I'm trying to solve the following problem about the uniform convergence of a series of functions involving the function arctan(x).
$$ \sum_{i=1}^\infty \left[1-\frac{n}{x^{2n}}\arctan\left(\frac{x^{2n}}{n}\right)\right] $$
I have to study the pointwise and uniform convergence of the series for $$x\neq0$$ I started to check where it is possible to have pointwise convergence by claculating this limit $$ \lim_{n \to \infty} \left[1-\frac{n}{x^{2n}}\arctan\left(\frac{x^{2n}}{n}\right)\right] $$ It turns out that for $ -1\le x \le 1 , x\neq 0 $ the limit is $ 0$ so it is possible to have pointwise convergence in $ [-1,1], x\neq 0$. For $ \vert x\vert \gt 1$ the limit is 1, so there's no way the series can have pointwise convergence in $ (-\infty , -1) \cup (1 , +\infty)$. I decided to check all of this stuff so I know where to look for uniform convergence better, and once I find it I will also know where the pointwise convergence is , correct?
I started by observing that $\left[1-\dfrac{n}{x^{2n}}\arctan\left(\dfrac{x^{2n}}{n}\right) \right]\ge 0 $ for $-1 \le x \le 1$ . The problem is that I can't find a way to bound this function with a sequence $Mn$ such that $ \sum\limits_{i=1}^\infty Mi $ converges. Any tips??
Using Taylor's theorem, $$ \frac{x^{2n}}{n}-\frac{x^{6n}}{3n^3} \le \arctan\left(\frac{x^{2n}}{n}\right)\le\frac{x^{2n}}n $$ so $$0\le\left[1-\frac{n}{x^{2n}}\arctan\left(\frac{x^{2n}}{n}\right)\right] \le1-\frac{n}{x^{2n}}\left(\frac{x^{2n}}{n}-\frac{x^{6n}}{3n^3}\right)= \frac{x^{4n}}{3n^2} \le \frac{1}{3n^2}.$$ If $ M_n=\frac{1}{3n^2} $ then $ \sum_{n=1}^\infty M_n$ converges thus the series converges uniformly.