# Discuss the convergence and the uniform convergence of a series

So the series to be discussed is $\sum (nx)^{-2}$ $\ (x\ne0)$.
And I think I can show that the series is pointwise convergent in its domain by comparison test with $\frac{1}{n^2}$.
But I'm not sure how to show that it's not uniformly convergent(I think). What I currently know is that Weierstrass M-test cannot be applied. And I feel that I have to use the Cauchy Criterion? But I'm not sure how to write the proof.

• Notice that $\sum (nx)^{-2}=ax^{-2}$, where $a=\sum n^{-2}=\pi^2/6$. – Mercy King Dec 28 '17 at 6:01
• That's definitely a nice shortcut! – Macrophage Dec 28 '17 at 7:08

Assume it were uniformly convergent. Then there exists some positive integer $N$ such that for all $x\ne 0$ we have $\left|\displaystyle\sum_{k=1}^{N}\dfrac{1}{k^{2}x^{2}}-\displaystyle\sum\dfrac{1}{n^{2}x^{2}}\right|<1$, then $\displaystyle\dfrac{1}{x^{2}}\sum_{k\geq N+1}\dfrac{1}{k^{2}}=\displaystyle\sum_{k\geq N+1}\dfrac{1}{k^{2}x^{2}}<1$, so $0<\displaystyle\sum_{k\geq N+1}\dfrac{1}{k^{2}}<x^{2}$ for all $x\ne 0$. Take $x\downarrow 0$ we get $\displaystyle\sum_{k\geq N+1}\dfrac{1}{k^{2}}=0$, a contradiction.
• Or one can choose $x=\sqrt{\displaystyle\sum_{k\geq N+1}\dfrac{1}{k^{2}}}\ne 0$ to get a contradiction at the end. – user284331 Dec 28 '17 at 5:43
• I don't quite get it. But $\displaystyle\sum_{k\geq N+1}\dfrac{1}{k^{2}}\geq\dfrac{1}{(N+1)^{2}}>0$, doesn't it? – user284331 Dec 28 '17 at 6:45
• The final part is due to this: If $x<\epsilon$ for all $\epsilon>0$, then $x\leq 0$. – user284331 Dec 28 '17 at 6:51
• I think you misunderstood my argument. No matter what, we must have $\displaystyle\sum_{k\geq N+1}\dfrac{1}{k^{2}}>0$, this is one point. Another point is that, because we have like, $\displaystyle\sum_{k\geq N+1}\dfrac{1}{k^{2}}<\epsilon$ for all $\epsilon>0$, then we have $\displaystyle\sum_{k\geq N+1}\dfrac{1}{k^{2}}\leq 0$, so this point contradicts with the previous one. – user284331 Dec 28 '17 at 6:53