Uniform convergence of sum in given interval Let $a>0$ and observe $$\ln(a) - \sum_{n=1}^{\infty} \frac{1}{n}a^{-n}x^n$$
Show that the above sum converges uniformly in $[-a,b]$ with $-a<b<a$
Here's my current approach: For $x \in (-a,b]$ we see that $$\left|\sum_{n=1}^{\infty} \frac{1}{n}a^{-n}x^n\right|\le\left|\sum_{n=1}^{\infty} a^{-n}x^n\right|=\left|\sum_{n=1}^{\infty} \left(\frac{x}{a}\right)^n\right|$$ which a geometric sequance, that do convergence since $\left(\frac{x}{a}\right)^n<1 \ \forall x$ in the given interval. Therefore, by the Weierstrass-M test, the original sum converges uniformly. But here is my question, is this correct, and if so, how can I show uniform convergence in the endpoint $-a$
In a previous task, I've shown that the sum converges pointwise in $[-a,a)$
 A: Put $f_n(x) = \ln (a) - \sum_{i=1}^{n} \frac{1}{n}a^{-n}x^n $. You've showed that $f_n$ converges uniformly in $(-a, b]$. On the other hand, \begin{align}\lim_{n \to \infty} f_n(-a)&=\ln(a) - \sum_{n=1}^{\infty}\frac{1}{n}a^{-n}(-a)^n  \\ &= \ln(a) + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \\&= \ln(a)+\ln(2)=\ln(2a)\end{align}
Now define $f : [-a, b] \to \mathbb{R}$ by $f(x)=\lim_{n\to \infty}f_n(x)$ for $-a < x \leq b$ and $f(-a)=\ln(2a)$. We shall show that $f_n$ converges uniformly to $f$ in $[-a, b]$.
Let $\epsilon>0$ be given. Since $f_n$ converges uniformly to $f$ in $(-a, b]$, there exists $N_1 \in \mathbb N$ such that $n \geq N_1$ implies $|f_n(x)-f(x)| < \epsilon $ for all $x\in (-a, b]$. Moreover, $\lim_{n \to \infty}f_n(-a) = f(-a)$ guarantees the existence of $N_2 \in \mathbb N$ such that $n \geq N_2$ implies $|f_n(-a) - f(-a)| < \epsilon$. Thus $n \geq \max(N_1, N_2)$ implies $|f_n(x) - f(x)| < \epsilon$ for all $x \in [-a, b]$. That is, $f_n$ converges uniformly to $f$ in $[-a, b]$.
