I was trying to prove that $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}x^2}{(1 + x^2)^n}$ converges uniformly on $\Bbb{R}$.
was trying to use $M$ test, in which I am trying to bound $|f_{n}(x)| < M_{n} $ such that $M_{n}$ converges uniformly.
I thought of using $|\frac{(-1)^{n}x^2}{(1+x^2)^n}| < |\frac{(-1)^n x^2}{(1 + nx^2)}| < |\frac{(-1)^n x^2}{nx^2}| < |\frac{(-1)^n}{n}|$ which converges absolutely and hence The series $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}x^2}{(1 + x^2)^n}$ converges absolutely for all $x$.
We can analyze the uniform convergence here without finding the sum explicitly. This is good practice for problems where the sum cannot be obtained in closed form.
The M-test is problematic because the series with absolute values of the terms
$$\sum_{n=1}^\infty \frac{x^2}{(1+x^2)^n}$$
is not uniformly convergent on $\mathbb{R}$.
However, on an interval like $[a,\infty)$ where $a > 0$ we have uniform convergence by the M-test , since for $n \geqslant 2$,
$$\left|\frac{(-1)^{n-1}x^2}{(1+x^2)^n}\right|= \frac{x^2}{(1+x^2)^n} \leqslant \frac{x^2}{\frac{n(n-1)}{2}x^4} \leqslant \frac{2}{n(n-1)a^2},$$
and $\sum_{n \geqslant 2} [n(n-1)]^{-1} = 1$.
Nevertheless, the alternating series is uniformly convergent on $\mathbb{R}$ by Dirichlet's test. Note that $\sum_{n=1}^m (-1)^{n-1}$ is uniformly bounded for all $m \in \mathbb{N}$ and $x \in \mathbb{R}$, and $x^2/(1+x^2)^n \to 0$ as $n \to \infty$ monotonically and uniformly (as shown by your $1/n$ bound).