Given the graph of a function , find a number such that if , then .
Table of contents
- 0. Functions7h 55m
- Introduction to Functions18m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms36m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation3h 18m
- 4. Applications of Derivatives2h 38m
- 5. Graphical Applications of Derivatives6h 2m
- 6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
- 7. Antiderivatives & Indefinite Integrals1h 26m
- 8. Definite Integrals4h 44m
- 9. Graphical Applications of Integrals2h 27m
- 10. Physics Applications of Integrals 3h 16m
- 11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 31m
- 12. Techniques of Integration7h 41m
- 13. Intro to Differential Equations2h 55m
- 14. Sequences & Series5h 36m
- 15. Power Series2h 19m
- 16. Parametric Equations & Polar Coordinates7h 58m
1. Limits and Continuity
Introduction to Limits
Multiple Choice
Determine the interval of convergence for the series . Express your answer using interval notation.
A
B
C
D
0 Comments
Verified step by step guidance1
Step 1: Recognize that the problem involves determining the interval of convergence for a power series. The series given is \( \sum_{n=1}^{\infty} \frac{(x-2)^n}{n} \). To find the interval of convergence, we use the ratio test or other convergence tests as appropriate.
Step 2: Apply the ratio test. The ratio test states that a series \( \sum a_n \) converges absolutely if \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \). Here, \( a_n = \frac{(x-2)^n}{n} \). Compute \( \left| \frac{a_{n+1}}{a_n} \right| \): \( \left| \frac{\frac{(x-2)^{n+1}}{n+1}}{\frac{(x-2)^n}{n}} \right| = \left| \frac{(x-2) \cdot n}{n+1} \right| \).
Step 3: Simplify the expression from Step 2. The limit becomes \( \lim_{n \to \infty} \left| (x-2) \cdot \frac{n}{n+1} \right| = \left| x-2 \right| \cdot \lim_{n \to \infty} \frac{n}{n+1} \). Since \( \lim_{n \to \infty} \frac{n}{n+1} = 1 \), the ratio test condition simplifies to \( \left| x-2 \right| < 1 \).
Step 4: Solve the inequality \( \left| x-2 \right| < 1 \) to find the interval of convergence. This inequality can be rewritten as \( -1 < x-2 < 1 \). Adding 2 to all sides gives \( 1 < x < 3 \). Thus, the interval of convergence is \( (1, 3) \).
Step 5: Check the endpoints \( x = 1 \) and \( x = 3 \) separately to determine whether the series converges at these points. Substitute \( x = 1 \) and \( x = 3 \) into the original series and use tests like the p-series test or direct evaluation to confirm that the series diverges at \( x = 1 \) and converges at \( x = 3 \). Therefore, the interval of convergence is \( (1, 3) \).
Related Videos
Related Practice
Multiple Choice
111
views

