TITLE: Is the sequence of functions $f_n(x) = x^n$ uniformly convergent on $[0,1)$?
QUESTION [5 upvotes]: Is the sequence of functions $f_n(x) = x^n$ uniformly convergent on $[0,1)$? I learned that $f_n(x)$ is not uniformly convergent on $[0,1]$ but what about on $[0,1)$? I think it does converge uniformly to $f(x)=0$ because I can always make $n$ big enough to fit $x^n$ to be within $\epsilon$ from $0$ for any $x$? Or is this not possible? I am not really sure on how to verify a sequence of functions is uniformly convergent on some domain. 
Also, does uniformly convergent series imply that the series is also absolutely convergent?

REPLY [3 votes]: No, if the sequence uniformly converged to 0, then by applying the definition with $\epsilon := \frac{1}{2}$ there must exist some $N$ such that for all $n \ge N$ and $x \in [0, 1)$, $|f_n(x)| < \frac{1}{2}$.  But that gives a contradiction, since if you set $n := N$ and $x := 2^{-1/N}$, then $n \ge N$ and $x \in [0, 1)$ but $|f_n(x)| = \frac{1}{2}$.