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.

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    $\begingroup$ Suppose $\lbrace S_{n}(x) \rbrace$ converges to $1+x^2$ pointwise. Then $S_{n}(0) = 0 \rightarrow 1$ as $n \rightarrow \infty$. But this is false. $\endgroup$ – UDAY PATEL Aug 17 '17 at 3:54
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    $\begingroup$ For the Dini's theorem to hold you need the limiting function to be continuous, the problem here is that you identified the limiting function incorrectly $\endgroup$ – clark Aug 17 '17 at 5:04
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    $\begingroup$ Also you know that the uniform limit of continuous functions is continuous $\endgroup$ – clark Aug 17 '17 at 5:05

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 $S$.


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