TITLE: Proving that a sequence of functions converge uniformly
QUESTION [6 upvotes]: I have a sequence of functions:
$$S_n(x) = x\frac{1-x^n}{1-x}, \text{ where } x \in (-1, 1)$$
It is known that:
$$\lim_{n \to \infty}S_n(x) = \frac{x}{1-x}$$
thus $S_n(x)$ converges pointwise. I am to prove that is does not converge uniformly. I tried to estimate:
$$\left|S_n(x) - \frac{x}{1-x}\right|$$ using an $x \in (-1, 1)$ but it didn't work.
I would appreciate any help.

REPLY [3 votes]: So, you have $$
\left\lvert S_n(x) - S(x)\right\rvert
=
\left\lvert \frac{1-x^n}{1-x} - \frac{1}{1-x}\right\rvert=
\left\lvert \frac{x^n}{1-x}\right\rvert
 $$
for all $x\in(-1,1)$.
Now, consider $x_n \stackrel{\rm def}{=} 1-\frac{1}{n}\in(-1,1)$. We have
$$
\sup_{x\in (-1,1)}\left\lvert S_n(x) - S(x)\right\rvert \geq 
\left\lvert S_n(x_n) - S(x_n)\right\rvert
=\left\lvert \frac{x_n^n}{1-x_n}\right\rvert
= \frac{(1-\frac{1}{n})^n}{\frac{1}{n}} = n \left(1-\frac{1}{n}\right)^n \xrightarrow[n\to\infty]{}\infty
$$
so $\sup_{x\in (-1,1)}\left\lvert S_n(x) - S(x)\right\rvert\not\xrightarrow[n\to\infty]{}0$, meaning there is no uniform convergence on $(-1,1)$.