Uniform convergence of the series $\sum(nx)^{-2}$ I am trying to check for uniform convergence of the series $\displaystyle\sum_{n=1}^\infty\dfrac{1}{(nx)^2}$ on $\mathbb R-\{0\}$
Here's what I've tried:
If $|x|\ge 1$ then $(nx)^2\geq n^2$ so that $\dfrac{1}{(nx)^2}\leq\dfrac{1}{n^2}$ and by Weirstrass M test the series is uniformly convergent. 
I am not sure how to proceed for $|x|<1$. One thing I observe is $f_n(1/n)=1$ but I'm not sure if that helps.
Is it possible to do it without knowing that $\displaystyle\sum_{n=1}^\infty\dfrac{1}{n^2}$ is $\dfrac{\pi^2}{6}$?
 A: We have that $$\sup_{x \neq 0}|\frac{1}{x^2}(\sum_{k=1}^N\frac{1}{n^2}-\frac{\pi^2}{6})|=|\sum_{k=1}^N\frac{1}{n^2}-\frac{\pi^2}{6}|\sup_{x \neq 0}\frac{1}{x^2}=+\infty$$
So the covergence is not uniform.
A: We'll show that there exists $\epsilon > 0$ such that for any $N \in \mathbb N$ we can choose $n,m \ge N$ such that $\sup\{|S_n(x) - S_m(x)| : x \in \mathbb R-\{0\}\} > \epsilon$, where $S_n(x) = \sum_{k=1}^n \frac{1}{x^2k^2}$
We'll choose $\epsilon$ later. Now consider $|S_{2n}(x) - S_n(x)|$ for some $n \in \mathbb N$.
$|S_{2n}(x) - S_n(x)| = \sum_{k=n+1}^{2n} \frac{1}{x^2k^2}$. Now if we take $x_n = \frac{1}{n}$, then we have $\sup\{|S_{2n}(x) - S_{n}(x)| : x \in \mathbb R-\{0\}\} > |S_{2n}(x_n) - S_n(x_n)| = n^2 \sum_{k=n+1}^{2n} \frac{1}{k^2} > n^2 \frac{n}{4n^2}= \frac{n}{4} \to \infty$.
So taking $\epsilon = 1$, and any $N \in \mathbb N$, we can choose $n = \max\{N,10\}, m=2n$ for it to work, which clearly means, the convergence cannot be uniform.
A: If the convergence was uniform, $\sup\limits_{x \in \mathbb R \setminus\{0\}} (nx)^{-2}$ would converge to $0$ with $n$. But this is not the case as this value is always greater than $1$.
