AP Calculus BC: 10.12 Lagrange Error Bound – Exam Style questions with Answer- FRQ

Question

The function g has derivatives of all orders for all real numbers. The Maclaurin series for g is given by \(g(x)=\sum_{n = 0}^{\infty }\frac{(-1)^{n}x^{n}}{2e^{n}+3}\) on its interval of convergence.
(a) State the conditions necessary to use the integral test to determine convergence of the series \(\sum_{n = 0}^{\infty }\frac{1}{e^{n}}\) Use the integral test to show that \(\sum_{n = 0}^{\infty }\frac{1}{e^{n}}\) converges.
(b) Use the limit comparison test with the series \(\sum_{n = 0}^{\infty }\frac{1}{e^{n}}\) to show that the series \(g(1)=\sum_{n = 0}^{\infty }\frac{(-1)^{n}}{2e^{n}+3}\) converges absolutely.
(c) Determine the radius of convergence of the Maclaurin series for g.
(d) The first two terms of the series \(g(1)=\sum_{n = 0}^{\infty }\frac{(-1)^{n}}{2e^{n}+3}\) are used to approximate g(1) . Use the alternating series error bound to determine an upper bound on the error of the approximation.

Answer/Explanation

Ans:

(a)

The function \(\frac{1}{e^{n}}\) is continuous, positive, and decreasing for n ≥ 0

\(\sum_{n = 0}^{\infty }\frac{1}{e^{n}}\) converges by integral test

\(\lim_{b\rightarrow \infty }\int_{0}^{b}\frac{1}{e^{n}}dn=\lim_{b\rightarrow \infty }\int_{0}^{b}e^{-n}dn\)

\(=\lim_{b\rightarrow \infty }\left [ -e^{-n} \right ]_{0}^{b}=\lim_{b\rightarrow \infty }\left [ -e^{-b} -(-e^{0})\right ]_{0}^{b}\)

= 0 + 1 = 1

(b)

\(g(1)=\sum_{n = 0}^{\infty }\frac{(-1)^{n}}{2e^{n}+3}\)           compare to     \(\sum_{n = 0}^{\infty }\frac{1}{e^{n}}\)    geometric   \(r = \frac{1}{e}<1\) converges

  absolute value of g(1)

\(\lim_{n\rightarrow \infty }\frac{\frac{1}{2e^{n}+3}}{\frac{1}{e^{n}}}=\lim_{n\rightarrow \infty }\frac{e^{n}}{2e^{n}+3}=\frac{1}{2}>0\)

\(\frac{1}{2e^{n}+3} > 0\)                                                                                  g(1) converges absolutely

\(\frac{1}{e^{n}} > 0\)                                                                                         by the limit comparison test


(c)

\(g(x)=\lim_{n\rightarrow \infty }\left | \frac{x^{n+1}}{2e^{n+1}+3} \cdot\frac{2e^{n}+3}{x^{n}} \right |=\left | \frac{x}{e} \right |\)

\(\left | \frac{x}{e} \right |<1\)

|x| < e

R = e

radius of convergence for g

(d)

approximation using first two terms

Error ≤ |third term of the series g(1)|

\(Error \leq \left | \frac{1}{2e^{2}+3} \right |\)

\(Error \leq \frac{1}{2e^{2}+3}\)

Scroll to Top