I want to prove Abel's theorem. There exist various versions of it but I am not sure which is the standard one. The one I have in mind claims:
If $\sum\limits_{k=0}^{\infty}c_k$ is a real valued convergent series, then $\sum\limits_{k=0}^{\infty}c_kx^k$ converges uniformly on $[0,1]$.
My approach:
As $\sum\limits_{k=0}^{\infty}c_k$ is convergent there exists a $n_0$ such that for all $n>n_0$ it holds: $|\sum\limits_{k=1}^{n}c_k-C|<\frac{\epsilon}{3}$, where $C:=\sum\limits_{k=1}^{\infty}c_k$. We define $C_n:=\sum\limits_{k=1}^{n}c_k$. Let be $x\in[0,1]$ arbitrarily chosen. Then, using summation by parts and adding $0=Cx^m-Cx^n-\sum\limits_{k=m}^{n-1}C(x^k-x^{k+1})$, yields:
$$\Big|\sum\limits_{k=m+1}^{n}c_kx^k\Big|=\Big|C_nx^n-C_mx^m+ \sum\limits_{k=m}^{n-1}C_k(x^k-x^{k+1})+0\Big| \\\leq\Big|(C_n-C)x^n\Big|+\Big|(C_m-C)x^m\Big|+ \Big|\sum\limits_{k=m}^{n-1}(C_k-C)(x^k-x^{k+1})\Big|\cdots $$ For $x\in[0,1]$ it holds $|x^k|\leq1$ for all $k$. Hence, $1$ is an upper bound. Moreover, we know that $x^k$ is monotonically decreasing on $[0,1]$. So we can conclude:
$$ \cdots <\frac{\epsilon}{3}+\frac{\epsilon}{3}+ \sum\limits_{k=m}^{n-1}\Big|(C_k-C)(x^k-x^{k+1})\Big|< \frac{2\epsilon}{3}+ \frac{\epsilon}{3}\sum\limits_{k=m}^{n-1}\Big|(x^k-x^{k+1})\Big| \\=\frac{2\epsilon}{3}+ \frac{\epsilon}{3}(x^m-x^{n})\leq \epsilon. $$ So the $n_0$ we defined at the beginning holds for all $x\in[0,1]$. This means that $\Vert \sum\limits_{k=m+1}^{n}c_kx^k \Vert_{\infty}<\epsilon$ and hence $\sum\limits_{k=1}^{\infty}c_kx^k$ converges unifmormly on $[0,1]$ due to the Cauchy criterion on uniform convergence. (Note, $\Vert\cdot\Vert_{\infty}$ is the supremum norm)
Is this correct? Do you have any suggestions on how to prove it faster?