Home / AP Calculus BC 10.5 Harmonic Series and p-Series – MCQs

AP Calculus BC 10.5 Harmonic Series and p-Series - MCQs - Exam Style Questions

No-CalcQuestion

What are all values of \(p\) for which the series \(\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{\,4p-1}}\) converges?
A. \(p\le \tfrac12\)
B. \(p\le 1\)
C. \(p> \tfrac12\)
D. \(p> 1\)
▶️ Answer/Explanation
Detailed solution
This is a \(p\)-series with exponent \( \alpha = 4p-1 \).
A \(p\)-series \( \sum \tfrac{1}{n^{\alpha}} \) converges iff \( \alpha > 1 \).
So require \(4p-1>1 \Rightarrow 4p>2 \Rightarrow p>\tfrac12\).
Correct: C

No-Calc Question

Which of the following series converge?
I. \( \displaystyle \sum_{k=1}^{\infty} \frac{1}{k^{2}+1} \)
II. \( \displaystyle \sum_{k=1}^{\infty} \frac{k}{k^{2}+1} \)
III. \( \displaystyle \sum_{k=1}^{\infty} \frac{k^{2}}{k^{2}+1} \)
(A) None
(B) I only
(C) I and II only
(D) I, II, and III
▶️ Answer/Explanation
Detailed solution

I: \( \dfrac{1}{k^{2}+1} \le \dfrac{1}{k^{2}} \) for \(k\ge1\). Since \( \sum \dfrac{1}{k^{2}} \) converges, by comparison I converges.

II: For large \(k\), \( \dfrac{k}{k^{2}+1} \sim \dfrac{1}{k}\). Since \( \sum \dfrac{1}{k} \) diverges and \( \dfrac{k}{k^{2}+1} \ge \dfrac{1}{2k}\) for big \(k\), II diverges (limit comparison with \(1/k\)).

III: \( \dfrac{k^{2}}{k^{2}+1} \to 1 \ne 0\). Terms do not go to \(0\) ⇒ series diverges.

Answer: (B) I only

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