Textbook Question
∑ (from n=1 to ∞) (1 / √(n + 1)) diverges
b. What should n be in order that the partial sum sₙ = ∑ (from i=1 to n) (1 / √(i + 1)) satisfies sₙ > 1000?
Ch. 10 - Infinite Sequences and Series
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- 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 Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 10, Problem 10.5.66a
Assume that bₙ is a sequence of positive numbers converging to 1/3. Determine if the following series converge or diverge.
a. ∑ (from n = 1 to ∞) [(bₙ₊₁ + bₙ) / n 4ⁿ]
Verified step by step guidance1
Identify the general term of the series: \( a_n = \frac{b_{n+1} + b_n}{n 4^n} \). Since \( b_n \) converges to \( \frac{1}{3} \), both \( b_n \) and \( b_{n+1} \) approach \( \frac{1}{3} \) as \( n \to \infty \).
Use the limit of \( b_n \) to approximate the behavior of the numerator for large \( n \): \( b_{n+1} + b_n \to \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \). This helps simplify the term for large \( n \).
Focus on the denominator \( n 4^n \), which grows very rapidly due to the exponential term \( 4^n \). This suggests the terms \( a_n \) decrease quickly.
Apply the Comparison Test or Limit Comparison Test by comparing \( a_n \) to a known convergent series, such as \( \frac{1}{4^n} \), since \( \frac{b_{n+1} + b_n}{n 4^n} \) behaves similarly to \( \frac{constant}{n 4^n} \).
Conclude about convergence: since \( \sum \frac{1}{4^n} \) converges (geometric series with ratio \( \frac{1}{4} < 1 \)) and \( \frac{1}{n} \) grows slowly, the given series converges by comparison.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Convergence of Infinite Series
An infinite series converges if the sequence of its partial sums approaches a finite limit. Understanding convergence involves analyzing the behavior of terms as n approaches infinity, ensuring the sum does not diverge to infinity or oscillate indefinitely.
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Limit of a Sequence
The limit of a sequence is the value that its terms approach as n becomes very large. Knowing that bₙ converges to 1/3 helps approximate terms in the series for large n, which is crucial for applying convergence tests effectively.
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Guided course
8:22Introduction to Sequences
Comparison and Root/Ratio Tests for Series
These tests help determine series convergence by comparing terms to known convergent series or analyzing the nth root or ratio of successive terms. Given the presence of 4ⁿ in the denominator, the ratio test is particularly useful to assess the exponential decay of terms.
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07:15Root Test
Related Practice
Textbook Question
(Continuation of Exercise 61.) Use the result in Exercise 61 to determine which of the following series converge and which diverge. Support your answer in each case.
a. ∑ (from n=2 to ∞) [1 / (n ln n)]
Textbook Question
The series
sec x = 1 + x²/2 + 5x⁴/24 + 61x⁶/720 + 277x⁸/8064 + ⋯
converges to sec x for −π/2 < x < π/2.
a. Find the first five terms of a power series for the function ln|sec x + tan x|. For what values of x should the series converge?
Textbook Question
Assume that the series ∑ aₙ(x − 2)ⁿ converges for x = −1 and diverges for x = 6. Answer true (T), false (F), or not enough information given (N) for the following statements about the series.
a. Converges absolutely for x = 1
Textbook Question
Quadratic Approximations The Taylor polynomial of order 2 generated by a twice-differentiable function f(x) at x = a is called the quadratic approximation of f at x = a. In Exercises 41–46, find the (a) linearization (Taylor polynomial of order 1)
f(x) = 1 / √(1 − x²)
Textbook Question
Intervals of Convergence
In Exercises 1–36, (a) find the series’ radius and interval of convergence.
∑ (from n = 1 to ∞) [ (√(n + 1) − √n)(x − 3)ⁿ ]