A reliable method
- Identify the expression playing the role of the geometric ratio.
- Write successive powers of that ratio.
- Solve its magnitude-less-than-1 condition.
- Check both boundary values before including endpoints.
Example 1
Write the first four terms and convergence interval of the geometric power series for \(\frac{1}{1 - 2 x}\) about x=0.
- Use \(\frac1{1-u}=1+u+u^2+u^3+\cdots\) only when \(|u|<1\).
- Here \(u=2 x\); the first four terms are \(8 x^{3} + 4 x^{2} + 2 x + 1\).
- Solve the strict convergence inequality. The interval is \(\left(- \frac{1}{2}, \frac{1}{2}\right)\). At either endpoint the terms fail to approach 0, so neither endpoint is included.
Answer: \(8 x^{3} + 4 x^{2} + 2 x + 1\); \(\left(- \frac{1}{2}, \frac{1}{2}\right)\)
Example 2
Write the first four terms and convergence interval of the geometric power series for \(\frac{1}{3 x + 1}\) about x=0.
- Use \(\frac1{1-u}=1+u+u^2+u^3+\cdots\) only when \(|u|<1\).
- Here \(u=- 3 x\); the first four terms are \(- 27 x^{3} + 9 x^{2} - 3 x + 1\).
- Solve the strict convergence inequality. The interval is \(\left(- \frac{1}{3}, \frac{1}{3}\right)\). At either endpoint the terms fail to approach 0, so neither endpoint is included.
Answer: \(- 27 x^{3} + 9 x^{2} - 3 x + 1\); \(\left(- \frac{1}{3}, \frac{1}{3}\right)\)
Common mistakes
- A polynomial truncation is not exactly the infinite sum.
- Other power series may behave differently at endpoints; this guide covers geometric series only.
Practice without the answer first
- Write the first four terms and convergence interval of the geometric power series for \(\frac{1}{1 - \frac{x}{2}}\) about x=0.
- Write the first four terms and convergence interval of the geometric power series for \(\frac{1}{1 - 4 x}\) about x=0.
- Write the first four terms and convergence interval of the geometric power series for \(\frac{1}{x + 1}\) about x=0.
Answers and step-by-step explanations
Practice 1
- Use \(\frac1{1-u}=1+u+u^2+u^3+\cdots\) only when \(|u|<1\).
- Here \(u=\frac{x}{2}\); the first four terms are \(\frac{x^{3}}{8} + \frac{x^{2}}{4} + \frac{x}{2} + 1\).
- Solve the strict convergence inequality. The interval is \(\left(-2, 2\right)\). At either endpoint the terms fail to approach 0, so neither endpoint is included.
Answer: \(\frac{x^{3}}{8} + \frac{x^{2}}{4} + \frac{x}{2} + 1\); \(\left(-2, 2\right)\)
Practice 2
- Use \(\frac1{1-u}=1+u+u^2+u^3+\cdots\) only when \(|u|<1\).
- Here \(u=4 x\); the first four terms are \(64 x^{3} + 16 x^{2} + 4 x + 1\).
- Solve the strict convergence inequality. The interval is \(\left(- \frac{1}{4}, \frac{1}{4}\right)\). At either endpoint the terms fail to approach 0, so neither endpoint is included.
Answer: \(64 x^{3} + 16 x^{2} + 4 x + 1\); \(\left(- \frac{1}{4}, \frac{1}{4}\right)\)
Practice 3
- Use \(\frac1{1-u}=1+u+u^2+u^3+\cdots\) only when \(|u|<1\).
- Here \(u=- x\); the first four terms are \(- x^{3} + x^{2} - x + 1\).
- Solve the strict convergence inequality. The interval is \(\left(-1, 1\right)\). At either endpoint the terms fail to approach 0, so neither endpoint is included.
Answer: \(- x^{3} + x^{2} - x + 1\); \(\left(-1, 1\right)\)
These original skill exercises are not official exam questions or a score prediction. Use the coverage notes on your exam hub to choose relevant practice.
