Express 0.314141414… as a ratio of two integers.
Table of contents
- 0. Functions7h 54m
- Introduction to Functions16m
- 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 34m
- 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
Series
Problem 10.3.79
Textbook Question
72–86. Evaluating series Evaluate each series or state that it diverges.
∑ (k = 2 to ∞) ln((k + 1)k⁻¹) / (ln k × ln(k + 1))
Verified step by step guidance1
First, rewrite the general term of the series to simplify the expression inside the summation. The term is given by \(\frac{\ln\left((k+1)k^{-1}\right)}{\ln k \times \ln(k+1)}\). Use the logarithm property \(\ln(a/b) = \ln a - \ln b\) to rewrite the numerator as \(\ln(k+1) - \ln k\).
Substitute the simplified numerator back into the term to get \(\frac{\ln(k+1) - \ln k}{\ln k \times \ln(k+1)}\). Then, separate this fraction into two parts: \(\frac{\ln(k+1)}{\ln k \times \ln(k+1)} - \frac{\ln k}{\ln k \times \ln(k+1)}\).
Simplify each part of the separated fraction. The first part simplifies to \(\frac{1}{\ln k}\) and the second part simplifies to \(\frac{1}{\ln(k+1)}\). So the general term becomes \(\frac{1}{\ln k} - \frac{1}{\ln(k+1)}\).
Recognize that the series is telescoping because each term is of the form \(a_k - a_{k+1}\), where \(a_k = \frac{1}{\ln k}\). Write out the first few terms explicitly to see the cancellation pattern.
Use the telescoping property to express the partial sum \(S_n = \sum_{k=2}^n \left( \frac{1}{\ln k} - \frac{1}{\ln(k+1)} \right)\) as \(\frac{1}{\ln 2} - \frac{1}{\ln(n+1)}\). Then analyze the limit of \(S_n\) as \(n \to \infty\) to determine whether the series converges or diverges.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Convergence and Divergence of Infinite Series
An infinite series converges if the sum of its terms approaches a finite limit as the number of terms grows indefinitely. Otherwise, it diverges. Determining convergence often involves applying tests that analyze the behavior of the terms or partial sums.
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Convergence of an Infinite Series
Properties of Logarithms
Logarithmic properties, such as ln(a/b) = ln(a) - ln(b) and ln(ab) = ln(a) + ln(b), help simplify complex expressions. Recognizing these can transform the series terms into simpler forms, making it easier to analyze or compare with known series.
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Change of Base Property
Comparison and Limit Comparison Tests
These tests compare a given series to a known benchmark series to determine convergence. The limit comparison test uses the limit of the ratio of terms from two series; if the limit is finite and positive, both series share the same convergence behavior.
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Limit Comparison Test
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