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AP Calculus BC 10.13 Radius and Interval of Convergence of Power Series - FRQs - Exam Style Questions

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No-Calc Question

The Maclaurin series for a function \(f\) is given by \[ \sum_{n=1}^{\infty}\frac{(n+1)\,x^{n}}{n^{2}6^{n}} \] and converges to \(f(x)\) for all \(x\) in the interval of convergence. It is known that the Maclaurin series for \(f\) has radius of convergence \(6\).

(a) Determine whether the Maclaurin series for \(f\) converges or diverges at \(x=6\).
Give a reason for your answer.

(b) It can be shown that \[ f(-3)=\sum_{n=1}^{\infty}\frac{(n+1)(-3)^{n}}{n^{2}6^{n}} =\sum_{n=1}^{\infty}\frac{n+1}{n^{2}}\!\left(-\tfrac12\right)^{n}, \] and that the first three terms of this series sum to \(S_{3}=-\dfrac{125}{144}\).
Show that \(\,\bigl|\,f(-3)-S_{3}\,\bigr|<\dfrac{1}{50}\).

(c) Find the general term of the Maclaurin series for \(f’\), the derivative of \(f\).
Find the radius of convergence of the Maclaurin series for \(f’\).

(d) Let \(g(x)=\displaystyle\sum_{n=1}^{\infty}\dfrac{(n+1)\,x^{2n}}{n^{2}3^{n}}\).
Use the ratio test to determine the radius of convergence of the Maclaurin series for \(g\).

Most-appropriate topic codes:

• TOPIC 10.13: Radius & Interval of Convergence of Power Series (testing endpoints; same radius after term-by-term differentiation) — parts (a,c). 
• TOPIC 10.10: Alternating Series Error Bound (bound the truncation error) — part (b). 
• TOPIC 10.15: Representing Functions as Power Series (operations such as term-by-term differentiation/integration) — part (c).
• TOPIC 10.8: Ratio Test for Convergence (determine radius for \(g\)) — part (d). 
▶️ Answer/Explanation
(a) Convergence at \(x=6\)
Substitute \(x=6\) into the series: \[ \sum_{n=1}^{\infty}\frac{(n+1)6^{n}}{n^{2}6^{n}} \;=\;\sum_{n=1}^{\infty}\frac{n+1}{n^{2}} \;=\;\sum_{n=1}^{\infty}\!\left(\frac{1}{n}+\frac{1}{n^{2}}\right). \] The harmonic series \(\sum \frac{1}{n}\) diverges, while \(\sum \frac{1}{n^{2}}\) converges, so the sum of the two diverges by comparison.
Conclusion: the Maclaurin series for \(f\) diverges at \(x=6\).
(Endpoint testing is part of finding an interval of convergence.)
(b) Alternating-series error bound at \(x=-3\)
At \(x=-3\), \[ f(-3)=\sum_{n=1}^{\infty}\frac{n+1}{n^{2}}\!\left(-\tfrac12\right)^{n} \quad\text{is alternating with}\quad a_n=\frac{n+1}{n^{2}}\!\left(\tfrac12\right)^{n}>0. \] The terms \(a_n\) decrease to \(0\) (since \(\tfrac{n+1}{n^{2}}\sim\tfrac{1}{n}\) and \((\tfrac12)^n\) shrinks exponentially), so the alternating series error bound applies. :contentReference[oaicite:6]{index=6}
The remainder after three terms satisfies \[ |f(-3)-S_3|\le a_{4} =\frac{5}{4^{2}}\!\left(\tfrac12\right)^{4} =\frac{5}{16}\cdot\frac{1}{16} =\frac{5}{256} <\frac{1}{50}. \] Thus \(\boxed{|\,f(-3)-S_3\,|<\tfrac{1}{50}}\).
(c) Series for \(f’\) and its radius
Differentiate term-by-term: \[ f(x)=\sum_{n=1}^{\infty}\frac{(n+1)}{n^{2}6^{n}}\,x^{n} \quad\Rightarrow\quad f'(x)=\sum_{n=1}^{\infty}\frac{(n+1)}{n^{2}6^{n}}\cdot n\,x^{\,n-1} =\sum_{n=1}^{\infty}\frac{(n+1)}{n\,6^{n}}\,x^{\,n-1}. \] The radius of convergence of a power series is unchanged by term-by-term differentiation, so the radius for \(f’\) is \(\boxed{6}\). 
(d) Radius for \(g(x)=\displaystyle\sum_{n=1}^{\infty}\frac{(n+1)\,x^{2n}}{n^{2}3^{n}}\)
General term: \(a_n=\dfrac{n+1}{n^{2}}\left(\dfrac{x^{2}}{3}\right)^{\!n}\).
Ratio test: \[ \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| =\lim_{n\to\infty}\frac{n+2}{n+1}\cdot\frac{n^{2}}{(n+1)^{2}}\cdot\left|\frac{x^{2}}{3}\right| =1\cdot 1\cdot\left|\frac{x^{2}}{3}\right| =\frac{|x|^{2}}{3}. \] Convergence when \(\dfrac{|x|^{2}}{3}<1\ \Rightarrow\ |x|<\sqrt{3}\).
Hence the radius of convergence of the Maclaurin series for \(g\) is \(\boxed{\sqrt{3}}\). 
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