Help with convergence of series of functions We have this series of functions
$\sum\limits_{n=1}^{\infty}nx^n(1-x)$
for $x$ ranging from $[0, 1]$. How do we determine whether it's uniformly convergent or not?
Thanks in advance.
 A: Hint :
$$...=\frac{(1-x)}{x}\sum_{n\geq1}nx^{n+1}=\frac{(1-x)}{x}\sum_{n\geq 0}\frac{d}{dx}(x^{n+1})=\frac{(1-x)}{x}\frac{d}{dx}\underbrace{\sum_{n\geq 0}x^{n+1}}_{=\frac{x}{1-x}}=\frac{(1-x)}{x}\frac{d}{dx}\left(\frac{x}{1-x}\right)=...$$
A: Let $\:F(x)=\sum_{n\in\mathbf N}f_n(x)=\underset{n\to\infty}{\text{lim}}F_n(x)\\ F_n(x)=\sum_{j=1}^nf_j(x),\quad f_j(x)=jx^j(1-x)\:\:\:\:\forall j\le n.$
Since $\:F(0)=F(1)\equiv0,\:$ let's focus exclusively on $\:I=[0,1]$\ $\{0,1\}=\left]0,1\right[$.
We have that $$\exists\large\epsilon\normalsize>0\:\:\:\forall N_{\epsilon}>0\:\:\:\exists n\ge N_{\epsilon}\:\:\exists x_n\in I \implies|f_n(x_n)-\underbrace 0_{f(x)}|\ge\large \epsilon\normalsize.$$
Choose $\:x_n=1-{\large{1\over n}}$
Then $$f_n(x_n)\overset{n\to\infty}{\longrightarrow} {1\over e}=\large \epsilon\normalsize >0.$$
Since $\:f_n\:$ does not converge uniformly on $\:I\subset\left[0,1\right]$, $\:\underset {n\ge N_{\epsilon}}{F_n}(x)\:$ is not an uniformly convergent series on this interval, so that the uniform convergence does not apply for our series on $\:[0,1].$
