# Uniform convergence of $(nx-n+1)x^n$

I'm learning about uniform convergence, and I wanted to ask if I'm properly using Cauchy's test for uniform convergence to prove the series $\sum_{n=1}^{\infty} (nx-n+1)x^n$ doesn't uniformly converge in $(-1,1)$ (I already know its 'convergence range' is $(-1,1)$):

The test gives us that, assuming there's uniform convergence, for every $l > 0$ we have an $N$ such that if $m,n>N$, $|\sum_{k=n}^{m} f_k(x)|<l$ for all $x$ in $(-1,1$). If we choose $n=m$ and $l=0.5$ we get that for all $n>N$, $|\sum_{k=n}^{m} f_k(x)|=|(nx-n+1)x^n|<0.5$. But as the limit of $(nx-n+1)x^n$ as $x$ approaches $1$ is $1$, this is a contradiction. So there's no uniform convergence.

Am I properly using the test? Is there an even simpler way to do do this (maybe using a different test)? Thanks!

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Looks like you have it right...

You can see that dissecting the series produces:

$$y =x\sum\limits_{n > 0} {n{x^n}} - \sum\limits_{n > 0} {n{x^n}} + \sum\limits_{n > 0} {{x^n}}$$

$$y =\left( {x - 1} \right)\frac{x}{{{{\left( {x - 1} \right)}^2}}} + \frac{1}{{1 - x}}$$

$$y = \frac{x}{{x - 1}} - \frac{1}{{x - 1}}$$

$$y = \frac{{x - 1}}{{x - 1}}$$

So this function is not continuous at $x=1$ (and the limit is $1$) and it is $1 \in [-1,0)\cup(0,1)$ but zero at $x=0$.

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@Peter, I don't think it's 0 at $x=0$, wouldn't it be $1$? –  david Feb 9 '12 at 23:00
!? $(n\cdot 0- n +1) \cdot 0^n = 0 \forall n$ –  Pedro Tamaroff Feb 9 '12 at 23:05
Oh, I made a typo. The series starts at $n=0$ in the original exercise ;). Thank you for your answer! –  david Feb 9 '12 at 23:07