Convergence of Power Series

Theorem 6.5.6   Given a power series $\sum_{n=0}^{\oo } c_n(x-a)^n$, there are exactly three possibilities:

1. The series conveges only when $x=a$.
2. The series conveges for all $x$.
3. There is an $R>0$ (called the ``radius of convergence'') such that $\sum_{n=0}^{\oo } c_n(x-a)^n$ converges for $\vert x-a\vert<R$ and diverges for $\vert x-a\vert>R$.

Example 6.5.7   For the power series $\sum_{n=0}^{\infty} x^n$, the radius $R$ of convergence is $1$.

Definition 6.5.8 (Radius of Convergence)   As mentioned in the theorem, $R$ is called the radius of convergence.

If the series converges only at $x=a$, we say $R=0$, and if the series converges everywhere we say that $R=\oo$.

The interval of convergence is the set of $x$ for which the series converges. It will be one of the following:

$$(a-R, a+R), \qquad [a-R, a+R), \qquad (a-R, a+R], \qquad [a-R, a+R]$$

The point being that the statement of the theorem only asserts something about convergence of the series on the open interval $(a-R, a+R)$. What happens at the endpoints of the interval is not specified by the theorem; you can only figure it out by looking explicitly at a given series.

Theorem 6.5.9   If $\sum_{n=0}^{\infty} c_n (x-a)^n$ has radius of convergence $R>0$, then $f(x) = \sum_{n=0}^{\infty} c_n (x-a)^n$ is differentiable on $(a-R, a+R)$, and

1. $$f'(x) = \sum_{n=1}^{\infty} n \cdot c_n (x-a)^{n-1}$$
2. $$\int f(x) dx = C + \sum_{n=0}^{\infty} \frac{c_n}{n+1}(x-a)^{n+1}$$,

and both the derivative and integral have the same radius of convergence as $f$.

Example 6.5.10   Find a power series representation for $f(x) = \tan^{-1}(x)$. Notice that

$$f'(x) = \frac{1}{1+x^2} = \frac{1}{1- (-x^2)} = \sum_{n=0}^{\oo } (-1)^n x^{2n},$$

which has radius of convergence $R=1$, since the above series is valid when $\vert-x^2\vert<1$, i.e., $\vert x\vert < 1$. Next integrating, we find that

$$f(x) = c + \sum_{n=0}^{\oo } (-1)^n \frac{x^{2n+1}}{2n+1},$$

for some constant $c$. To find the constant, compute $c = f(0) = \tan^{-1}(0) = 0$. We conclude that

$$\tan^{-1}(x) = \sum_{n=0}^{\oo } (-1)^n \frac{x^{2n+1}}{2n+1}.$$

Example 6.5.11   We will see later that the function $f(x) = e^{-x^2}$ has power series

$$e^{-x^2} = 1 - x^{2} + \frac{1}{2}x^{4} - \frac{1}{6}x^{6} + \cdots.$$

Hence

$$\int e^{-x^2} dx = c + x - \frac{1}{3}x^{3} + \frac{1}{10}x^{5} - \frac{1}{42}x^{7} + \cdots.$$

This despite the fact that the antiderivative of $e^{-x^2}$ is not an elementary function (see Example ).