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.
 A: Hint: $(1-x)x^n=x^n-x^{n+1}$. Just use the partial sums, then everything gets simple as these are partial sums of the geometric series.
A: 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.
A: Though it is a overkill but i will mention it as alternative approach :
Dini's Theorem : If $f_n$ is a sequence of real valued function converging pointwise to a continuous limit function f on a compact set $S$ and if $f_n(x) \ge f_{n+1}(x)$ for each $x\in S$ and for each $n=1,2,3..$  then $f_n \to f $ uniformly .
In your problem set $f_n=\sum_{1}^{n} (1-x)x^n=x-x^{n+1}$ . $ \lim f_n(x)=0$ for all $x\in [0,1]$ ,a compact set . Hence by dini's theorem $f_n\to f$ uniformly .
