Interval of convergence and integration of a power series The $\arctan(x)$ can be expanded as a MacLaurin series starting from the integral
$$\arctan(x) = \int \frac{1}{1 + x^2} \mathrm{d}x$$
and using
$$\frac{1}{1 + x^2} = \sum_{n = 0}^{\infty} (-1)^n x^{2n}$$
as suggested in this answer. This series converges for $x \in (-1,1)$, but, after integration, it can be shown that the resulting MacLaurin series
$$\arctan(x) = \sum_{n = 0}^{\infty} (-1)^n \frac{x^{2n + 1}}{2n + 1}$$
converges for $x \in \left[ -1, 1 \right]$.
The integration of a series is possible only when the series is evaluated within its interval of convergence, $x \in (-1,1)$: the MacLaurin series for $x = \pm 1$ shouldn't even be considered, because the above result for $\arctan(x)$ should not be available!


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*Why instead, here two more points ($x = \pm 1$) can be added to the interval of convergence?


Integration does not change the radius of convergence of a series. 


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*What are then the differences between the interval of convergence and the radius of convergence? Do $(-1,1)$ and $\left[ -1, 1 \right]$ correspond to the same radius of convergence?



I read also this question, answer and comments am I am not familiar with Cauchy-Hadamard Radius Formula.
A comment here states (given an interval of convergence $(a - R, a + R)$):

The issue of convergence at the points $x= a ± R$ is independent of the convergence within the interval $(a−R,a+R)$.

My questions above essentially are: why?

In general, you won't be able to say things about endpoints of an interval after integrating.
In this case, however, we can just check that the sum converges at the endpoints directly; if $x = \pm 1$, then $x^{2n+1} = x$, so the sum becomes
$$x \sum_{n=0}^\infty \frac{(-1)^n}{2n+1},$$
which converges because $\frac{(-1)^n}{2n+1}$ is an alternating series with $\frac{1}{2n+1}$ decreasing and going to $0$.
