Proving that a sequence of functions converge uniformly 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.

First, we simplify the expression.  Note that:
\begin{align}
\left|\frac{x(1-x^n)}{1-x} - \frac{x}{1-x}\right| &= \left|\frac{x(1-x^n) -x}{1-x}\right| \\
&=\left|\frac{-x^{n+1}}{1-x}\right| \\
&= \frac{x}{1-x}\cdot x^{n}
\end{align}
We want $\left|S_n(x) - \frac{x}{1-x} \right| < \epsilon$ for given $\epsilon$ and $n \ge N$ for some large enough $N$.  For this to be the case, we would need an $N$ such that, for all $n\ge N$:
\begin{align*}
&\frac{x}{1-x}x^n < \epsilon \\
\iff & x^n < \epsilon\frac{1-x}{x} \\
\iff & n > \frac{\ln\left(\epsilon\frac{1-x}{x}\right)}{\ln x} = \frac{\ln \left(\epsilon(1-x)\right) - \ln x}{\ln x}
\end{align*}
But, as $x\to 1$ from the left, $\frac{\ln \left(\epsilon(1-x)\right) - \ln x}{\ln x} \to \infty$.  So, no such $N$ exists.
Therefore, $S_n(x)$ does not converge uniformly.
