AP Calculus BC 10.8 Ratio Test for Convergence - MCQs - Exam Style Questions
No-Calc Question
(B) Three
(C) Four
(D) Five
▶️ Answer/Explanation
Alternating series error bound: \(|R_N|\le a_{N+1}\), where \(a_n=\dfrac{1}{(2n-1)!}\).
Check the first omitted term:
\(a_2=\dfrac{1}{3!}=\dfrac{1}{6}\approx0.1667\) (too large)
\(a_3=\dfrac{1}{5!}=\dfrac{1}{120}\approx0.00833\) (still too large)
\(a_4=\dfrac{1}{7!}=\dfrac{1}{5040}\approx0.000198<0.001\) ✅
Therefore summing the first \(N=3\) terms makes the remainder \(<0.001\).
✅ Answer: (B)
No-Calc Question
Which of the following series converge?
I. \(\displaystyle \sum_{n=1}^{\infty}\frac{10^{n}}{n!}\)
II. \(\displaystyle \sum_{n=1}^{\infty}e^{-n}\)
III. \(\displaystyle \sum_{n=1}^{\infty}\frac{2^{n}}{n(n+1)}\)
(B) II only
(C) I and II only
(D) I, II, and III
▶️ Answer/Explanation
I: Ratio test \(\dfrac{a_{n+1}}{a_{n}}=\dfrac{10^{n+1}/(n+1)!}{10^{n}/n!}=\dfrac{10}{n+1}\to 0<1\) ⇒ convergent.
II: Geometric with ratio \(r=e^{-1}\) ⇒ convergent.
III: Compare with \(\dfrac{2^{n}}{n^{2}}\) or use ratio: \(\dfrac{a_{n+1}}{a_{n}}\to 2>1\) ⇒ divergent.
So I and II converge.
✅ Answer: (C)