Prove non-uniform convergence of $f_n(x) = 1+x+\cdots+x^n$ Background:
I have, in a previous problem, shown that $f_n(x) = 1+x+\cdots+x^n$ converges pointwise toward $f(x) = \frac1{1-x}$ given $x\in(-1, 1)$.
Problem:
Show that the convergence is not uniform on $(-1, 1)$, but that it is uniform on every sub-interval $(-a, a)$ where $a\in(0,1)$.
My progress:
I'm having a hard time seeing the difference between $(-1, 1)$ and $(-a, a) : a\in(0, 1)$. It seems to me like the extreme case of the latter, where $a\to1$ would give the exact same interval as the former. 
Barring that, I haven't been able to find a way to start the problem. Any help appreciated!
 A: Hint
Let
$$\mu_n(x)=|f_n(x)-f(x)|=\frac{|x^{n+1}|}{1-x}$$
and we have uniform convergence of $(f_n)$ to $f$ on a set $A$ if $\sup\limits_{x\in A}\mu_n(x)\xrightarrow{n\to\infty}0.$
Can you take it from here?
A: If $x\in(-a,a)$, then $|x|<a$ and $|x|^n<a^n$. Since $\sum_{n=0}^\infty a^n<\infty$, Weierstrass $M$-test implies that $\sum_{n=0}^\infty x^n$ converges uniformly on $(-a,a)$.
To prove that the convergence is not uniform on $(-1,1)$, I will show that it does not satisfy the uniform Cauchy's criterium: for any $N\in\mathbb{N}$
$$
\sum_{n=N+1}^{2N}x^n\ge N\,x^{2N},\quad 0\le x<1.
$$
Thus
$$
\sup_{x\in(-1,1)}\Bigl|\sum_{n=N+1}^{2N}x^n\Bigr|\ge N.
$$
A: Let $S_n(x)=\sum_{k=0}^n x^k$. If $m>n$ and $r\in(0,1)$,
$$|S_m(x)-S_n(x)|\leq \sum_{k=n+1}^n|x|^k\leq r^k.$$
You know that for $x=r$ the series is absolutely convergent and thus a Cauchy sequence. Therefore $(S_n)$ is a sequence of continuous function on $[-r,r]$ and a Cauchy sequence for the norm of the maximum. Therefore, $(S_n)$ converge uniformly to a function $f$. Now take $a\in (0,1)$. Then, there exist $r>0$ such that $(-a,a)\subset [-r,r]\subset (-1,1)$. Since the $(S_n)$ converge uniformly on $[-r,r]$, it converge uniformly on $(-a,a)$.
A: We have
$$f_n(x) = \sum_{k=0}^n x^k = \frac{1-x^{n+1}}{1-x}, $$
which clearly converges to $$f(x)=\frac1{1-x}$$ for $x\in(-1,1)$.
But for any $M$, we can choose $n$ such that $1-\frac1n<\frac1{2M}$, so
$$\sup_{x\in(-1,1)}f_n(x)\geqslant f_n\left(\frac1n\right)=\frac{1-\left(\frac 1n\right)^{n+1}} {1-\frac1n}>2M\left(1-\left(\frac 1n\right)^{n+1}\right)>M $$
(since $\left(\frac1n\right)^{n+1}<\frac12$ for $n\geqslant 2$). It follows that $f_n$ does not converge uniformly.
However, if $0<a<1$ then 
$$|f_n(x)| = \frac{|1-x^{n+1}|}{|1-x|}\leqslant \frac1{1-a}, $$
so that $f_n$ converges uniformly on $(-a,a)$.
A: The partial sums $S_n(x)=\sum_{k=0}^n x^k$ are polynomials, bounded in any finite interval. The sum  $f(x)=1/(1-x)$ is bounded in $(-a,a)$, $a>1$ but unbounded in $(-1,1)$. Bottom line: $f-S_n$ is unbounded in $(-1,1)$ and can't be uniformly small in this interval.
A: If $f_n$ are uniformly continuous functions on a set $A$ and converge uniformly to $f$ then $f$ is uniformly continuous on $A$. 
