Power Series
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 6.1: “Power Series and Functions” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/6-1-power-series-and-functions |
| Textbook used in class | Stewart, Calculus, Section 11.8: “Power Series” (Examples 1, 2, 3, 4) |
Opening Scenario
A polynomial is a finite sum of terms $c_0 + c_1 x + c_2 x^2 + \cdots + c_n x^n$. A power series extends this idea to infinitely many terms. If the series converges for a given $x$, it defines a function value at that $x$. The set of $x$-values where the series converges is its interval of convergence, and the key question is: how large is that interval?
Quick Reference
Power series centered at $a$: \[ \sum_{n=0}^{\infty} c_n (x - a)^n = c_0 + c_1(x-a) + c_2(x-a)^2 + \cdots \]
For $a = 0$ (centered at the origin): $\sum c_n x^n$.
Convergence theorem. Exactly one of three things happens:
- The series converges only at $x = a$ (radius $R = 0$).
- The series converges for all $x$ (radius $R = \infty$).
- The series converges for $|x - a| < R$ and diverges for $|x - a| > R$, for some $R > 0$.
In case 3, the convergence at the endpoints $x = a \pm R$ must be checked separately.
Finding $R$: Apply the Ratio Test to get $\lim |c_{n+1}/c_n| \cdot |x - a| < 1$, giving $R = \lim |c_n/c_{n+1}|$. (Or use the Root Test: $R = 1/\lim |c_n|^{1/n}$.)
Key Concepts
1. The Ratio Test for Power Series
To find where a power series converges, apply the Ratio Test treating $x$ as a fixed parameter.
Example 1. Find the radius of convergence of $\displaystyle\sum_{n=0}^{\infty} \frac{x^n}{n!}$. (Stewart 11.8, Example 1.)
\[ \left|\frac{a_{n+1}}{a_n}\right| = \frac{|x|^{n+1}}{(n+1)!} \cdot \frac{n!}{|x|^n} = \frac{|x|}{n+1} \to 0 \text{ for every } x. \]
Since $L = 0 < 1$ for every real $x$, the series converges absolutely for all $x$.
Boxed answer: $R = \infty$. (This is the power series for $e^x$, which converges everywhere.)
Example 2. Find the radius and interval of convergence of $\displaystyle\sum_{n=0}^{\infty} n!\, x^n$. (Stewart 11.8, Example 2.)
\[ \left|\frac{a_{n+1}}{a_n}\right| = (n+1)|x| \to \infty \text{ for every } x \neq 0. \]
$L = \infty > 1$ for all $x \neq 0$. At $x = 0$ the series is $1 + 0 + 0 + \cdots = 1$: converges.
Boxed answer: $R = 0$; the series converges only at $x = 0$.
Example 3. Find the radius and interval of convergence of $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n x^n}{n}$. (Stewart 11.8, Example 3.)
\[ \left|\frac{a_{n+1}}{a_n}\right| = \frac{|x|^{n+1}}{n+1} \cdot \frac{n}{|x|^n} = \frac{n}{n+1}|x| \to |x|. \]
$L = |x| < 1$ means the series converges for $|x| < 1$, so $R = 1$.
Check endpoints:
- $x = 1$: $\sum (-1)^n/n$, the alternating harmonic series. Converges (AST).
- $x = -1$: $\sum (-1)^n(-1)^n/n = \sum 1/n$, the harmonic series. Diverges.
Boxed answer: Radius $R = 1$; interval of convergence $[-1, 1)$ (includes $x = -1$... wait, at $x = -1$: $(-1)^n(-1)^n = (-1)^{2n} = 1$, so the series is $\sum 1/n$ which diverges. At $x = 1$: $\sum (-1)^n/n$ converges). Interval is $(-1, 1]$.
Wait, let me recheck. Series is $\sum_{n=1}^{\infty} \frac{(-1)^n x^n}{n}$. At $x=1$: $\sum \frac{(-1)^n}{n}$, converges (AST). At $x=-1$: $\sum \frac{(-1)^n (-1)^n}{n} = \sum \frac{(-1)^{2n}}{n} = \sum \frac{1}{n}$, diverges.
Boxed answer: $R = 1$; interval of convergence $(-1, 1]$.
2. Power Series Centered at $a \neq 0$
Example 4. Find the radius and interval of convergence of $\displaystyle\sum_{n=0}^{\infty} \frac{(x-3)^n}{n}$ (with the convention that the $n=0$ term is $1$). Actually use $\displaystyle\sum_{n=1}^{\infty} \frac{(x-3)^n}{n}$.
By the ratio test, $|{(x-3)^{n+1}/(n+1)}\cdot n/(x-3)^n| = \frac{n}{n+1}|x-3| \to |x-3|$. Converges for $|x-3| < 1$, so $R = 1$. Center is $a = 3$; the interval of convergence is $(2, 4]$ (endpoints require separate checks: at $x = 4$, $\sum 1/n$ diverges; at $x = 2$, $\sum (-1)^n/n$ converges).
Boxed answer: $R = 1$; interval $(2, 4]$.
confusing the radius $R$ with the interval. The radius is the half-width $R$; the interval of convergence is the full range $a - R < x < a + R$, possibly with one or both endpoints included. Saying “the radius of convergence is $(-1, 1]$” misuses the term. $R$ is a single non-negative number.
Common Errors Summary
| Error | Correction |
|---|---|
| Forgetting to check endpoints | The ratio test is inconclusive at $|x-a| = R$; check each endpoint separately |
| Treating $R$ as the interval | $R$ is a number, not an interval; state the interval explicitly |
| Ignoring the center $a$ | The interval is centered at $a$; $|x - a| < R$, not $|x| < R$ |
Leveled Practice
Problem 1. Find the radius of convergence of $\displaystyle\sum_{n=0}^{\infty} \frac{x^n}{2^n}$.
Show answer
$|a_{n+1}/a_n| = |x|/2 \to |x|/2$. Converges for $|x|/2 < 1$, i.e. $|x| < 2$. $R = 2$.
Problem 2. Find the interval of convergence of $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n (x-2)^n}{\sqrt{n}}$.
Show answer
$|a_{n+1}/a_n| = \frac{\sqrt{n}}{\sqrt{n+1}}|x-2| \to |x-2|$. $R = 1$, centered at $2$.
At $x = 3$ ($|x-2| = 1$): $\sum (-1)^n/\sqrt{n}$, converges by AST. At $x = 1$ ($|x-2| = 1$): $\sum (-1)^n(-1)^n/\sqrt{n} = \sum 1/\sqrt{n}$, diverges ($p = 1/2$).
Interval: $(1, 3]$.