Proving uniform convergence of an infinite series using complex function Consider an infinite series
$$f_r(t) = \frac{4}{\pi}\sum_{n=0}^{\infty}\frac{\sin((2n+1)t)}{2n+1}r^{2n+1}$$
where $|t|< \pi$. Let $\delta \in (0, \pi/2)$ and show that $\lim_{r \uparrow 1}f_r(t) = 1$ uniformly on $[\delta, \pi-\delta]$.
Attempt:
Consider $g(z) = \sum_{n=0}^{\infty}\frac{1}{2n+1}z^{2n+1}$. This series converges normally in the open disk $U_1(0)$. It implies by Weierstrass majorant test that this series of functions converges absolutely and locally uniformly. Note that $f_r(t) = \operatorname{Im}(g(z))$, so I can interchange the limit and the sum operator, which yields
$$\lim_{r\uparrow 1}f_r(t) = \frac{4}{\pi}\sum_{n=0}^{\infty}\frac{\sin((2n+1)t)}{2n+1}.$$
But $f_r(t)$ should have converged to a constant $1$. Am I missing a step? Did I make a mistake in my reasoning? Is
$$\frac{4}{\pi}\sum_{n=0}^{\infty}\frac{\sin((2n+1)t)}{2n+1} = 1$$
on $[\delta, \pi - \delta]$?
 A: Write this as $\displaystyle f_r(t) = \sum_{n=0}^\infty a_n(t) r^n$ where $a_{2n}(t) = 0$ and $\displaystyle a_{2n+1}(t) = \frac{4}{\pi} \frac{\sin (2n+1)t}{2n+1}$, and note that
$$f_1(t) = \frac{4}{\pi} \sum_{n=0}^\infty\frac{\sin (2n+1)t}{2n+1} = 1 \text{ for } t\in [\delta,\pi - \delta],$$
as this is the Fourier series for a square wave.
What matters in proving that $\lim_{r \to 1-}f_r(t) = f_1(t)  = 1$ is the uniform convergence of the series representation of  $f_r(t)$ for $t \in [\delta,\pi - \delta]$, which follows from the Dirichlet test.  The proof is a generalization of that of Abel's limit theorem.
Using the Cauchy product formula, we have
$$(1-r)^{-1}\sum_{n=0}^\infty a_n(t) r^n = \sum_{n=0}^\infty r^n\sum_{n=0}^\infty a_n(t) r^n = \sum_{n=0}^\infty \sum_{k=0}^n a_k(t) r^k r^{n-k} = \sum_{n=0}^\infty S_n(t) r^n, $$
where $S_n(t) = \sum_{k=0}^n a_k(t)$.
Thus, $f_r(t) = (1-r)\sum_{n=0}^\infty S_n(t) r^n$, and
$$f_r(t) - f_1(t) = (1-r)\sum_{n=0}^\infty S_n(t) r^n - f_1(t) (1-r)\sum_{n=0}^\infty r^n = (1-r)\sum_{n=0}^\infty[S_n(t)-f_1(t)]r^n$$
Since $S_n(t) \to f_1(t)$ uniformly, given $\epsilon > 0$ there exists $N_\epsilon\in \mathbb{N}$ such that $|S_n(t) - f_1(t)| < \epsilon/2$ when $n \geqslant N_\epsilon$ for all $t \in [\delta, \pi - \delta]$.
Hence, for $0 < r < 1$ and with $M = \sup\{|S_n(t) - f_1(t)|: n=0,\ldots, N_\epsilon-1, \,\, t\in[\delta,\pi-\delta]\}$,
$$\begin{align} |f_r(t) - f_1(t) | &\leqslant (1-r)\sum_{n=0}^{N_\epsilon-1}|S_n(t)-f_1(t)|r^n +  (1-r)\sum_{n=N_\epsilon}^\infty|S_n(t)-f_1(t)|r^n \\ &\leqslant (1-r)N_\epsilon M +  \frac{\epsilon}{2}(1-r)\sum_{n= N_\epsilon}^\infty r^n \\ &\leqslant (1-r)N_\epsilon M + \frac{\epsilon}{2} (1-r) \frac{r^{N_\epsilon}}{1- r} \\ &\leqslant (1-r)N_\epsilon M + \frac{\epsilon}{2} \end{align}$$
Taking $\eta = \epsilon / (2N_\epsilon M)$, we have $|f_r(t) - 1| = |f_r(t) - f_1(t)| < \epsilon$ when $1- \eta < r <1 $ for all $t \in [\delta,\pi - \delta]$. Since $\eta$ does not depend on $t$, the convergence is uniform.
