AP Calculus BC 10.6 Comparison Tests for Convergence - Exam Style Questions - MCQs - New Syllabus
Question
Which of the following statements about the series \(\displaystyle \sum_{n=1}^{\infty}\frac{n^{2}+n}{n^{3}+3n^{2}-1}\) is true?
A) The series converges because \(\displaystyle \lim_{n\to\infty}\frac{n^{2}+n}{n^{3}+3n^{2}-1}=0\).
B) The series converges by limit comparison to the series \(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^{3}}\).
C) The series diverges by the ratio test.
D) The series diverges by limit comparison to the series \(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}\).
B) The series converges by limit comparison to the series \(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^{3}}\).
C) The series diverges by the ratio test.
D) The series diverges by limit comparison to the series \(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}\).
▶️ Answer/Explanation
Detailed solution
Compare with \(\displaystyle \sum \frac{1}{n}\) (harmonic). Compute the limit:
\[ \lim_{n\to\infty} \frac{\frac{n^{2}+n}{n^{3}+3n^{2}-1}}{\frac{1}{n}} =\lim_{n\to\infty}\frac{n^{3}+n^{2}}{n^{3}+3n^{2}-1}=1\ (\neq 0,\ \text{finite}). \]
By the limit comparison test, the given series has the same behavior as the harmonic series, which diverges. Hence it diverges by limit comparison to \(\displaystyle \sum \frac{1}{n}\).
✅ Correct: D
Question
Using the comparison test, to which of the following series should \(\displaystyle \sum_{k=1}^{\infty}\frac{2+\cos k}{k^{3}}\) be compared to determine its convergence or divergence?
(A) \(\displaystyle \sum_{k=1}^{\infty}\frac{3}{k}\)
(B) \(\displaystyle \sum_{k=1}^{\infty}\frac{2}{k^{2}}\)
(C) \(\displaystyle \sum_{k=1}^{\infty}\frac{3}{k^{3}}\)
(D) \(\displaystyle \sum_{k=1}^{\infty}\frac{1}{k^{4}}\)
(B) \(\displaystyle \sum_{k=1}^{\infty}\frac{2}{k^{2}}\)
(C) \(\displaystyle \sum_{k=1}^{\infty}\frac{3}{k^{3}}\)
(D) \(\displaystyle \sum_{k=1}^{\infty}\frac{1}{k^{4}}\)