Let $g: (-1, 1) \to \mathbb{R}$ be the function $g(x) : = x/(1-x)$. Let $f^{(n)}:(-1, 1) \to \mathbb{R}$ be the function $f^{(n)}(x) : = x^n$. Show that the partial sums $\sum_{n=1}^N f^{(n)}$ converges pointwise as $N \to \infty$ to $g$, but does not converge uniformly to $g$, on the open interval $(-1, 1)$. What would happen if we replaced the open interval $(-1, 1)$ with the closed interval $[-1, 1]$?
Let $F^{(N)} = \sum_{n=1}^N f^{(n)}$. I can show that $F^{(N)}$ pointwise converges to $g$ on $(-1, 1)$, but not on $[-1, 1]$.
For uniform convergence, we first have that $|F^{(N)}(x) - g(x)| = |\frac1{1-x}||{x^{N+1}}|$. Consider the sequence $x_N = 1- \frac1N \in (-1, 1)$ for all $N >1$. Then, $|F^{(N)}(x_N) - g(x_N)| = |N||1-\frac1N|^{N+1}$. Which $\epsilon$ should I pick so that it is less than $|N||1-\frac1N|^{N+1}$ for all $N>1$? How can I finish the proof?
We have $$\sup_{x\in(-1,1)}\left|g(x)-\sum_{n=1}^Nx^n\right|=\sup_{x\in(-1,1)}\left|\frac{x^{N+1}}{1-x}\right|=\infty $$ (think about the limit of $x^{N+1}/(1-x)$ as $x\to 1^-$) which means we do not have uniform convergence on $(-1,1)$. As for the closed interval $[-1,1]$, we don't even have pointwise convergence let alone uniform. Simply take $x=1$: the partials sums $\sum_{n=1}^N 1=N$ clearly diverge.