Uniform convergence in the endpoints of an interval

Study the pointwise and uniform convergence of the series $$\sum_{n=1}^\infty\dfrac{4^n}{n^2}\dfrac1{(1+x^2)^n}$$

I'm doing this exercise and I'm not sure about the following:

What I've done is the change of variable $t^2=(1+x^2)^{-1}$ to write the series as a power series centered in $t_0=0$. $$\sum_{n=1}^\infty\dfrac{4^n}{n^2}t^{2n}$$ Now, by root test is easy to find the intervals of pointwise convergence: $$t^2\leq\dfrac{1}{4}\iff t\in\left[-\frac12,\frac12\right]\iff x\in(-\infty,\sqrt{3}]\cup[\sqrt{3},\infty)$$ (In the endpoints the series becomes $\sum_{n=1}^\infty\frac1{n^2}$, which clearly converges.)

Now, there's a theorem that says that the convergence is uniform in the sets $$\{t\in\mathbb R:|t|\leq r<R\},\quad \forall 0\leq r<R$$

where $R=\frac12$ is the radius of convergence. So we can conclude that the series converges uniformly (respect $t$) in the intervals $[-r,r]$, $r<R$. But can't we say that the uniform convergence is in $[-R,R]$, since it converges in this endpoints?

Generally, if we have a power series

$$S(x) = \sum_{n=0}^\infty a_n x^n$$

(with centre $0$ for simpler notation) whose radius of convergence is $R \in (0,+\infty)$, a majorisation by a geometric series shows that the series is absolutely and uniformly convergent on any interval $[-r,r]$ for $0 < r < R$. When we admit complex arguments, the same majorisation shows absolute and uniform convergence on every closed disk $\overline{D_r(0)} = \{ z\in \mathbb{C} : \lvert z\rvert \leqslant r\}$ for $0 < r < R$:

Since the radius of convergence is $R$, the series converges for $x = c = \frac{r+R}{2}$. The necessary condition for convergence of a series that the terms converge to $0$ then yields a bound

$$\lvert a_n c^n\rvert \leqslant M$$

for all $n$, and hence for $\lvert z\rvert \leqslant r$ we have

$$\lvert a_n z^n\rvert = \lvert a_n\rvert\cdot \lvert z\rvert^n \leqslant \lvert a_n\rvert r^n = \lvert a_n c^n\rvert \biggl(\frac{r}{c}\biggr)^n \leqslant M\biggl(\frac{r}{c}\biggr)^n,$$

which yields

\begin{align} \Biggl\lvert S(z) - \sum_{n = 0}^N a_n z^n\Biggr\rvert &= \Biggl\lvert \sum_{n = N+1}^\infty a_n z^n\Biggr\rvert\\ &\leqslant \sum_{n = N+1}^\infty \lvert a_n z^n\rvert\\ &\leqslant \sum_{n = N+1}^\infty M\biggl(\frac{r}{c}\biggr)^n\\ &= \frac{M\bigl(\frac{r}{c}\bigr)^{N+1}}{1 - \frac{r}{c}} \end{align}

simultaneously for all $z$ with $\lvert z\rvert \leqslant r$. The bound tends to $0$ as $N\to \infty$, showing the uniform convergence on $\overline{D_r(0)}$.

Since polynomials are continuous on all of $\mathbb{R}$ resp. $\mathbb{C}$, if the series converges uniformly on the open interval $(-R,R)$ resp. the open disk $D_R(0)$ - we do not assume that the series converges at either endpoint of the interval or at any boundary point of the disk - it in fact converges uniformly on the closed interval $[-R,R]$ resp. the closed disk $\overline{C_R(0)}$, in particular, the series then converges at both endpoints, resp. at every boundary point of the disk. For, if $S_N(x)$ denotes the $N^{\text{th}}$ partial sum of the series,

$$S_N(x) = \sum_{n = 0}^N a_n x^n,$$

by the continuity of the $S_N$ on all of $\mathbb{R}$ (or $\mathbb{C}$) we have

$$\sup \bigl\{ \lvert S_N(x) - S_K(x)\rvert : \lvert x\rvert < R\bigr\} = \sup \bigl\{ \lvert S_N(x) - S_K(x)\rvert : \lvert x\rvert \leqslant R\bigr\},$$

and so $\lvert S_N(x) - S_K(x)\rvert \leqslant \varepsilon$ for all $N,K \geqslant N_\varepsilon$ and all $x$ with $\lvert x\rvert < R$ implies the same bound for all $x$ with $\lvert x\rvert \leqslant R$.

A word of warning may be in order here: If the power series converges uniformly on $[-R,R]$, that does not imply that it converges uniformly on $D_R(0)$, and it does not imply convergence of the series at any other boundary point of the disk than $R$ and $-R$.

So we have the necessary condition that the series converges at the endpoints of the interval of convergence resp. at the boundary points of the disk of convergence for uniform convergence on the whole interval/disk of convergence, and then the convergence is also uniform on the closure of the interval/disk.

