Let $g : (−1, 1) → R$ be the function $g(x ) := \frac{x}{(1−x)}$. With the notation as in $(b)$ show that the partial sums $\Sigma _{n=1}^N f^n$ converges pointwise as $N → ∞$ 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 $[−1, 1]$?
Notation from part $(b)$ is the following. For each integer $n ≥ 1$, let $f^n : (−1,1) → R$ be the function $f^n(x) := x^n$.
On
We know that $$\sum_{n=1}^{N}f^n=x+x^2+...+x^N=x\dfrac{x^n-1}{x-1}$$therefore for any $x\in(-1,1)$ we have $$|\sum_{n=1}^{N}f^n(x)-g(x)|=|\dfrac{x^{N+1}}{1-x}|=\dfrac{|x|^{N+1}}{1-x}$$for convergence we must use the definition$$\forall\epsilon>0\qquad,\qquad \exists M\qquad,\qquad N>M\to|\sum_{n=1}^{N}f^n(x)-g(x)|<\epsilon$$which yields to$$\dfrac{|x|^{M}}{1-x}<\epsilon\to \\|x|^{M}<\epsilon(1-x)$$since $-1<x<1$ we have $0<1-x<2$ therefore$$|x|^{M}<\epsilon(1-x)<2\epsilon$$and finally$$M>\log_{\frac{1}{|x|}}^{2\epsilon}$$and$$N=\log_{\frac{1}{|x|}}^{2\epsilon}$$so there exists some $N$ for any $x$ but dependent to $\epsilon$ therefore the set of functions tend to $g(x)$ but not uniformly.
the red curve is $g(x)$ and blues are the functions $\sum_{n=1}^{N}f^n(x)$
For closed interval $[-1,1]$, then $f_{n}(1)=1\rightarrow 1$ as $n\rightarrow\infty$. But $f_{n}(-1)=(-1)^{n}$ does not converge. The $g$ is not defined at $x=1$. Even so, the convergence is not uniform if we set $h(x)=\dfrac{x}{1-x}$ for $x\in(-1,1)$, $h(1)=1$, and consider the convergence on $(-1,1]$.
About the issue of uniform convergence: Assume that it were, then \begin{align*} \left|\sum_{n=1}^{N}f_{n}(x)-g(x)\right|<1,~~~~x\in(-1,1),~~~~N\geq L \end{align*} for big $L>0$, then \begin{align*} \dfrac{x-x^{L+1}}{1-x}>\dfrac{x}{1-x}-1,~~~~x\in(-1,1), \end{align*} taking $x\rightarrow 1^{-}$ and using L'Hopital on the left, one gets $\dfrac{x-x^{L+1}}{1-x}\rightarrow(L+1)$ by the right blows up, a contradiction.