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Make up an infinite series of nonzero terms whose sum is
b. −3
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Ch. 10 - Infinite Sequences and Series
Hass15th EditionThomas' CalculusISBN: 9780137616077
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TOC.tableOfContents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions490
- Ch. 8 - Techniques of Integration399
- Ch. 9 - First-Order Differential Equations66
- Ch. 10 - Infinite Sequences and Series132
- Ch. 11 - Parametric Equations and Polar Coordinates58
Capitolo 10, Problema 10.4.15
Limit Comparison Test
In Exercises 9–16, use the Limit Comparison Test to determine if each series converges or diverges.
∑ (from n=2 to ∞) 1 / ln n
(Hint: Limit Comparison with ∑ (from n=2 to ∞) (1/n))
Guida verificata passo dopo passo1
Identify the given series: \( \sum_{n=2}^{\infty} \frac{1}{\ln n} \). We want to determine if this series converges or diverges using the Limit Comparison Test.
Choose a comparison series that is easier to analyze. The hint suggests using \( \sum_{n=2}^{\infty} \frac{1}{n} \), which is a well-known divergent harmonic series.
Set up the Limit Comparison Test by computing the limit \( L = \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{1}{\ln n}}{\frac{1}{n}} = \lim_{n \to \infty} \frac{n}{\ln n} \).
Analyze the limit \( L \). Since \( n \) grows faster than \( \ln n \), this limit tends to infinity, which is a positive number (not zero or infinite in the sense of the test's conditions).
Interpret the result: Because the limit \( L \) is infinite and the comparison series \( \sum \frac{1}{n} \) diverges, the Limit Comparison Test tells us that the original series \( \sum \frac{1}{\ln n} \) also diverges.
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Limit Comparison Test
The Limit Comparison Test is used to determine the convergence or divergence of an infinite series by comparing it to a second series with known behavior. It involves taking the limit of the ratio of the nth terms of the two series. If the limit is a positive finite number, both series either converge or diverge together.
Video consigliato:
Percorso guidato
07:45
Limit Comparison Test
Behavior of the Harmonic Series
The harmonic series ∑ 1/n is a well-known divergent series. Understanding its divergence is crucial when using it as a comparison series in the Limit Comparison Test. Since it diverges, any series that behaves similarly (in terms of term size) will also diverge.
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Percorso guidato
04:30P-Series and Harmonic Series
Properties of the Natural Logarithm Function
The natural logarithm function ln(n) grows slowly as n increases, but it still tends to infinity. Recognizing how 1/ln(n) compares to 1/n helps in applying the Limit Comparison Test. Since ln(n) grows slower than n, 1/ln(n) decreases more slowly than 1/n, affecting convergence.
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Percorso guidato
06:21Properties of Functions
Pratica correlata
Domanda del libro di testo
In Exercises 57–82, use any method to determine whether the series converges or diverges. Give reasons for your answer.
∑ (from n = 1 to ∞) [3ⁿ / n³]
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Domanda del libro di testo
Error Estimation
In Exercises 49–52, estimate the magnitude of the error involved in using the sum of the first four terms to approximate the sum of the entire series.
1 / (1 + t) = ∑ (from n = 0 to ∞) [(-1)ⁿ tⁿ],0 < t < 1
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Domanda del libro di testo
Applying the Integral Test
Use the Integral Test to determine if the series in Exercises 1–12 converge or diverge. Be sure to check that the conditions of the Integral Test are satisfied.
∑ (from n = 2 to ∞) ln(n²) / n
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Domanda del libro di testo
Telescoping Series
In Exercises 39–44, find a formula for the nth partial sum of the series and use it to determine if the series converges or diverges. If a series converges, find its sum.
∑ (from n = 1 to ∞) [ (1/n) − (1/(n + 1)) ]
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Domanda del libro di testo
Intervals of Convergence
In Exercises 1–36, (a) find the series’ radius and interval of convergence. For what values of x does the series converge (b) absolutely, (c) conditionally?
∑ (from n = 0 to ∞) (2x)ⁿ
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