Power Series
Definition 27.1. Let \( t \) be a fixed real number. A power series (expanded about \( t \)) is an infinite series of the form \[ \sum_{n=0}^{\infty} a_n (x - t)^n \] where \( \{a_n\}_{n=0}^{\infty} \) is a sequence and \( x \) is a real number. \([(x - t)^0\) is defined to be 1.]
Theorem 27.2
Theorem 27.2. Let \( \sum_{n=0}^{\infty} a_n (x - t)^n \) be a power series. Let \[ L = \limsup_{n \to \infty} |a_n|^{\frac{1}{n}}. \] Let \[ R = \begin{cases} 0 & \text{if } L = \infty, \\ \frac{1}{L} & \text{if } 0 < L < \infty, \\ \infty & \text{if } L = 0. \end{cases} \] Then the power series \( \sum_{n=0}^{\infty} a_n (x - t)^n \) converges absolutely if \( |x - t| < R \) and diverges if \( |x - t| > R \).
Proof. Suppose that $\{ |a_n|^{\frac{1}{n}} \}$ is unbounded. Then $L = \infty$ by Definition 21.2, and $R = 0$. If $|x - t| > 0$, then $|a_n|^{\frac{1}{n}} > \frac{1}{|x - t|}$ for infinitely many positive integers $n$. ($\{ |a_n|^{\frac{1}{n}} \}$ is unbounded.) Thus we have $|a_n(x-t)^n| > 1$ for infinitely many positive integers $n$, so that $\sum_{n=0}^{\infty} a_n(x-t)^n$ diverges. (($\because$) If not, then $\lim_{n \to \infty} a_n(x-t)^n = 0$, so that $1 \leq 0$. $\bigotimes$)
Suppose that $\{|a_n|^{\frac{1}{n}} \}$ is bounded. Then $\limsup_{n \to \infty} |a_n|^{\frac{1}{n}} = L$ is a real number. Assume that $L > 0$. By the root test, $\sum_{n=0}^{\infty} a_n(x-t)^n$ converges if $\limsup_{n \to \infty} |a_n(x-t)^n|^{\frac{1}{n}} < 1$ and diverges if $\limsup_{n \to \infty} |a_n(x-t)^n|^{\frac{1}{n}} > 1$.
Note that $$\limsup_{n \to \infty} |a_n(x-t)^n|^{\frac{1}{n}} = \limsup_{n \to \infty} |a_n|^{\frac{1}{n}} |x-t| \\ = |x-t| \limsup_{n \to \infty} |a_n|^{\frac{1}{n}} = |x-t|L$$ by Theorem 20.8. Thus the power series $\sum_{n=0}^{\infty} a_n(x-t)^n$ converges absolutely if $|x-t| < R$ and diverges if $|x-t| > R$.
Assume that $L = 0$. Then for any rean number $x$, $$\limsup_{n \to \infty} |a_n(x-t)^n|^{\frac{1}{n}} = \limsup_{n \to \infty} |a_n|^{\frac{1}{n}} |x-t| \\ = |x-t| \limsup_{n \to \infty} |a_n|^{\frac{1}{n}} = |x-t|L = 0$$ by Theorem 20.8. Thus the power series $\sum_{n=0}^{\infty} a_n(x-t)^n$ converges absolutely for any $x$ by the root test. $\blacksquare$
Exercise 27.2
Exercise 27.2. Let \( \sum_{n=0}^{\infty} a_n (x - t)^n \) be a power series. Let \[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|. \] Let \[ R = \begin{cases} 0 & \text{if } L = \infty, \\ \frac{1}{L} & \text{if } 0 < L < \infty, \\ \infty & \text{if } L = 0. \end{cases} \] Then the power series \( \sum_{n=0}^{\infty} a_n (x - t)^n \) converges absolutely if \( |x - t| < R \) and diverges if \( |x - t| > R \).
Radius of Convergence
Definition 27.3. The value \( R \) of Theorem 27.2 is called the radius of convergence of the power series \( \sum_{n=0}^{\infty} a_n (x - t)^n \).