Ch. 10 - Sequences and Infinite Series

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 10 - Sequences and Infinite Series

Problem 10.2.79

Chapter 10, Problem 10.2.79

Growth rates of sequences
Use Theorem 10.6 to find the limit of the following sequences or state that they diverge.

{n¹⁰⁰⁰ / 2ⁿ}

Verified step by step guidance

Identify the sequence given: $a_n = \frac{n^{1000}}{2^n}$.

Recall Theorem 10.6, which states that for sequences of the form $\frac{n^p}{r^n}$ where $p$ is a positive integer and $r > 1$, the exponential term in the denominator grows faster than any polynomial term in the numerator.

Apply this theorem to the sequence: since \$2^n$ grows exponentially and $n^{1000}$ grows polynomially, the denominator grows faster than the numerator as $n \to \infty$.

Conclude that the limit of the sequence $a_n$ as $n$ approaches infinity is 0 because the denominator dominates the numerator.

Therefore, $\lim_{n \to \infty} \frac{n^{1000}}{2^n} = 0$.

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Key Concepts

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

Limits of Sequences

A limit of a sequence describes the value that the terms of the sequence approach as the index goes to infinity. Understanding how to evaluate these limits helps determine whether a sequence converges to a finite number or diverges to infinity or does not settle at all.

Growth Rates of Functions

Comparing growth rates involves analyzing how fast different functions increase as their input becomes very large. For example, exponential functions like 2^n grow faster than any polynomial function like n^1000, which is crucial in determining the behavior of sequences involving these terms.

Theorem 10.6 (Limit Comparison of Sequences)

Theorem 10.6 typically refers to a result that helps find limits of sequences by comparing them to known sequences with established limits. It often involves using dominant terms or applying limit laws to simplify complex sequences and determine their convergence or divergence.

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Textbook Question

45–63. Absolute and conditional convergence Determine whether the following series converge absolutely, converge conditionally, or diverge.

∑ (k = 1 to ∞) (−1)ᵏ · k / (2k + 1)

Textbook Question

72–86. Evaluating series Evaluate each series or state that it diverges.

∑ (k = 2 to ∞) ln((k + 1)k⁻¹) / (ln k × ln(k + 1))

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Textbook Question

46–53. Decimal expansions

Write each repeating decimal first as a geometric series and then as a fraction (a ratio of two integers).

53.0.00952̅ = 0.00952952…

Textbook Question

Is it possible for an alternating series to converge absolutely but not conditionally?

Textbook Question

Periodic doses

Suppose you take 200 mg of an antibiotic every 6 hr. The half-life of the drug (the time it takes for half of the drug to be eliminated from your blood) is 6 hr. Use infinite series to find the long-term (steady-state) amount of antibiotic in your blood. You may assume the steady-state amount is finite.

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Textbook Question

40–62. Choose your test Use the test of your choice to determine whether the following series converge.

∑ (k = 2 to ∞) (5lnk) / k

Growth rates of sequencesUse Theorem 10.6 to find the limit of the following sequences or state that they diverge.{n¹⁰⁰⁰ / 2ⁿ}

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Key Concepts

Limits of Sequences

Growth Rates of Functions

Theorem 10.6 (Limit Comparison of Sequences)

Growth rates of sequences
Use Theorem 10.6 to find the limit of the following sequences or state that they diverge.

{n¹⁰⁰⁰ / 2ⁿ}