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Ch. 10 - Infinite Sequences and Series
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 10 - Infinite Sequences and SeriesProblem 10.4.54
Chapter 10, Problem 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)

Verified step by step guidance
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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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

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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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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Direct Comparison Test
Related Practice
Textbook Question

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)]

Textbook Question

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]

Textbook Question

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

Textbook Question

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]

Textbook Question

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

(tan⁻¹x)²

Textbook Question

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))