Show that if $\sum_{k=0}^\infty a_kx^k$ has interval of convergence $[-1,1]$, then it converges uniformly on $[-1,1]$

Question

Show that if $\sum_{k=0}^\infty a_kx^k$ has interval of convergence $[-1,1]$, then it converges uniformly on $[-1,1]$.

Hint: Let $R_n=a_n+a_{n+1}+a_{n+2}+\dots$ and

$R_n(x)=a_nx^n+a_{n+1}x^{n+1}+a_{n+2}x^{n+2}+\dots$

Then $R_n(x)=(R_n-R_{n+1})x^n+(R_{n+1}-R_{n+2})x^{n+1}+(R_{n+2}-R_{n+3})x^{n+2}+\dots$

My attempt: I understand that this is Abel's Theorem and proofs exist that do not necessarily use this hint, but I'm interested in following this line of reasoning.

We want to show that $\sum_{k=0}^\infty a_kx^k$ converges uniformly, i.e. $\forall\epsilon>0$, $\exists N$ s.t. $\forall n>N$, $\forall x$,

$|\sum_{k=0}^na_kx^k-\sum_{k=0}^\infty a_kx^k|<\epsilon$, i.e. $|R_n(x)|<\epsilon$.

I know that since $\sum_{k=0}^\infty a_k$ converges, we have that for every $\epsilon>0,$ $\exists M$ such that if $n>M$, $|R_n|<\epsilon$.

My trouble now lies in bounding the sum $|R_n(x)|$. I've tried expanding and setting $x=1$, but it hasn't worked.

Any help appreciated!

Answer 1

Here is a subtlety that is worth to notice. Consider the power series

$$ \log(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} x^n. $$

Although the interval of convergence is $(-1, 1]$, the hint also applies to this series and thus proves that it converges uniformly on $[0, 1]$. On the other hand, there is no way this series converges uniformly on all of its interval of convergence. This shows that you have to establish the uniform convergence on $[0, 1]$ and on $[-1, 0]$ separately!

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For the proof, let $n < m$ and use summation by parts to write

\begin{align*} \sum_{k=n}^{m} a_k x^k &= \sum_{k=n}^{m} (R_{k} - R_{k+1}) x^k \\ &= R_n x^n - R_{m+1}x^m + \sum_{k=n+1}^{m} R_{k} (x^k - x^{k-1}) \end{align*}

For $0 \leq x \leq 1$, letting $m \to \infty$ yields

$$ R_n(x) = R_n x^n + \sum_{k=n+1}^{\infty} R_{k} (x^k - x^{k-1}). $$

Now using the fact that $x^k \leq x^{k-1}$, it follows that

$$ |R_n(x)| \leq |R_n| + \sum_{k=n+1}^{\infty} |R_{k}| (x^{k-1} - x^k). $$

Thus if $\epsilon > 0$ and $N = N(\epsilon)$ is chosen so that $|R_n| < \epsilon/2$ whenever $n \geq N$, then

$$ |R_n(x)| < \frac{\epsilon}{2} + \sum_{k=n+1}^{\infty} \frac{\epsilon}{2} (x^{k-1} - x^k) = \frac{\epsilon}{2} + \frac{\epsilon}{2} x^n \leq \epsilon. $$

This proves the uniform convergence of the power series $\sum_{n=0}^{\infty} a_n x^n$ on $[0, 1]$. Applying this result to the power series $\sum_{n=0}^{\infty} (-1)^n a_n x^n $ instead, we also find that $\sum_{n=0}^{\infty} a_n x^n$ converges uniformly on $[-1, 0]$. Therefore the claim follows.

Answer 2

Hint: You can use summation by parts (Abel summation):$$\sum_{n=1}^Nu_n(v_n-v_{n-1}) = (u_Nv_N-u_1v_0)+\sum_{n=0}^{N-1}v_n(u_n-u_{n-1})$$.

So first, you need to truncate the $R_n(x)$, or simply use the above formula to $$R_n^N(x)= (R_n-R_{n+1})x^n+...+(R_{N-1}-R_{N})x^{N-1}$$.