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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 10.6.58
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]
Guida verificata passo dopo passo1
Identify the given series: \( \sum_{n=1}^{\infty} \frac{3^n}{n^3} \). This is an infinite series with terms \( a_n = \frac{3^n}{n^3} \).
Recognize that the numerator \(3^n\) is an exponential function and the denominator \(n^3\) is a polynomial function. Generally, exponential growth dominates polynomial growth as \(n\) becomes large.
Consider applying the Root Test or Ratio Test, which are effective for series involving exponential terms. For the Ratio Test, compute \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).
Calculate \( \frac{a_{n+1}}{a_n} = \frac{3^{n+1} / (n+1)^3}{3^n / n^3} = 3 \cdot \frac{n^3}{(n+1)^3} \). Then find the limit \( L = \lim_{n \to \infty} 3 \cdot \frac{n^3}{(n+1)^3} \).
Evaluate the limit \( L \). If \( L > 1 \), the series diverges; if \( L < 1 \), the series converges; if \( L = 1 \), the test is inconclusive. Use this conclusion to determine the behavior of the series.
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Infinite Series and Convergence
An infinite series is the sum of infinitely many terms. Determining convergence means checking if the sum approaches a finite limit as the number of terms grows. If the series converges, its partial sums get closer to a specific value; if it diverges, the sums grow without bound or oscillate.
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Percorso guidato
06:52
Convergence of an Infinite Series
Ratio Test
The Ratio Test is used to determine the convergence of series with positive terms, especially those involving exponentials or factorials. It examines the limit of the absolute value of the ratio of consecutive terms. If this limit is less than 1, the series converges absolutely; if greater than 1, it diverges.
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Percorso guidato
05:35Ratio Test
Comparison of Growth Rates (Exponential vs Polynomial)
In series terms, exponential functions grow faster than any polynomial function as n approaches infinity. When the numerator grows exponentially and the denominator polynomially, the terms often do not approach zero, leading to divergence. Recognizing this helps in quickly assessing series behavior.
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09:29Exponential Growth & Decay
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Finding Terms of a Sequence
Each of Exercises 1–6 gives a formula for the nth term aₙ of a sequence {aₙ}. Find the values of a₁, a₂, a₃, and a₄.
aₙ = 1 / n!
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In Exercises 15–22, determine if the geometric series converges or diverges. If a series converges, find its sum.
1 − (2/e) + (2/e)² − (2/e)³ + (2/e)⁴ − …
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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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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))
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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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