# Isn't Abel's limit theorem trivial?

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?

• 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. – Wojowu Aug 13 at 12:45
• 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. – Jyrki Lahtonen Aug 13 at 12:50
• @Wojowu I suggest that you post your comment as an answer. – José Carlos Santos Aug 13 at 13:13
• @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. – gtoques Aug 13 at 13:51
• @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. – lzralbu Aug 13 at 15:32