Question: If $\sum_{n=0}^\infty A_n$ converges, does the series $\sum A_n x^n$ converge uniformly on $[0,1]$?

Pointwise convergence is easy enough to see, and intuitively I think the series should converge uniformly as well, but I'm having trouble showing that in proof. Indeed, if we fix $\varepsilon > 0$ and procure a $N$ by the convergence of $\sum A_n$, it is not universally true that $|\sum_{n=q}^p A_n x^n| \leq |\sum_{n=q}^p A_n| < \varepsilon$ whenever $p \geq q \geq N$. I also tried to use the statement $|\sum_N^\infty A_n x^n| < |\sum_N^\infty x^n|$ past some point where $|A_n| < 1$, but this only gives pointwise convergence.

Perhaps we can break up the interval in $[0, \xi]$ and $[\xi, 1]$, the first interval converging uniformly using the geometric property and the second somehow else?

  • 1
    $\begingroup$ Yes, the series converges uniformly, but this is not entirely straightforward. $\endgroup$ – Andrés E. Caicedo Sep 26 '15 at 19:33
  • $\begingroup$ The series $\displaystyle\sum_{n=0}^\infty A_n x^n$ converges uniformly on the interval whose endpoints are $\pm(1-\varepsilon)$. $\endgroup$ – Michael Hardy Sep 26 '15 at 19:38
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    $\begingroup$ @MichaelHardy Right, because we can use the geometric argument in that case. Is this enough to conclude that it converges uniformly on $[0,1]$? $\endgroup$ – MCT Sep 26 '15 at 19:39
  • $\begingroup$ @Winther If $A_n \ge 0$ then this is just the Weierstrass M-test. I think the interesting part of the question is when $A_n$ does not always have the same sign :). $\endgroup$ – Erick Wong Sep 26 '15 at 19:54
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    $\begingroup$ Yes, the convergence is uniform on $[0,1]$. A proof is contained in this answer. $\endgroup$ – Daniel Fischer Sep 26 '15 at 20:07

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 $[0,1]$.


We may use Abel's theorem itself here.

$ \sum A_n $ is uniformly convergent (as it is free from 'x').

And sequence {$ {x^n} $} is monotonically decreasing and bounded in (0,1).

Therefore, by Abel's theorem, $\sum A_nx^n$ is uniformly convergent.


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