Interval of convergence and integration of a power series

Question

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?

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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?

Answer 1

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 $|z|<R$.
• The series $\sum_{n=0}^{\infty} |a_nz^n|$ diverges to $\infty $ for $|z|>R$.

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 $|z|<R$ 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.

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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.

Answer 2

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 $0$.