If the series converges absolutely at one endpoint of the interval of convergence resp. at one boundary point of the disk of convergence, then the series converges absolutely and uniformly on the closure of the interval/disk of convergence. The assumption here is that for some $z_0$ with $\lvert z_0\rvert = R$ we have

$$\sum_{n = 0}^\infty \lvert a_n z_0^n\rvert < +\infty.$$

But $\lvert a_n z_0^n\rvert = \lvert a_n\rvert\cdot \lvert z_0\rvert^n = \lvert a_n\rvert R^n$, and thus we have the uniform majorisation

$$\lvert S(z) - S_N(z)\rvert = \Biggl\lvert \sum_{n = N+1}^\infty a_n z^n\Biggr\rvert \leqslant \sum_{n = N+1}^\infty \lvert a_n\rvert R^n$$

for all $z$ with $\lvert z\rvert \leqslant R$.

Things are less obvious if the convergence at an end point of the interval/a boundary point of the disk is conditional and not absolute. To further simplify the notation, we can assume that the radius of convergence is $1$, that the boundary point is $1$, and that the sum of the series at $1$ is $0$. Abel's theorem asserts that we have

$$\lim_{z\to 1} S(z) = S(1)$$

under these circumstances, provided that we restrict $z$ to lie in a region $R_M = \{ z \in \mathbb{C} : \lvert z\rvert < 1, \lvert 1-z\rvert \leqslant M(1-\lvert z\rvert)\}$ for some constant $M\in (0,+\infty)$ (such regions are also called Stolz regions or Stolz angles). Hence we have continuity of the sum function of the series on the augmented Stolz region $R_M \cup \{1\}$, which is a necessary condition for the uniform convergence of the series on the Stolz region.

Abel's theorem does not mention uniform convergence of the power series, but its usual proof quickly yields the uniform convergence on the Stolz region. As in the usual proof of Abel's theorem, we let $s_n$ denote the partial sums of the series, $s_n = \sum_{k=0}^n a_k$, and find - setting $s_{-1} = 0$ -

$$S(z) = \sum_{n = 0}^\infty a_n z^n = \sum_{n = 0}^\infty (s_n - s_{n-1}) z^n = \sum_{n = 0}^\infty s_n z^n - \sum_{m = 0}^\infty s_m z^{m+1} = (1-z)\sum_{n = 0}^\infty s_n z^n$$

for $\lvert z\rvert < 1$. We make the same construction for the partial sums of $S$. Setting $s_n^N = s_{\min \{n,N\}}$, we have

$$S_N(z) = \sum_{n = 0}^N a_n z^n = (1-z)\sum_{n = 0}^\infty s_n^N z^n,$$

and hence in our Stolz region, with $K_N = \sup \{ \lvert s_n\rvert : n \geqslant N\}$,

\begin{align} \lvert S(z) - S_N(z)\rvert &= \lvert 1-z\rvert\cdot\Biggl\lvert \sum_{n = 0}^\infty (s_n - s_n^N) z^n\Biggr\rvert\\ &= \lvert 1-z\rvert\cdot\Biggl\lvert \sum_{n = N+1}^\infty (s_n - s_N)z^n\Biggr\rvert\\ &\leqslant \lvert 1-z\rvert \sum_{n = N+1}^\infty \lvert s_n - s_N\rvert\cdot \lvert z^n\rvert\\ &\leqslant \lvert 1-z\rvert \sum_{n = N+1}^\infty (\lvert s_n \rvert + \lvert s_N\rvert)\lvert z\rvert^n\\ &\leqslant \lvert 1-z\rvert\cdot 2K_N\sum_{n = N+1}^\infty \lvert z\rvert^n\\ &= 2K_N \lvert z\rvert^{N+1}\frac{\lvert 1-z\rvert}{1-\lvert z\rvert}\\ &\leqslant 2MK_N. \end{align}

Since by assumption $s_n \to 0$, for any $\varepsilon > 0$ we can find an $N_\varepsilon$ such that

$$K_{N_\varepsilon} \leqslant \frac{\varepsilon}{2M},$$

and a fortiori $K_N \leqslant \frac{\varepsilon}{2M}$ for $N > N_\varepsilon$, which shows the uniform convergence of the series on $R_M$. By continuity, that implies the uniform convergence on $R_M \cup \{1\}$.

Hence, in the case of real arguments, we have a simple summary:

• If a power series with centre $0$ and radius of convergence $R \in (0,+\infty)$ converges absolutely at $R$ or at $-R$, it converges absolutely and uniformly on $[-R,R]$.
• If a power series with centre $0$ and radius of convergence $R \in (0,+\infty)$ converges (conditionally) at $R$ (at $-R$), it converges uniformly on every interval $[-r,R]$ (on every interval $[-R,r]$) where $0 < r < R$. The convergence is absolute on $(-R,R)$, but not necessarily at the endpoint(s). If the series converges at both endpoints, the convergence is uniform on $[-R,R]$.

For complex arguments, we only have the simple characterisation when we have absolute convergence at one (and hence all) point on the boundary circle. Then the series converges absolutely and uniformly on the closed disk $\overline{D_R(0)}$.

In the case of conditional but not absolute convergence at a point $z_0$ on the boundary, we have uniform convergence on an augmented Stolz region with vertex at $z_0$. But even if the series is conditionally convergent at all boundary points, the convergence is in general not uniform on the whole disk $D_R(0)$. As mentioned in this Math Overflow question, Sierpiński has constructed a power series that converges at every point on the unit circle, but whose sum function is discontinuous on the unit circle (at some points of the unit circle at least, I don't know if it is discontinuous at every point of the boundary circle). That is incompatible with uniform convergence.