AP Calculus 10.3 The nth Term Test for Divergence - Exam Style Questions - MCQs - New Syllabus
No-CalcQuestion
Which of the following series diverges by the \(n\)th-term test?
(A) \(\displaystyle \sum_{n=1}^{\infty}\frac{n^{2}+\sin n}{3n^{2}+1}\)
(B) \(\displaystyle \sum_{n=1}^{\infty}\frac{\cos(n\pi)}{n}\)
(C) \(\displaystyle \sum_{n=1}^{\infty}\frac{n^{2}}{n^{3}+1}\)
(D) \(\displaystyle \sum_{n=1}^{\infty}\frac{1}{3n+1}\)
(B) \(\displaystyle \sum_{n=1}^{\infty}\frac{\cos(n\pi)}{n}\)
(C) \(\displaystyle \sum_{n=1}^{\infty}\frac{n^{2}}{n^{3}+1}\)
(D) \(\displaystyle \sum_{n=1}^{\infty}\frac{1}{3n+1}\)
▶️ Answer/Explanation
The \(n\)th-term test says a series \(\sum a_n\) diverges if \(\lim_{n\to\infty} a_n\neq 0\).
\(\displaystyle a_n=\frac{n^{2}+\sin n}{3n^{2}+1}\to \frac{1}{3}\neq 0\), so (A) diverges by the test.
For (B), \(\cos(n\pi)/n=(-1)^n/n\to 0\); for (C), \(n^2/(n^3+1)\sim 1/n\to 0\); for (D), \(1/(3n+1)\to 0\).
✅ Correct: (A)
\(\displaystyle a_n=\frac{n^{2}+\sin n}{3n^{2}+1}\to \frac{1}{3}\neq 0\), so (A) diverges by the test.
For (B), \(\cos(n\pi)/n=(-1)^n/n\to 0\); for (C), \(n^2/(n^3+1)\sim 1/n\to 0\); for (D), \(1/(3n+1)\to 0\).
✅ Correct: (A)
No-Calc Question
Consider the series \(\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^n}{\cos\!\left(\tfrac{1}{n}\right)}\). Which of the following statements is true?
(A) The series converges absolutely.
(B) The series converges conditionally.
(C) The series diverges.
(D) It cannot be determined from the information given.
(B) The series converges conditionally.
(C) The series diverges.
(D) It cannot be determined from the information given.
▶️ Answer/Explanation
Correct answer: (C) The series diverges
As \(n\to\infty\), \(\cos\!\big(\tfrac{1}{n}\big)\to 1\), hence \(a_n=\dfrac{(-1)^n}{\cos(1/n)}\to (-1)^n\), which does not approach \(0\).
Since \(a_n\nrightarrow 0\), the series fails the nth-term test and therefore diverges.