First question:
I want to show that this series : $$\sum _{k=0}^{\infty }\:\frac{x^2}{\left(1+x^2\right)^k}$$
converges uniformly in $[a,\infty)$ for $a>0$.
I thought about the M test and i get : $\frac{x^2}{\left(1+x^2\right)^k}\:\le \frac{x^2}{\left(x^2\right)^k}\:=\frac{1}{\left(x^2\right)^{k-1}}\le \frac{1}{\left(a^2\right)^{k-1}}$
therefore, the function-sum converge. It is good enough?
Second question: For the sum above, how can i show that the sum does not converge uniformly in $[-a,a]$ for any $a>0$?
For the second question: One way is to show the convergence is not uniform on $[-a,a]$, $a>0$ is to show that the limit function is not continuous at $0$.
You can explicitly evaluate the limit function $f$:
$f(0)=0$.
For $x\ne0$, $f(x)=1+x^2$. (Use the formula for the sum of a Geometric series here.)
For your first question, your approximation is too loose. What if $|a|<1$? Then the relevant sum in the $M$-test does not converge.
But, you can use the formula for the sum of a Geometric series to show $\sum\limits_{k=n}^\infty {x^2\over(1+x^2)^k}$ is uniformly small for large $n$ (I calculated the sum is ${1\over (1+x^2)^{n-1}}\le {1\over (1+a)^{n-1}}$ for $x\in[a,\infty)$).