1
$\begingroup$

I'm reading about Abel's limit theorem, and I don't get why the result isn't practically trivial. Below, I will state the theorem and my very, very simple argument. I would appreciate if someone could let me know where I'm going wrong.

$\mathbf{Theorem}$: If $\sum_{n=1}^{\infty} a_{n}$ exists, then $\sum_{n=1}^{\infty} a_{n}x^{n}$ converges uniformly for $x \in [0,1]$.

"$\textit{Proof}$" (incorrect): Obviously, the key thing to prove above is that we can include the point $x = 1$ in the interval on which the series converges uniformly. For $x < 1$, uniform convergence follows from the theorem on power series. Now, my bad argument goes as follows:

We know that there is $N_{1} \in \mathbb{N}$ s.t. $\sum_{n=N_{1}}^{\infty} a_{n}x^{n} < \varepsilon$ for $x <1$.

Additionally, we have $N_{2} \in \mathbb{N}$ s.t. $\sum_{n=N_{2}}^{\infty} a_{n} < \varepsilon$ (i.e. the above power series with $x=1$).

From this, why is it incorrect to conclude that we have uniform convergence on $[0,1]$ by picking $N \geq $max$(N_{1}, N_{2})$, i.e. why do we need Abel's theorem?

$\endgroup$
  • 7
    $\begingroup$ The theorem on power series only gives uniform convergence for $|x|\leq a$ where $a$ is some number strictly smaller than the radius of convergence. A power series needn't converge uniformly inside its disc of convergence. $\endgroup$ – Wojowu Aug 13 at 12:45
  • 1
    $\begingroup$ And if you assume that $a_n\ge0$ for all $n$, then the Weierstrass $M$-test gives uniform convergence on $[0,1]$ without further ado. We only ever need Abel, when the signs vary, I think. Mind you for the alternating power series that you see in calculus ($\arctan x$, $\ln(1+x)$) their uniform convergence on $[0,1]$ follows from Leibniz test on alternative series similarly without needing Abel. $\endgroup$ – Jyrki Lahtonen Aug 13 at 12:50
  • 1
    $\begingroup$ @Wojowu I suggest that you post your comment as an answer. $\endgroup$ – José Carlos Santos Aug 13 at 13:13
  • 2
    $\begingroup$ @Wojowu Got it. Always thought there was something strange about saying that the series converges uniformly or all $|x| \leq a$ for all $a < 1$ instead of simply saying that the series converges uniformly for all $x <1$ - now I know the difference. $\endgroup$ – gtoques Aug 13 at 13:51
  • 1
    $\begingroup$ @gtoques You could restate it as uniform convergence on compact subsets of the (open) disc of convergence. I think it makes the condition of being uniformly away from the boundary more clear, although less explicit. $\endgroup$ – lzralbu Aug 13 at 15:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.