AP Calculus BC 10.10 Alternating Series Error Bound- MCQs -Exam Style Questions
Question
The alternating series \( \displaystyle \sum_{n=1}^{\infty} (-1)^{n} a_{n} \) converges to \(S\) and satisfies \( a_{n} > a_{n+1} > 0 \) for all \( n \ge 1 \). Selected values are \[ a_{2}=0.2000,\qquad a_{3}=0.1835,\qquad a_{4}=0.1715,\qquad a_{5}=0.1622. \] What is the smallest integer \(k\) such that \[ \left|\, S – \sum_{n=1}^{k} (-1)^{n} a_{n} \right| \;<\; 0.1667 \, ? \]
(B) \(4\)
(C) \(5\)
(D) \(6\)
▶️ Answer/Explanation
Correct answer: (B) \(4\)
Alternating–series remainder (Leibniz estimate): for partial sums \(S_k=\displaystyle\sum_{n=1}^{k}(-1)^n a_n\), \[ |S – S_k| \;=\; |R_k| \;\le\; a_{k+1}, \] provided \(a_n\) is positive, decreasing, and \(\lim a_n=0\) (all satisfied by the given data).
We need the smallest \(k\) with \[ a_{k+1} \;<\; 0.1667 \;\approx\; \frac{1}{6}. \] Check successive candidates:
\(k=2:\; a_{3}=0.1835 > 0.1667\) \(\Rightarrow\) fails.
\(k=3:\; a_{4}=0.1715 > 0.1667\) \(\Rightarrow\) fails.
\(k=4:\; a_{5}=0.1622 < 0.1667\) \(\Rightarrow\) succeeds.
Therefore the minimal \(k\) is \[ \boxed{k=4}. \] Note. Using the exact threshold \( \tfrac{1}{6}\approx 0.166\overline{6} \) gives the same conclusion since \( 0.1715 > \tfrac{1}{6} \) but \( 0.1622 < \tfrac{1}{6}. \)
Calc-Ok Question
(B) \(1/7\)
(C) \(1/14\)
(D) \(1/16\)
▶️ Answer/Explanation
For an alternating series with positive decreasing terms \(a_n\), the remainder after \(N\) terms satisfies \(|R_N|\le a_{N+1}\).
Here \(a_n=\dfrac{1}{2n+2}\) ⇒ with \(N=6\): \(a_{7}=\dfrac{1}{2(7)+2}=\dfrac{1}{16}\).
✅ Answer: (D)