sequences and series - Prove that $xmapsto sum^{infty}_{n=0}f_n(x)=sum^{infty}_{n=0}frac{x^2}{(1+x^2)^n}$ does not converge uniformly on $[-1,1]$

For each $n,$ we define \begin{align} f_n:\Bbb{R}\to \Bbb{R} \end{align}
\begin{align} x\mapsto f_n(x)=\frac{x^2}{(1+x^2)^n}\end{align}
We consider the function
\begin{align} f:\Bbb{R}\to \Bbb{R} \end{align}

\begin{align} x\mapsto f(x)=\sum^{\infty}_{n=0}f_n(x)=\sum^{\infty}_{n=0}\frac{x^2}{(1+x^2)^n}\end{align}

I want to show that the series does not converge uniformly on $[-1,1]$ but I'm finding it difficult to do that.

First of all, I considered the Weierstrass M test.

MY TRIAL

\begin{align}\left|f_n(x)\right|=\left|\frac{x^2}{(1+x^2)^n}\right|\leq \frac{1}{(1+x^2)^n} ,\;\;\forall \;x\in[-1,1],\;\forall\;n\in\Bbb{N}\end{align}

I'm also thinking that the $\beta_n=\sup\limits_{x\in[-1,1]}|\sum^{n}_{i=0}f_i(x)-\sum^{\infty}_{i=0}f_i(x)|$ approach could be very helpful too!

Answer

The convergence is not uniform.

For $x \ne 0$ this is a geometric series:

$$\sum^{\infty}_{n=0}\frac{x^2}{(1+x^2)^n} = x^2\cdot \frac{1}{1-\frac1{1+x^2}}= x^2+1$$

and for $x = 0$ the sum is $0$.

Therefore

$$f(x) = \begin{cases}
x^2+1, &\text{ if } x \ne 0\\
0, &\text{ if }x = 0
\end{cases}$$

Since $f$ is not continuous, the convergence cannot be uniform.