I'm asked to prove whether the following function-series
$\sum_{n=1}^\infty f_n$ on $E$ with values in $Y$ converges pointwise/uniformly/absolutely.
$$E = [0, 1), Y= \Bbb{R}$$ $$f_n = x^n$$
What I have tried so far:
According to the Weierstrass M-Test $f_n$ is not convering absolutetly (if I didn't produce any mistakes.)
$$ \sum_{k=0}^{n}x^k=\frac{1-x^{n+1}}{1-x}\underset{n \rightarrow +\infty}{\rightarrow}\frac{1}{1-x} $$ because $\left|x\right|<1$. Hence it converges pointly. Furthermore it is the same for $\left|x\right|$ because $\left|x^k\right|=\left|x\right|^k$. Hence it is absolutely convergente.
$$ \sum_{k=p+1}^{N}x^k=x^{p+1}\frac{1-x^{N-p}}{1-x} $$ So $$ \sum_{k=p+1}^{+\infty}x^k=x^{p+1}\frac{1}{1-x}$$ For $x \in \left[a,b\right], \ a<b$ with $\left(a,b\right) \in \left[0,1\right[^2$ we have $$ \left\|\sum_{k=p+1}^{+\infty}x^k\right\|_{\infty,\left[0,b\right]} \leq b^{p+1}\underset{p \rightarrow +\infty}{\rightarrow}0 $$ So it also converges uniformly on every compact set of $\left[0,1\right[$.