I want to prove that the following function series is uniform convergent in $\mathbb{R}$:
$$ \sum_{n=0}^\infty\bigg(\frac{-1}{1+x^2}\bigg)^n\cdot x^2 $$
I've already proved that this series is pointwise convergent but I'm kind of stuck over the epsilon-argument for the uniform case.
On
This is always an alternating series, because $\dfrac{-1}{1+x^2}<0$, and the absolute values are monotone decreasing, because $\dfrac1{1+x^2}<1$. So the remainder of the $n$-th partial sum is less than the last term, i.e. $\le x^2\left(\dfrac1{1+x^2}\right)^n=\left(1-\dfrac1{1+x^2}\right)\left(\dfrac1{1+x^2}\right)^{n-1}$, and it's easy to see (with the AM-GM-inequality, say) that this is $\le\dfrac1{n-1}\left(\dfrac{n-1}n\right)^n$, uniformly in $x$.
You can explicitly compute the partial sums of the series (since it is a geometric series). Also, uniform convergence for $|x| \geq 2$ is very easy: just use M-test. For $|x|\leq 2$ uniform convergence of the series is equivalent to the fact that $\frac {x^{2}} {(1+x^{2})^{n}} \to 0$ uniformly. [This follows by the computation of the partial sums]. Let $\epsilon >0$ and choose m such that $(1+\epsilon)^{-m} <\epsilon$. For $|x|\leq \epsilon$ we have $\frac {x^{2}} {(1+x^{2})^{n}} \leq |x|^{2} \leq \epsilon ^{2}$ and for $\epsilon <|x|\leq 2$ we have $\frac {x^{2}} {(1+x^{2})^{n}} \leq 4 \frac 1 {(1+\epsilon)^{n}} <4\epsilon$ if $n \geq m$.