Domanda del libro di testo
12–24. Limits of sequences Evaluate the limit of the sequence or state that it does not exist.
aₙ = ((3n² + 2n + 1) · sin(n)) / (4n³ + n) (Hint: Use the Squeeze Theorem.)
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Ch. 10 - Sequences and Infinite Series
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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TOC.tableOfContents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 10, Problema 10.R.47
42–76. Convergence or divergence Use a convergence test of your choice to determine whether the following series converge.
∑ (from k = 1 to ∞)(7 + sin k) / k²
Guida verificata passo dopo passo1
Identify the given series: \( \sum_{k=1}^{\infty} \frac{7 + \sin k}{k^2} \). We want to determine if this series converges or diverges.
Note that \( \sin k \) is bounded between -1 and 1, so the numerator \( 7 + \sin k \) is bounded between 6 and 8. This means the terms behave roughly like \( \frac{\text{constant}}{k^2} \) for large \( k \).
Recall the p-series test: \( \sum_{k=1}^{\infty} \frac{1}{k^p} \) converges if \( p > 1 \). Here, since the denominator is \( k^2 \), which corresponds to \( p = 2 > 1 \), the series \( \sum \frac{1}{k^2} \) converges.
Use the Comparison Test by comparing \( \frac{7 + \sin k}{k^2} \) with \( \frac{8}{k^2} \) (since 8 is an upper bound for the numerator). Since \( \sum \frac{8}{k^2} \) converges, and \( 0 \leq \frac{7 + \sin k}{k^2} \leq \frac{8}{k^2} \), the original series converges by the Comparison Test.
Conclude that the series \( \sum_{k=1}^{\infty} \frac{7 + \sin k}{k^2} \) converges absolutely because it is bounded by a convergent p-series with \( p = 2 \).
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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. If the sum does not approach a finite value, the series diverges. Understanding this distinction is fundamental to analyzing series behavior.
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Percorso guidato
06:52
Convergence of an Infinite Series
Comparison Test for Series Convergence
The Comparison Test involves comparing a given series to a second series whose convergence behavior is known. If the terms of the given series are smaller than those of a convergent series, it also converges. This test is useful when terms are positive and can be bounded.
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Percorso guidato
09:25Direct Comparison Test
Behavior of the p-Series
A p-series has the form ∑ 1/k^p and converges if and only if p > 1. Since the given series has terms involving 1/k², recognizing this helps determine convergence by comparing to a known convergent p-series with p=2.
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Percorso guidato
04:30P-Series and Harmonic Series
Pratica correlata
Domanda del libro di testo
42–76. Convergence or divergence Use a convergence test of your choice to determine whether the following series converge.
∑ (from k = 1 to ∞)5ᵏ / 2²ᵏ⁺¹
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Domanda del libro di testo
12–24. Limits of sequences Evaluate the limit of the sequence or state that it does not exist.
aₙ = (–1)ⁿ (3n³ + 4n) / (6n³ + 5)
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Domanda del libro di testo
42–76. Convergence or divergence Use a convergence test of your choice to determine whether the following series converge.
∑ (from k = 1 to ∞)3 / (2 + eᵏ)
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Domanda del libro di testo
Explain why or why not
Determine whether the following statements are true and give an explanation or counterexample.
a.The terms of the sequence {aₙ} increase in magnitude, so the limit of the sequence does not exist.
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Domanda del libro di testo
12–24. Limits of sequences Evaluate the limit of the sequence or state that it does not exist.
aₙ = 8ⁿ / n!
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