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Ch. 10 - Sequences and Infinite Series
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoBriggs 3rd EditionCh. 10 - Sequences and Infinite SeriesProblema 10.8.83
Capitolo 10, Problema 10.8.83

11–86. Applying convergence tests Determine whether the following series converge. Justify your answers.
∑ (from j = 2 to ∞)1 / (j ln¹⁰j)

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Identify the series given: \( \sum_{j=2}^{\infty} \frac{1}{j (\ln j)^{10}} \). This is a positive term series, so we can consider convergence tests suitable for positive series.
Recognize that the series resembles a p-series with an additional logarithmic factor in the denominator. Since \( j \) grows linearly and \( (\ln j)^{10} \) grows slower than any power of \( j \), we should consider the Integral Test for convergence.
Set up the Integral Test by considering the integral \( \int_{2}^{\infty} \frac{1}{x (\ln x)^{10}} \, dx \). If this improper integral converges, then the series converges; if it diverges, the series diverges.
Make the substitution \( u = \ln x \), which implies \( du = \frac{1}{x} dx \). This transforms the integral into \( \int_{\ln 2}^{\infty} \frac{1}{u^{10}} \, du \).
Evaluate the integral \( \int_{\ln 2}^{\infty} u^{-10} \, du \). Since this is an integral of a power function \( u^{-p} \) with \( p = 10 > 1 \), it converges. Therefore, by the Integral Test, the original series converges.

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Convergence of Infinite Series

An infinite series converges if the sequence of its partial sums approaches a finite limit. Understanding convergence is essential to determine whether the sum of infinitely many terms results in a finite value or diverges to infinity or oscillates.
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Percorso guidato
06:52
Convergence of an Infinite Series

Integral Test for Convergence

The integral test compares a series to an improper integral of a related function. If the integral of f(x) from some point to infinity converges, then the series ∑f(n) also converges, and vice versa. This test is especially useful for series with positive, decreasing terms.
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Percorso guidato
07:51
Choosing a Convergence Test

Behavior of Logarithmic Functions in Series

Logarithmic functions grow slowly, and their powers affect convergence rates. In series like ∑ 1/(j (ln j)^p), the exponent p determines convergence: the series converges if p > 1 and diverges if p ≤ 1, highlighting the importance of understanding how logarithmic terms influence series behavior.
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Graphs of Logarithmic Functions
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Domanda del libro di testo

The first ten terms of the sequence {(1 + 1/10ⁿ)^10ⁿ}∞ ₙ₌₁ are rounded to 8 digits right of the decimal point (see table). Make a conjecture about the limit of the sequence.

n an

1 2.59374246

2 2.70481383

3 2.71692393

4 2.71814593

5 2.71826824

6 2.71828047

7 2.71828169

8 2.71828179

9 2.71828204

10 2.71828203

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32–49. Choose your test Use the test of your choice to determine whether the following series converge absolutely, converge conditionally, or diverge.

∑ (from k = 1 to ∞) 2ᵏ k! / kᵏ

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54–69. Telescoping series

For the following telescoping series, find a formula for the nth term of the sequence of partial sums {Sₙ}. Then evaluate limₙ→∞ Sₙ to obtain the value of the series or state that the series diverges.


65. ∑ (k = 1 to ∞) (1 / √(k + 1) – 1 / √(k + 3))

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"21–26. Recurrence relations Write the first four terms of the sequence {aₙ} defined by the following recurrence relations.

aₙ₊₁ = 3aₙ-12; a₁ = 10

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48–63. Choose your test Determine whether the following series converge or diverge using the properties and tests introduced in Sections 10.3 and 10.4.

∑ (k = 2 to ∞) 1 / eᵏ

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11–27. Alternating Series Test Determine whether the following series converge.

∑ (k = 0 to ∞) (−1)ᵏ / (2k + 1)

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