AP Calculus BC 10.13 Radius and Interval of Convergence of Power Series - MCQs - Exam Style Questions
Question
What is the radius of convergence of the series \[ \sum_{n=0}^{\infty}\frac{3n\,x^{n}}{(n+2)^{2}}\ ? \]
A) \( \tfrac{1}{3} \)
B) \( 1 \)
C) \( 3 \)
D) \( \infty \)
B) \( 1 \)
C) \( 3 \)
D) \( \infty \)
▶️ Answer/Explanation
Detailed solution
Let \(a_{n}=\dfrac{3n}{(n+2)^{2}}\). For the power series \(\sum a_{n}x^{n}\), the ratio test gives \[ \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right|\,|x| =\lim_{n\to\infty}\frac{3(n+1)}{(n+3)^{2}}\cdot\frac{(n+2)^{2}}{3n}\,|x| =|x|. \] Convergence requires \(|x|<1\), so the radius of convergence is \(R=1\).
✅ Correct: B) \(1\)
Calc-Ok Question
The power series \(\displaystyle \sum_{n=0}^{\infty} a_n (x+1)^n\) converges at \(x=3\). Which of the following must be true?
(A) \(\displaystyle \sum_{n=0}^{\infty} a_n (x+1)^n\) converges at \(x=-5\)
(B) \(\displaystyle \sum_{n=0}^{\infty} a_n (x+1)^n\) diverges at \(x=-5\)
(C) \(\displaystyle \sum_{n=0}^{\infty} a_n (x+1)^n\) converges at \(x=-4\)
(D) \(\displaystyle \sum_{n=0}^{\infty} a_n (x+1)^n\) diverges at \(x=4\)
(B) \(\displaystyle \sum_{n=0}^{\infty} a_n (x+1)^n\) diverges at \(x=-5\)
(C) \(\displaystyle \sum_{n=0}^{\infty} a_n (x+1)^n\) converges at \(x=-4\)
(D) \(\displaystyle \sum_{n=0}^{\infty} a_n (x+1)^n\) diverges at \(x=4\)
▶️ Answer/Explanation
Detailed solution
The series is centered at \(x=-1\). Since it converges at \(x=3\), the radius \(R\) satisfies \(R\ge |3-(-1)|=4\).
Any \(x\) with \(|x+1|<4\) must be inside the interval of convergence, so the series must converge at such points. For \(x=-4\), \(|-4+1|=3<4\), hence convergence is guaranteed.
Behavior at \(|x+1|=4\) (e.g., \(x=-5\) or \(x=3\)) depends on the coefficients and cannot be concluded in general.
✅ Answer: (C)