Uniform convergence of $\sum\limits_{n=1}^{\infty}(1-x)x^n$ I want to study the uniform convergence of $\sum\limits_{n=1}^{\infty}(1-x)x^n  \rm{~~for}~~ x$ on $[0, 1]$. This is my attempt:
First I study convergence on $[0, 1)$: by Ratio test sum is convergent when $|x|< 1$. Hence, $\sum\limits_{n=1}^{\infty}(1-x)x^n$ on $[0, 1)$. When $x = 1$, $f_n(x) = 0$ and sum of series is convergent. Thus, sum of functional series is convergent on $[0, 1]$.
Is my reasoning correct, and if so can it be considered as a rigorous proof? Thanks.

You proved pointwise convergence, not uniform convergence. To study the question of uniform convergence, it is more convenient to start at $n=0$ (this does not affect the answer).  Then your partial sums have a closed form $\frac{(1-x)(1-x^{n+1})}{1-x}=1-x^{n+1}$, and the question is reduced to the question of the uniform convergence of this sequence of functions. But $x^{n+1}$ is not uniformly convergent on the unit interval.  Hence your sequence does not converge uniformly.
