The series which has been used here is of a special kind called power series of the form $$\sum_{n=0}^{\infty}a_nx^n$$ and such series have associated with them a number $R\geq 0$ called radius of convergence. The series is convergent if $|x|<R$ and divergent if $|x|>R$. The case for $|x|=R$ has to be handled for each series individually. Also there are cases when $R=\infty$ so that the series converges for any given value of $x$.
A key property of such series is that they can be integrated or differentiated term by term inside the region of convergence $|x|<R$ (with new series having same radius of convergence). However there can be cases (like in current question) where the given series is convergent only when $|x|<R$ but the new series after integration also converges for $|x|=R$.
It is much simpler to handle your specific series directly by noting that $$\arctan x=\int_{0}^{x}\frac{dt}{1+t^2}=\int_{0}^{x}\left(1-t^2+t^4-\dots+(-1)^{n-1}t^{2n-2}+(-1)^{n}\frac{t^{2n}}{1+t^2}\right)\,dt$$ which leads to $$\arctan x=x-\frac{x^3}{3}+\frac{x^5}{5}-\cdots +(-1)^{n-1}\frac{x^{2n-1}}{2n-1}+(-1)^nR_n\tag{1}$$ where $$R_n=\int_{0}^{x}\frac{t^{2n}}{1+t^2}\,dt$$ The identity $(1)$ holds for all $x\in\mathbb {R} $ but the expression $R_{n} \to 0$ as $n\to\infty $ only when $|x|\leq 1$ (prove this!) and hence the identity $$\arctan x=\sum_{n=0}^{\infty} (-1)^n\frac{x^{2n+1}}{2n+1}$$ is valid only when $|x|\leq 1$.
