Uniform convergence of the series : $ \sum\frac{1}{n^2 x^2}$, x is not zero.

Question

If $|x|\ge 1$ , then the series converges uniformly by Weierstrass M- test as $\frac{1}{n^2 x^2}$ $\le$ $\frac{1}{n^2}$.

It seems that the series should also converge uniformly if $ |x| <1$. But how to prove it ?

please suggest.

Answer 1

It will converge uniformly on $|x|\ge r$ for any $r>0$, but converges non-uniformly on $|x|>0$. To see this, note that $$\left|\sum_{n=N}^\infty\frac1{x^2n^2}\right|=\frac1{|x|^2}\left|\sum_{n=N}^\infty\frac1{n^2}\right|$$ is not bounded on $|x|>0$ for any $N$, so convergence cannot be uniform, there.

Answer 2

The series is not defined at x=0. However, the series does not even converge uniformly for $0<|x|<1$. Just choose $\varepsilon=1$, then for any $n$ we choose $x=1/\sqrt{n}$. Then $$|f(1/\sqrt{n})-f_n(1/\sqrt{n})|=n\sum_{k=n}^\infty\frac1{k^2}>\frac n{n}=1\,.$$