AP Calculus BC 10.7 Alternating Series Test for Convergence - MCQs - Exam Style Questions
No-Calc Question
(B) The series converges absolutely.
(C) The series converges, but there is not enough information to determine if it converges absolutely.
(D) There is not enough information to determine whether the series converges or diverges.
▶️ Answer/Explanation
Alternating Series Test also requires \(a_n\) to be eventually decreasing; only \(a_n>0\) and \(a_n\to 0\) are given.
Without monotonic decrease, the series may converge or diverge (e.g., by rearranging bumps). Hence we cannot decide.
✅ Answer: (D)
No-Calc Question
(B) \(\displaystyle \sum_{n=1}^{\infty} (-1)^n\!\left(\frac{n+1}{2n}\right)\)
(C) \(\displaystyle \sum_{n=1}^{\infty} (-1)^n\!\left(\frac{n^{2}}{3\sqrt{n}}\right)\)
(D) \(\displaystyle \sum_{n=1}^{\infty} (-1)^n\!\left(\frac{2\sqrt{n}}{n}\right)\)
▶️ Answer/Explanation
Use the nth-term test and the Alternating Series Test (AST).
(A) \(\displaystyle a_n=(-1)^n\!\left(\frac{1}{n}-1\right)\). Then \(a_n\to (-1)^n\cdot(-1)\) (does not approach \(0\)) ⇒ diverges by nth-term test.
(B) \(\displaystyle a_n=(-1)^n\!\left(\frac{1}{2}+\frac{1}{2n}\right)\to \pm\frac{1}{2}\neq 0\) ⇒ diverges by nth-term test.
(C) \(\displaystyle a_n=(-1)^n\frac{n^{2}}{3\sqrt{n}}=\frac{1}{3}(-1)^n n^{3/2}\); terms do not go to \(0\) ⇒ diverges (indeed blows up in magnitude).
(D) \(\displaystyle a_n=(-1)^n\frac{2\sqrt{n}}{n}=(-1)^n\frac{2}{\sqrt{n}}\). Here \(\frac{2}{\sqrt{n}}\downarrow 0\) and is eventually decreasing, so by AST the series converges (conditionally). Note that the absolute series \(\sum \frac{2}{\sqrt{n}}\) diverges (p-series with \(p=\tfrac{1}{2}\)).
✅ Answer: (D)