Proving series is not uniform convergent by contradiction

Question

I'm trying to show that the series $\displaystyle\sum_{n=0}^{\infty} \frac{n^2x}{1+n^4x^2}$ is not uniformly convergent on the interval $(0,1]$ and I'm trying to do it by assuming that a sequence of partial sums is uniformly Cauchy and getting a contradiction. So under the assumption it's Cauchy

$\forall \epsilon > 0$, $\exists N \in \mathbb{N}$ such that $\forall n > m > N$ and $\forall x \in (0,1]$

$\Bigg|\displaystyle\sum_{k=m+1}^{n} \frac{k^2x}{1+k^4x^2}\Bigg| < \epsilon$

So clearly I need to find an $\epsilon$ this isn't true for, but I can't see where to go at all.

Many thanks!

Answer 1

I suggest a different road. The series is clearly convergent for all $x\in[0,1]$. Let's call $f(x)$ its sum. If it were uniformly convergent, then $f$ would be continuous. Clearly $f(0)=0$, but for all $n\in\mathbb{N}$ $$ f(1/n^2)>\frac12, $$ and $f(x)$ does not converge to $0$ as $x\to0$.

Answer 2

Actually I think I've worked it out with the help of Julián. From

$\Bigg|\displaystyle\sum_{k=m+1}^{n} \frac{k^2x}{1+k^4x^2}\Bigg| < \epsilon$

If we let $x = \displaystyle\frac{1}{(m+1)^2} \in (0,1] \ \ \forall k$ and set $\epsilon = \frac{1}{4}$ we get

$\Bigg|\displaystyle\sum_{k=m+1}^{n} \frac{k^2x}{1+k^4x^2}\Bigg| \geq \frac{1}{2} > \epsilon$

Hence the sum is not uniformly continuous on the interval $(0,1]$