If ∑aₙ is a convergent series of positive terms, prove that ∑sin(aₙ) converges.
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
14. Sequences & Series
Convergence Tests
Multiple Choice
Use the divergence test to determine if the following series diverge or state that the test is inconclusive.
A
Divergent
B
Convergent
C
Inconclusive
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Verified step by step guidance1
Step 1: Recall the divergence test (also known as the nth-term test for divergence). This test states that if the limit of the nth term of a series, a_n, as n approaches infinity does not equal 0, or if the limit does not exist, then the series diverges. If the limit equals 0, the test is inconclusive.
Step 2: Identify the nth term of the given series. Here, the nth term is a_n = sin(nπ/2).
Step 3: Analyze the behavior of sin(nπ/2) as n increases. Notice that the sine function oscillates periodically. Specifically, sin(nπ/2) takes on the values 0, 1, 0, -1, and repeats in this pattern as n increases.
Step 4: Since sin(nπ/2) does not settle to a single value as n approaches infinity (it oscillates), the limit of a_n = sin(nπ/2) does not exist.
Step 5: By the divergence test, since the limit of a_n does not exist, the series ∑_{n=1}^{∞} sin(nπ/2) diverges.
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