Use the divergence test to determine if the following series diverge or state that the test is inconclusive.
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
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- 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
14. Sequences & Series
Convergence Tests
Multiple Choice
Explain why the integral test does not apply to the series.
A
The series is increasing
B
Not continuous at x=1
C
The series is decreasing
D
The series is not positive
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Verified step by step guidance1
Step 1: Recall the conditions for the integral test. The integral test applies to a series ∑aₙ if the corresponding function f(x) is positive, continuous, and decreasing for x ≥ 1.
Step 2: Analyze the given series ∑n=0∞n/(n²−1). The term aₙ = n/(n²−1) corresponds to the function f(x) = x/(x²−1).
Step 3: Check the continuity of f(x). The function f(x) = x/(x²−1) is undefined at x = 1 because the denominator becomes zero (x²−1 = 0). This means f(x) is not continuous at x = 1, violating one of the conditions for the integral test.
Step 4: Check if the series is positive. For x > 1, the numerator x is positive, but the denominator x²−1 is also positive. However, for x < 1, the denominator x²−1 becomes negative, making f(x) negative. Thus, the series is not positive for all x ≥ 1, violating another condition for the integral test.
Step 5: Check if the series is decreasing. While the series may appear to decrease for large values of n, the integral test requires the function to be strictly decreasing for all x ≥ 1. Since the series fails the other two conditions (continuity and positivity), the integral test cannot be applied regardless of whether the series is decreasing.
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