Power Series Summary
A power series is an infinite series of the form:
where \( a_n \) are coefficients, \( c \) is the center of the series, and \( x \) is the variable.
Concepts Covered
- Convergence of Power Series: Determining the radius of convergence \( R \) such that the series converges for \( |x-c| < R \) and diverges for \( |x-c| > R \).
- Combining Power Series: Adding, subtracting, or multiplying series term-by-term within the interval of convergence.
- Differentiating and Integrating Power Series: Term-by-term operations valid within the radius of convergence.
Power Series Convergence Examples
Example 1: Basic Geometric-type Series
\[ \sum_{k=0}^{\infty} \frac{(-1)^k(x-2)^k}{4^k} \]Solution
Step 1: Apply Ratio Test
\[ L = \lim_{k\to\infty} \left|\frac{a_{k+1}}{a_k}\right| = \lim_{k\to\infty} \left|\frac{(x-2)^{k+1}}{4^{k+1}} \cdot \frac{4^k}{(x-2)^k}\right| = \frac{|x-2|}{4} \]Step 2: Determine Radius of Convergence
\[ \frac{|x-2|}{4} < 1 \implies |x-2| < 4 \implies R = 4 \]Step 3: Check Endpoints
At \( x = -2 \): \[ \sum_{k=0}^\infty \frac{(-1)^k(-4)^k}{4^k} = \sum_{k=0}^\infty 1 \text{ (diverges)} \] At \( x = 6 \): \[ \sum_{k=0}^\infty \frac{(-1)^k(4)^k}{4^k} = \sum_{k=0}^\infty (-1)^k \text{ (diverges)} \]
Interval of Convergence: \( (-2, 6) \)
Radius of Convergence: 4
Example 2: Factorial Denominator Series
\[ \sum_{n=0}^{\infty} \frac{x^n}{n!} \]Solution
Step 1: Apply Ratio Test
\[ L = \lim_{n\to\infty} \left|\frac{x^{n+1}}{(n+1)!} \cdot \frac{n!}{x^n}\right| = \lim_{n\to\infty} \frac{|x|}{n+1} = 0 \]Conclusion: Since \( L = 0 < 1 \) for all \( x \)
Interval of Convergence: \( (-\infty, \infty) \)
Radius of Convergence: \( \infty \)
Example 3: Alternating Root Test Series
\[ \sum_{k=1}^{\infty} (-1)^k \frac{(x+1)^k}{\sqrt{k}} \]Solution
Step 1: Apply Root Test
\[ L = \lim_{k\to\infty} \sqrt[k]{\left|\frac{(x+1)^k}{\sqrt{k}}\right|} = |x+1| \lim_{k\to\infty} k^{-1/(2k)} = |x+1| \]Step 2: Determine Radius
\[ |x+1| < 1 \implies R = 1 \]Step 3: Check Endpoints
At \( x = -2 \): \[ \sum_{k=1}^\infty \frac{1}{\sqrt{k}} \text{ (diverges, p-series)} \] At \( x = 0 \): \[ \sum_{k=1}^\infty \frac{(-1)^k}{\sqrt{k}} \text{ (converges conditionally)} \]
Interval of Convergence: \( [-2, 0) \)
Radius: 1
Power Series Convergence Examples with Detailed Solutions
Example 4: Squared Term Series
\[ \sum_{n=1}^{\infty} \frac{x^n}{n^2 3^n} \]Solution
Step 1: Apply Ratio Test
\[ L = \lim_{n\to\infty} \left|\frac{x^{n+1}}{(n+1)^2 3^{n+1}} \cdot \frac{n^2 3^n}{x^n}\right| = \frac{|x|}{3} \lim_{n\to\infty} \left(\frac{n}{n+1}\right)^2 = \frac{|x|}{3} \]Step 2: Determine Radius
\[ \frac{|x|}{3} < 1 \implies |x| < 3 \implies R = 3 \]Step 3: Check Endpoints
At \( x = 3 \): \[ \sum_{n=1}^\infty \frac{1}{n^2} \text{ (converges, p-series)} \] At \( x = -3 \): \[ \sum_{n=1}^\infty \frac{(-1)^n}{n^2} \text{ (absolutely converges)} \]
Interval of Convergence: \( [-3, 3] \)
Radius of Convergence: 3
Example 5: Polynomial Denominator Series
