If $\sum A_n$ converges, does $\sum A_n x^n$ converge uniformly on $[0,1]$?

Solution 1:

The answer is yes. We use partial summation (used to prove Abel theorem) to get an estimate of Cauchy sum.

Let $B_n=\sum_{k=m}^n A_k$. So by Cauchy Criterion, for any $\epsilon>0$, there is a $N$ such that $$ |B_n|<\epsilon \quad\text{whenever }\quad n,m>N\tag1 $$ We have \begin{align} \sum_{k=m}^n A_kx^k&=\sum_{k=m}^n (B_k-B_{k-1})x^k \\ &=\sum_{k=m}^n B_kx^k -\sum_{k=m}^n B_{k-1}x^k \\ &=\sum_{k=m}^{n-1} B_k(x^k-x^{k+1})+B_nx^n\tag{2} \end{align} Note: $B_{m−1}=0$.

Since $\lim_{n\to\infty}x^n$ exists and $\:x^n \downarrow$ on $[0,1]$, for all $k>0$ and $x\in[0,1]$ we have $$ x^k-x^{k+1}\geqslant0\: $$ Since for all $k>m$, $-\epsilon<B_k<\epsilon$ $$ |B_k(x^k-x^{k+1})|<\epsilon(x^k-x^{k+1})\tag3 $$ So for all $n,m>N-1$ and $x\in[0,1]$, by $(1)$, $(2)$ and $(3)$ there is \begin{align} \left|\sum_{k=m}^n A_kx^k\right|&\leqslant\sum_{k=m}^{n-1} |B_k(x^k-x^{k+1})|+|B_nx^n| \\ &\leqslant\sum_{k=m}^{n-1} \epsilon\:(x^k-x^{k+1})+\epsilon \:x^n \\ &=\epsilon \:(x^m-x^n+x^n) \\ &=\epsilon \:x^m \\ &\leqslant \epsilon \end{align} So by Cauchy Criterion, $\sum_{k=1}^{\infty} A_kx^k$ converges uniformly on $[0,1]$.