3.3 Power Series, Radius and Interval of Convergence

3.3 Power Series, Radius and Interval of Convergence

Power series link discrete series to continuous analytic functions. On the GRE Mathematics Subject Test, questions test your ability to determine the exact radius and interval of convergence, analyze endpoint behavior, and differentiate or integrate series to evaluate numerical sums.

Power Series Foundations and Convergence Trichotomy

A power series centered at $c$ has the form: $\sum_{n=0}^\infty c_n (x - c)^n = c_0 + c_1(x - c) + c_2(x - c)^2 + \cdots$ where $c_n \in \mathbb{R}$ are coefficients and $c \in \mathbb{R}$ is the center.

For every power series, exactly one of three cases holds:

1. The series converges only at $x = c$ ($R = 0$).
2. The series converges absolutely for all $x \in \mathbb{R}$ ($R = \infty$).
3. There exists a finite $R > 0$ (radius of convergence) such that the series converges absolutely for $|x - c| < R$, diverges for $|x - c| > R$, and may converge or diverge at endpoints $|x - c| = R$.

Finding the Radius of Convergence

Two primary formulas compute $R$:

1. Ratio Limit Formula

If the limit of successive coefficient ratios exists: $R = \lim_{n \to \infty} \left|\frac{c_n}{c_{n+1}}\right| = \frac{1}{\lim_{n \to \infty} \left|\frac{c_{n+1}}{c_n}\right|}$

2. The Cauchy-Hadamard Theorem

The universal formula applying to all coefficient sequences is: $\frac{1}{R} = \limsup_{n \to \infty} |c_n|^{1/n}$ with conventions $1/0 = \infty$ and $1/\infty = 0$. Cauchy-Hadamard is essential when coefficients have periodic gaps or alternating growth rates (e.g., $c_n = 5^n$ for even $n$, $2^n$ for odd $n$).

Endpoint Analysis and Interval of Convergence

The interval of convergence is the set of $x \in \mathbb{R}$ where the series converges: $(c - R, c + R), \quad [c - R, c + R), \quad (c - R, c + R], \quad [c - R, c + R]$

• Endpoint Rule: Ratio and Root tests always yield 1 at $x = c \pm R$. Substitute $x = c \pm R$ into the series and test the resulting numerical series using $p$-series, Leibniz, or comparison tests.

Power Series with Transformed Arguments

When a series features an argument like $\sum c_n (a x - b)^k$:

1. Determine base radius $R_{\text{base}}$ of $\sum c_n u^n$.
2. Set $|ax - b|^k < R_{\text{base}} \implies |ax - b| < (R_{\text{base}})^{1/k}$.
3. Solve for $x$: $|x - b/a| < \frac{(R_{\text{base}})^{1/k}}{|a|}$. New center is $b/a$, and new radius is $R_{\text{new}} = \frac{(R_{\text{base}})^{1/k}}{|a|}$.

Term-by-Term Differentiation and Integration

Let $f(x) = \sum_{n=0}^\infty c_n (x - c)^n$ have radius $R > 0$. Within $(c - R, c + R)$, $f(x)$ is infinitely differentiable and integrable term-by-term:

• Derivative: $f'(x) = \sum_{n=1}^\infty n c_n (x - c)^{n-1}$
• Antiderivative: $\int f(x),dx = C + \sum_{n=0}^\infty \frac{c_n}{n+1} (x - c)^{n+1}$

Invariance of Radius vs. Endpoint Sensitivity

• Radius Invariance: Differentiation and integration preserve the exact radius $R$.
• Endpoint Warning: Endpoint convergence can change. Differentiation can destroy endpoint convergence (e.g., $\sum \frac{x^n}{n}$ converges at $x = -1$, but derivative $\sum x^{n-1}$ diverges). Integration can gain endpoint convergence.

Abel's Theorem on Boundary Continuity

If $\sum c_n (x - c)^n$ has radius $R > 0$ and converges at $x = c + R$, then: $\lim_{x \to (c + R)^-} \sum_{n=0}^\infty c_n (x - c)^n = \sum_{n=0}^\infty c_n R^n$ Abel's Theorem guarantees continuity extends to convergent endpoints, justifying evaluation of series sums via limits (e.g., $\sum \frac{(-1)^{n-1}}{n} = \lim_{x \to 1^-} \ln(1+x) = \ln 2$).

Summary Comparison Table: Power Series Properties

Worked Problem: Interval of Convergence with Mixed Endpoints

Problem: Find the exact interval of convergence for $\sum_{n=1}^\infty \frac{(2x - 1)^n}{n \cdot 3^n}$.

Solution:

1. Apply ratio test: $\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| = \frac{|2x - 1|}{3} \lim \frac{n}{n+1} = \frac{|2x - 1|}{3}$.
2. Convergence requires $\frac{|2x - 1|}{3} < 1 \implies |2x - 1| < 3 \implies -1 < x < 2$. The center is $c = 1/2$ and radius is $R = 3/2$.
3. Test $x = -1$: $2(-1) - 1 = -3$. The series becomes $\sum_{n=1}^\infty \frac{(-3)^n}{n \cdot 3^n} = \sum_{n=1}^\infty \frac{(-1)^n}{n}$, which converges by Leibniz.
4. Test $x = 2$: $2(2) - 1 = 3$. The series becomes $\sum_{n=1}^\infty \frac{3^n}{n \cdot 3^n} = \sum_{n=1}^\infty \frac{1}{n}$, which diverges.
5. The interval of convergence is $[-1, 2)$.

GRE Exam Traps & Fast Test-Taking Strategies

• Trap 1: Omitting the scaling coefficient in radius. For $\sum c_n (ax - b)^n$, calculating base radius without dividing by $|a|$ gives an incorrect interval. Always divide by $|a|$.
• Trap 2: Assuming identical endpoint behavior. It is common for one endpoint to converge conditionally and the other to diverge. Test both endpoints independently.
• Trap 3: Applying ratio test to lacunary series. When series contain powers like $x^{2n}$ or oscillating coefficients, use the Cauchy-Hadamard formula directly.