\[ \sum_{k=1}^{\infty} \frac{(x+3)^k}{k^2 + 1} \]Solution
Step 1: Apply Root Test
\[ L = \lim_{k\to\infty} \sqrt[k]{\frac{|x+3|^k}{k^2 + 1}} = |x+3| \lim_{k\to\infty} (k^2 + 1)^{-1/k} = |x+3| \]Step 2: Determine Radius
\[ |x+3| < 1 \implies R = 1 \]Step 3: Check Endpoints
At \( x = -4 \): \[ \sum_{k=1}^\infty \frac{(-1)^k}{k^2 + 1} \text{ (converges absolutely)} \] At \( x = -2 \): \[ \sum_{k=1}^\infty \frac{1}{k^2 + 1} \text{ (converges)} \]
Interval of Convergence: \( [-4, -2] \)
Radius: 1
Example 6: Alternating Logarithmic Series
\[ \sum_{n=2}^{\infty} \frac{(-1)^n (x-1)^n}{n \ln n} \]Solution
Step 1: Apply Ratio Test
\[ L = \lim_{n\to\infty} \left|\frac{(x-1)^{n+1}}{(n+1)\ln(n+1)} \cdot \frac{n \ln n}{(x-1)^n}\right| = |x-1| \lim_{n\to\infty} \frac{n \ln n}{(n+1)\ln(n+1)} = |x-1| \]Step 2: Determine Radius
\[ |x-1| < 1 \implies R = 1 \]Step 3: Check Endpoints
At \( x = 0 \): \[ \sum_{n=2}^\infty \frac{1}{n \ln n} \text{ (diverges, integral test)} \] At \( x = 2 \): \[ \sum_{n=2}^\infty \frac{(-1)^n}{n \ln n} \text{ (converges conditionally)} \]
Interval of Convergence: \( (0, 2] \)
Radius: 1
Example 7: Exponential-Composite Series
\[ \sum_{n=1}^{\infty} \frac{n^n x^n}{5^{2n}} \]Solution
Step 1: Apply Root Test
\[ L = \lim_{n\to\infty} \sqrt[n]{\frac{n^n |x|^n}{25^n}} = \lim_{n\to\infty} \frac{n |x|}{25} = \infty \text{ for } x \neq 0 \]Conclusion: Only converges at center
Interval of Convergence: \( \{0\} \)
Radius: 0
Example 8: Recursive Sequence Series
\[ \sum_{k=1}^{\infty} \frac{(2x)^k}{k^k} \]Solution
Step 1: Apply Root Test
\[ L = \lim_{k\to\infty} \sqrt[k]{\frac{(2|x|)^k}{k^k}} = \lim_{k\to\infty} \frac{2|x|}{k} = 0 \]Conclusion: Converges for all real numbers
Interval of Convergence: \( (-\infty, \infty) \)
Radius: \( \infty \)
Example 9: Trigonometric Numerator Series
\[ \sum_{n=1}^{\infty} \frac{\sin(n) x^n}{n^3} \]Solution
Step 1: Use Comparison Test
\[ \left|\frac{\sin(n) x^n}{n^3}\right| \leq \frac{|x|^n}{n^3} \]Step 2: Apply Root Test to Absolute Series
\[ L = \lim_{n\to\infty} \sqrt[n]{\frac{|x|^n}{n^3}} = |x| \]Step 3: Determine Radius
\[ |x| < 1 \implies R = 1 \]Step 4: Check Endpoints
At \( x = \pm 1 \): \[ \sum_{n=1}^\infty \frac{|\sin(n)|}{n^3} \text{ converges by comparison to } \sum 1/n^3 \]
Interval of Convergence: \( [-1, 1] \)
Radius: 1
Example 10: Combined Operations Series
\[ \sum_{m=2}^{\infty} \frac{m(x+2)^m}{5^m \sqrt{m}} \]Solution
Step 1: Apply Ratio Test
\[ L = \lim_{m\to\infty} \left|\frac{(m+1)(x+2)^{m+1}}{5^{m+1}\sqrt{m+1}} \cdot \frac{5^m \sqrt{m}}{m(x+2)^m}\right| = \frac{|x+2|}{5} \]Step 2: Determine Radius
\[ \frac{|x+2|}{5} < 1 \implies |x+2| < 5 \implies R = 5 \]Step 3: Check Endpoints
At \( x = -7 \): \[ \sum_{m=2}^\infty \frac{m(-5)^m}{5^m \sqrt{m}} = \sum_{m=2}^\infty \frac{(-1)^m m^{1/2}}{5^{m-1}} \text{ (diverges)} \] At \( x = 3 \): \[ \sum_{m=2}^\infty \frac{m(5)^m}{5^m \sqrt{m}} = \sum_{m=2}^\infty \frac{\sqrt{m}}{1} \text{ (diverges)} \]
Interval of Convergence: \( (-7, 3) \)
Radius: 5