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:

$\displaystyle (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. $ \displaystyle f'(x) = \sum_{n=1}^{\infty} n \cdot c_n (x-a)^{n-1}$
  2. $ \displaystyle \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

$\displaystyle 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

$\displaystyle 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

$\displaystyle \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

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

Hence

$\displaystyle \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 [*]).

William Stein 2006-03-15