x)^2}$
Examine series $\sum\limits_{n=1}^\infty \frac{(-1)^n}{(n-x)^2}$ on uniform convergence.
Let's check if $u'_n(x)$ is uniformly converges to $0$. I claim that it doesn't hold as
$a_n = \sup\limits_{x\not = n} \frac{1}{(n-x)^2} = \infty$ , because we can get $x$ arbitary close to $n$, hence we don't have uniform convergency. But answer in my book is vice versa. I can't get it where am I wrong?