How to show the series $\sum\limits_{n=0}^\infty \frac{{x^2}}{(1+x^2)^n}$ is uniformly convergent over $[-1,1]$.
My try: say $S_n(x) = \sum\limits_{k=0}^n \frac{{x^2}}{(1+x^2)^k}$ and {$S_n(x)$} is a sequence of monotonic increasing continuous functions and converges to $(1+x^2)$. So by Dini's theorem we can say $\sum\limits_{n=0}^\infty \frac{{x^2}}{(1+x^2)^n}$ will be uniformly convergent over $[-1,1]$.
If I went wrong anywhere please mention it.
The series is not uniformly convergent. Consider the partial sum: $$S_n(x) = \sum_{k=0}^n \frac{x^2}{(1+x^2)^k} = 1+x^2-\frac{1}{(1+x^2)^n}$$ for all $x\in [-1, 1]$. Let $$S(x) = \begin{cases} 1+x^2,& x\neq 0 \\ 0,& x = 0 \end{cases}$$ be the pointwise limit of $S_n$. Then, $$\|S-S_n\|_{\infty} = \left\|\frac{1}{(1+x^2)^n}\right\|_{\infty} = 1$$ for all $n$, as $\lim_{x\to 0} \frac{1}{(1+x^2)^n} = 1$ for all $n$. As $\|S-S_n\|_{\infty}\not\to 0$, $S_n$ does not converge uniformly to $S$.