How do I prove that $f_n(x)=\frac{1-x^n}{1-x}$ converges pointwise and uniformly?

Question

I'm trying to check if the function sequence $f_n(x)=\frac{1-x^n}{1-x}$, $x \in (-1,1)$ converges pointwise and uniformly.

If I had the sequence of $g_n(x)= \frac{1-x^n}{1-x^n}$ and $x \in (-1,1)$ i know that $\lim_{n \to \infty } g_n(x) = 1 $ but now how the denominator's $x$ behaves?

Thank you!

Answer 1

$\dfrac{1 - x^n}{1 - x} = x^{n - 1} + x^{n - 2} + .. + 1$, so $f_n(x)$ is the partial sum of a geometric series. One can then show that $\{f_n\}$ converges pointwise and $f_n(x) \rightarrow \sum_{k = 0}^{\infty} x^k = \dfrac{1}{1-x}$.

Answer 2

It converges pointwise to $\frac1{1-x}$ because, for $\lvert x\rvert<1$, $$\lim_{n\to\infty} x^n=0$$

Precisation: Specifically the denomiantor is constant in $n$. So, as much as the pointwise limit is concerned, $$\lim_{n\to\infty} \frac{1-x^n}{1-x}=\frac{\lim\limits_{n\to\infty}(1-x^n)}{1-x}$$

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It does not converge uniformly to $\frac1{1-x}$ on $(-1,1)$ because, for all $n$, $$\sup_{x\in(-1,1)}\left\lvert f_n(x)-\frac1{1-x}\right\rvert=\sup_{x\in(-1,1)} \frac{\lvert x\rvert^n}{1-x}\stackrel{x=-1+\frac1n}{\ge}\frac{\left(1-\frac1n\right)^n}{1/n}\stackrel{n\to\infty}{\longrightarrow}\left[\frac {e^{-1}}0\right]=\infty$$