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Ch. 10 - Infinite Sequences and Series
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoHass 15th EditionCh. 10 - Infinite Sequences and SeriesProblema 10.4.54
Capitolo 10, Problema 10.4.54

Determining Convergence or Divergence
Which of the series in Exercises 17–56 converge, and which diverge? Use any method, and give reasons for your answers.
∑ (from n=1 to ∞) 1 / (1 + 2² + 3² + ⋯ + n²)

Guida verificata passo dopo passo
1
First, recognize the general term of the series: \( a_n = \frac{1}{1^2 + 2^2 + 3^2 + \cdots + n^2} \).
Recall the formula for the sum of the first \( n \) squares: \( 1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6} \). Substitute this into the term to get \( a_n = \frac{1}{\frac{n(n+1)(2n+1)}{6}} = \frac{6}{n(n+1)(2n+1)} \).
Analyze the behavior of \( a_n \) for large \( n \). Since \( n(n+1)(2n+1) \) is a cubic polynomial in \( n \), \( a_n \) behaves roughly like \( \frac{6}{2n^3} = \frac{3}{n^3} \) as \( n \to \infty \).
Compare the series \( \sum a_n \) to the p-series \( \sum \frac{1}{n^p} \) with \( p = 3 \). Since \( p > 1 \), the p-series converges.
By the Comparison Test, since \( a_n \) behaves like \( \frac{3}{n^3} \) and \( \sum \frac{1}{n^3} \) converges, the original series \( \sum_{n=1}^\infty \frac{1}{1^2 + 2^2 + \cdots + n^2} \) also converges.

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Infinite Series and Convergence

An infinite series is the sum of infinitely many terms. Determining whether such a series converges means checking if the sum approaches a finite limit as the number of terms grows indefinitely. If the sum does not approach a finite value, the series diverges.
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Convergence of an Infinite Series

Sum of Squares Formula

The sum of the first n squares is given by the formula 1² + 2² + ⋯ + n² = n(n + 1)(2n + 1)/6. This formula helps simplify the denominator in the given series, making it easier to analyze the behavior of the terms as n becomes large.
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06:11
Introduction to Riemann Sums

Comparison Test for Series

The comparison test involves comparing the terms of a given series to those of a known benchmark series. If the terms of the given series are smaller than those of a convergent series, it also converges; if larger than those of a divergent series, it diverges. This test helps determine convergence without finding the exact sum.
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Percorso guidato
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Direct Comparison Test
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 = 0 to ∞) [(2n + 3)(2ⁿ + 3) / (3ⁿ + 2)]

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 = 2 to ∞) [(3n)! / (n!)³]

Domanda del libro di testo

Determining Convergence or Divergence

Which of the series in Exercises 17–56 converge, and which diverge? Use any method, and give reasons for your answers.

∑ (from n=1 to ∞) (1 + cos n) / n²

Domanda del libro di testo

Determining Convergence or Divergence

In Exercises 17–46, use any method to determine whether the series converges or diverges. Give reasons for your answer.

∑(from n=3 to ∞) [2n² / n²ⁿ]

Domanda del libro di testo

Find the first four nonzero terms in the Maclaurin series for the functions in Exercises 31–38.

(tan⁻¹x)²

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 ∞) tan(n^(1/n))