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 $|x|&lt; 1$. 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.
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.
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 .