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!


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. 


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?
On the request of asker I am writing a full blown answer based on my comments.

Let us then note that a series of the form $$\sum_{n=0}^{\infty} a_nz^n$$ is called a power series in variable $z$. Here $a_n, z$ are complex numbers.

A fundamental feature of such a series is the existence of a non-negative real number $R$ called the radius of convergence and it has the following properties:


The series $\sum_{n=0}^{\infty} a_nz^n$ converges absolutely for $|z|&lt;R$. 
The series $\sum_{n=0}^{\infty} |a_nz^n|$ diverges to $\infty $ for $|z|&gt;R$.


There exist special and useful cases when $R=\infty$ and then the series converges everywhere in the complex plane. Functions represented by such series are called entire functions.

The region $|z|&lt;R$ is called the circle of convergence. In case of real variable $z$ this gets replaced by interval of convergence $(-R, R) $.

A power series can be always be integrated as well as  differentiated term by term to lead to another power series with the same radius of convergence.

The convergence of a power series at boundary points $|z|=R$ requires special analysis on a case by case basis and there are no general theorems without any additional hypotheses to confirm convergence or divergence at these points.



Some textbooks define a power series in the form $\sum a_n(z-a) ^n$ and to handle these we can just replace $z$ with $z-a$ everywhere in previous part of this answer.
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 $0$.