# Proving series is not uniform convergent by contradiction

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!

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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 $x\to0$.

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Thanks for the alternative method. I've actually edited my main post with an idea for the proof for contradiction, the question doesn't specify it must be done via contradiction (although it is hinted) but I'd still like to know how to do it. Once again thanks for your help though (and I'll take your advice if I can't do it via contradiction easily). –  Noble. Feb 27 '13 at 18:26
The argument can be used, as you have done in your answer, to show that the Cauchy criterion is not satisfied. –  Julián Aguirre Feb 28 '13 at 11:24

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 $(0,1]$

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