The sequence is: $$f_n=x^n-x^{n+1}=x^n(1-x),\quad x\in[0,1]$$ I need to check whether it converges uniformly, but I'm having doubts about the answer for $x\in(0,1)$. Below is what I've tried.
I apply Cauchy's covergence test: \begin{align*} &\text{for x=0,x=1}\quad \lim_{n\to\infty}\sqrt[n]{x^n(1-x)}=0\\ &\text{for x}\in(0,1)\quad \lim_{n\to\infty}(x\sqrt[n]{(1-x)})=x<1 \end{align*} A series of 0's converges uniformly, and to check the rest of the inverval I evaluate: $$\sup_{[0,1]}(x^n-x^{n+1})=\sup_{[0,1]}(\Big(\frac{n}{n+1}\Big)^n(1-\frac{n}{n+1})$$ because $f_n^{max}$ is reached when $x=x_{max}=\frac{n}{n+1}$. Now, since $x_{max}<1$, $$\sup_{[0,1]}(\Big(\frac{n}{n+1}\Big)^n(1-\frac{n}{n+1})\leq \sup_{[0,1]}((1-\frac{n}{n+1})$$ And the problem here is that the majorant series clearly diverges, which gives no information concerning convergence of the series in question.
What am i doing wrong and where do I proceed from here? Can I argue that the last inequality is strict and say that uniform convergence does follow from it?
It is controlled by \begin{align*} \left(1-\dfrac{1}{n+1}\right)^{n}\left(1-\dfrac{n}{n+1}\right)=\left(1-\dfrac{1}{n+1}\right)^{n}\dfrac{1}{n+1}. \end{align*} We know that \begin{align*} \left(1-\dfrac{1}{n+1}\right)^{n}\rightarrow e^{-1}, \end{align*} so for large $n$, \begin{align*} \left(1-\dfrac{1}{n+1}\right)^{n}\left(1-\dfrac{n}{n+1}\right)\leq 2e^{-1}\dfrac{1}{n+1}, \end{align*} which tends to zero as $n\rightarrow\infty$.