Why is $\sum_{k = 0}^{\infty}(-x^{2})^{k}$ not uniformly convergent on $(-1,1)$ but it is on $[-r,r] \subset (-1,1)$? Clarification Note: I'm aware there are similar postings about the same idea, but I'm attempting to develop understanding of the concept through my explanation of what is happening.
Why is $\sum_{k = 0}^{\infty}(-x^{2})^{k}$ not uniformly convergent on $(-1,1)$  but it is on $[-r,r] \subset (-1,1)$?. I'm trying to understand precisely why we cannot use the M-Test in such situations.
From the notes I took in class we demonstrated this by showing that the series is not uniformly Cauchy.
Proof
Let $S_{n} = \sum_{k = 0}^{n}(-x^{2})^{k}$
Therefore: $$\|S_{n+m}(x) - S_{n}(x)\|_{\infty} = \sup_{x \in (-1,1)}\Bigg|\sum_{k = n}^{n+m}(-x^{2})^{k}\Bigg|$$
If we let $$x_{0} := \Bigg(\frac{1}{2}\Bigg)^{\frac{1}{2N}} \in (-1,1)$$ and $n + m \leq N$, where N is the integer that satifies the Cauchy sequence criterion for $n,m \geq N$.
Then $$\sup_{x \in (-1,1)}\Bigg|\sum_{k = n}^{n+m}(-x^{2})^{k}\Bigg| \geq \Bigg|\sum_{k = n}^{n+m}(-1)^{K}\Bigg(\frac{1}{2}\Bigg)^{\frac{k}{N}}\Bigg|$$
Interpretation:
So if I am understanding this properly, this shows that we can always choose a $N > 0$ such that if $n + m \geq N$ then $$x_{0} := \Bigg(\frac{1}{2}\Bigg)^{\frac{1}{2N}} \in [-r,r]$$, this is because $\frac{1}{2N} > \frac{1}{n+m}$ and in which case our series would converge uniformly. But on the other hand if we choose$N$ such that $n+m < N$ then $$x_{0} := \Bigg(\frac{1}{2}\Bigg)^{\frac{1}{2N}} \notin [-r,r]$$ but is in $(-1,1)$ because $\frac{1}{2N} < \frac{1}{n+m}$.  As such there will always be an $\epsilon$ such that $$\|S_{n+m}(x) - S_{n}(x)\|_{\infty} > \epsilon$$.  But I thought the negation of uniformly Cauchy in this case would be:
"There exists $\epsilon > 0$ such that for all $N > 0$, $\|S_{n+m}(x) - S_{n}(x)\|_{\infty} > \epsilon$ for all $x \in (-1,1)$ and $n, m > N$" ?  So how would I be able to choose my $N$ such that $n+m < N$ ?

We will use the following result to prove that the series $\sum_{k=0}^\infty(-x^2)^k$ does not converge uniformly.  
Result: Let $D$ be a subset of a metric space $(X,d)$, and a series of function $\sum{f_k}$ be uniformly convergent on $D$ to a function $f$. Let $x_0$ be a limit point of $D$ and $\lim_{x\rightarrow x_0}f_k(x)=a_k$. Then the series $\sum{a_k}$ converges, $\lim_{x\rightarrow x_0}f(x)$ exists and $\lim_{x\rightarrow x_0}f(x)=\sum{a_k}$.
We know that for $x\in(-1,1)$, $\sum_{k=0}^\infty(-x^2)^k=\frac{1}{1+x^2}$ pointwise. Now if the convergence is uniform, then as $1$ is a limit point of $(-1,1)$ and $\lim_{x\rightarrow 1}(-x^2)^k=(-1)^k$, so by the stated result $\sum_k(-1)^k$ exists, which is not true.    
To prove the second part, observe that if $x\in[-r,r]\subset(-1,1)$, then $|(-x^2)^k|\leq r^{2k}$ and $\sum_kr^{2k}$ converges as $0\leq r<1$. Therefore by so called Weierstrass' M-test, $\sum_{k=0}^\infty(-x^2)^k$ converges uniformly on $[-r,r]\subset(-1,1)$.
