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Ch. 10 - Sequences and Infinite Series
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243당신이 사용하는 게 아니라요?교과서 변경
10장, 문제 10.R.1g

Explain why or why not
Determine whether the following statements are true and give an explanation or counterexample.


g.The series ∑ (from k = 1 to ∞) (k² / (k² + 1)) converges.

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1
Identify the general term of the series: \( a_k = \frac{k^2}{k^2 + 1} \).
Examine the behavior of \( a_k \) as \( k \to \infty \). Calculate the limit \( \lim_{k \to \infty} a_k = \lim_{k \to \infty} \frac{k^2}{k^2 + 1} \).
Since the degrees of numerator and denominator are the same, the limit is the ratio of the leading coefficients, which is \( 1 \). So, \( \lim_{k \to \infty} a_k = 1 \).
Recall the Divergence Test (also called the Test for Divergence): if \( \lim_{k \to \infty} a_k \neq 0 \), then the series \( \sum a_k \) diverges.
Because \( \lim_{k \to \infty} a_k = 1 \neq 0 \), the series \( \sum_{k=1}^\infty \frac{k^2}{k^2 + 1} \) does not converge; it diverges.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Definition of Series Convergence

A series converges if the sequence of its partial sums approaches a finite limit as the number of terms goes to infinity. If the partial sums do not approach a finite value, the series diverges. Understanding this helps determine whether the given infinite sum converges or not.
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가이드 코스
06:52
Convergence of an Infinite Series

Term Test for Divergence

If the terms of a series do not approach zero as k approaches infinity, the series must diverge. This is a quick test to check convergence; if the limit of the general term is not zero, the series cannot converge.
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가이드 코스
05:44
Divergence Test (nth Term Test)

Behavior of the General Term (k² / (k² + 1))

Analyzing the limit of the term k² / (k² + 1) as k approaches infinity shows it approaches 1, not zero. Since the terms do not tend to zero, this indicates the series ∑ (k² / (k² + 1)) diverges by the term test.
추천 영상:
가이드 코스
06:11
Introduction to Riemann Sums