Is the sequence of functions $f_n(x) = x^n$ uniformly convergent on $[0,1)$? 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? 
 A: 
METHODOLOGY $1$:

Note that for $x=1-1/n$, $x^n\to e^{-1}$.  
Hence, $\displaystyle \sup_{x\in [0,1)} x^n\ne 0$ and the sequence does not converge to $0$ uniformly.


METHODOLOGY $2$:

Alternatively, take $\epsilon=1/4$.  Then, for all $N\ge 1$ there exists a number $x=1-1/n$ and a number $n>N$ such that 
$$x^n=\left(1-\frac1n\right)^n\ge \epsilon=\frac14$$
since $\left(1-\frac1n\right)^n$ increases monotonically.
Therefore, from the definition of the negation of uniform convergence we conclude that $x^n$ does not uniformly to $0$.


SHOWING UNFORM CONVERGENCE FOR $\displaystyle x\in [0,r], 0<r<1$

We can show that $x^n$ converges uniformly on any compact subset of $[0,1)$.  Take $x\in [0,r]$ for $0<r<1$.  Then, we have for all $\epsilon>0$
$$x^n\le r^n<\epsilon$$
whenever $n>\max\left(1,\log(\epsilon)/\log(r)\right)$.
A: 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 $|f_n(x)| = \frac{1}{2}$.
