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.
 A: The convergence is not uniform on $(-1,1)$, but it is uniform on closed subsets thereof.  We first examine the convergence on the closed subinterval $[-r,r]$ for $0<r<1$.  After that, we show that the convergence fails to be uniform on $(-1,1)$

Let $\epsilon>0$ be given.  Then, we have for $x\in [-r,r]$ for $0<r<1$
$$\begin{align}
\left|S_n(x)-\frac{x}{1-x}\right|&=\frac{|x|^{n+1}}{|1-x|}\\\\
&\le \frac{r^{n+1}}{1-r}\\\\
&<\epsilon
\end{align}$$
whenever $n>\frac{\log(1-r)+\log(\epsilon)}{\log(r)}-1$

Take $\epsilon=1/2$.  Then, for all $N$ we take $x=(1/2)^{1/(n+1)}$ for $0<r<1$ and any $n>N$.  Then, we find that 
$$\begin{align}
\left|S_n(x)-\frac{x}{1-x}\right|&=\frac{|x|^{n+1}}{|1-x|}\\\\
&= \frac{1/2}{1-(1/2)^{1/n+1}}\\\\
&\ge 1/2\\\\
&=\epsilon
\end{align}$$
which negates the uniform convergence on the open interval $(-1,1)$
A: I think you are on the right way, note that:
$$ \left|S_n(x)- \frac{x}{1-x} \right|=\left| -\frac{x^{n+1}}{1-x} \right| \ge | x^{n+1} | > 0 $$
for every $x \in (0,1) $ and every $n \in \mathbb{N}$. Taking least upper bounds basically gives the solution.
A: 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 $(-1,1)$.
A: Note: If a sequence of functions is bounded by a sequence $M_n$, and it converges uniformly to a function $f$, then $f$ is bounded. 
Suppose then that 
$$
S_n(x)=\frac{x(1-x^n)}{1-x}\stackrel{\text{uniformly}}{\rightarrow} \frac{x}{1-x}
$$
Then, as $S_n(x)$ is a finite sum, it is bounded for any given $n$.
However, we have that $\frac{x}{1-x}$ is unbounded on $(-1,1)$, which implies the convergence cannot be uniform.
A: 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.
