AP Calculus BC 10.9 Determining Absolute or Conditional Convergence - MCQs - Exam Style Questions
No-Calc Question
Consider the series \[ \sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt[3]{n}} \quad\text{and}\quad \sum_{n=1}^{\infty}\frac{(-1)^n}{n^3}. \] Which of the following statements is true?
(A) Both series converge absolutely.
(B) Both series converge conditionally.
(C) \(\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt[3]{n}}\) converges absolutely, and \(\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^n}{n^3}\) converges conditionally.
(D) \(\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt[3]{n}}\) converges conditionally, and \(\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^n}{n^3}\) converges absolutely.
(B) Both series converge conditionally.
(C) \(\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt[3]{n}}\) converges absolutely, and \(\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^n}{n^3}\) converges conditionally.
(D) \(\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt[3]{n}}\) converges conditionally, and \(\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^n}{n^3}\) converges absolutely.
▶️ Answer/Explanation
Detailed solution
For \(\displaystyle \sum \frac{(-1)^n}{\sqrt[3]{n}}\): alternating, terms decrease to \(0\) ⇒ converges by the Alternating Series Test.
Absolute series \(\displaystyle \sum \frac{1}{\sqrt[3]{n}}\) is a \(p\)-series with \(p=\tfrac13<1\) ⇒ diverges.
So this series converges conditionally.
For \(\displaystyle \sum \frac{(-1)^n}{n^3}\): absolute series \(\displaystyle \sum \frac{1}{n^3}\) has \(p=3>1\) ⇒ converges.
So this series converges absolutely.
✅ Answer: (D)
No-Calc Question
Which of the following series are conditionally convergent?
I. \(\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n}}{n!}\)
II. \(\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n}}{\sqrt{n}}\)
III. \(\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n}}{n+1}\)
II. \(\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n}}{\sqrt{n}}\)
III. \(\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n}}{n+1}\)
(A) I only
(B) II only
(C) II and III only
(D) I, II, and III
(B) II only
(C) II and III only
(D) I, II, and III
▶️ Answer/Explanation
Detailed solution
I. \(\sum \frac{(-1)^n}{n!}\): converges absolutely since \(\sum \frac{1}{n!}\) converges (factorial growth). ⇒ not conditional.
II. \(\sum \frac{(-1)^n}{\sqrt{n}}\): alternating with decreasing terms \(\to 0\) ⇒ converges by AST; absolute series \(\sum \frac{1}{\sqrt{n}}\) diverges (p = \(1/2\)). ⇒ conditionally convergent.
III. \(\sum \frac{(-1)^n}{n+1}\): alternating with decreasing terms \(\to 0\) ⇒ converges by AST; absolute series \(\sum \frac{1}{n+1}\) diverges (harmonic). ⇒ conditionally convergent.
✅ Answer: (C)