Home / AP Calculus BC: 10.7 Alternating Series Test  for Convergence – Exam Style questions with Answer- FRQ

AP Calculus BC: 10.7 Alternating Series Test  for Convergence – Exam Style questions with Answer- FRQ

Question

(a) Topic- 10.4 Integral Test for Convergence

(b) Topic-10.7 Alternating Series Test for Convergence

(c) Topic-10.8 Ratio Test for Convergence

(d) Topic-10.10 Alternating Series Error Bound

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

\(\textbf{6(a)}\)
\(
\frac{1}{e^x} \text{ is continuous, positive, and decreasing for } x \geq 0, \text{ so the conditions are met for the integral test.}
\)
\(
\int_{0}^{\infty} \frac{1}{e^x} \, dx = \lim_{b \to \infty} \int_{0}^{b} e^{-x} \, dx = \lim_{b \to \infty} \left[-\frac{1}{e^x}\right]_0^b
\)
\(
= \lim_{b \to \infty} \left[-\frac{1}{e^b} + 1\right] = -[0 – 1] = 1.
\)
Since the integral converges, then the series converges.
\(\textbf{6(b)}\)
\(
\sum_{n=0}^\infty \frac{(-1)^n}{2e^{n}+3} \text{ is an alternating series whose terms decrease in absolute value to } 0, \text{ so it converges.}
\)
\(
\sum_{n=0}^\infty \left|\frac{(-1)^n}{2e^n+3}\right| = \sum_{n=0}^\infty \frac{1}{2e^n+3}.
\)
Using the limit comparison test with $\sum_{n=0}^\infty \frac{1}{e^n}$, which converges:
\(
\lim_{n \to \infty} \frac{\frac{1}{2e^n+3}}{\frac{1}{e^n}} = \lim_{n \to \infty} \frac{e^n}{2e^n+3} = \frac{1}{2} > 0 \text{ and finite,}
\)
so $\sum_{n=0}^\infty \frac{1}{2e^n+3}$ converges, which implies
\(
\sum_{n=0}^\infty \left|\frac{(-1)^n}{2e^n+3}\right| \text{ converges.}
\)
Therefore, $\sum_{n=0}^\infty \frac{(-1)^n}{2e^n+3}$ converges absolutely.
\(\textbf{6(c)}\)
Using the ratio test to determine the radius of convergence:
\(
\lim_{n \to \infty} \left| \frac{x^{n+1}}{2e^{n+1} + 3} \cdot \frac{2e^n + 3}{x^n} \right|
= \lim_{n \to \infty} \left| \frac{x^n x}{2e^{n+1} + 3} \cdot \frac{2e^n + 3}{x^n} \right|
= \lim_{n \to \infty} \left( \frac{2e^n + 3}{2e^{n+1} + 3} \right)|x|
\)
\(
= \frac{1}{e}|x| < 1 \implies |x| < e.
\)
Thus:
\(
-x < 0 < e \quad \text{or} \quad -e < x < e \quad \text{so the radius of convergence is } e.
\)
\(\textbf{6(d)}\)
From part (b), we know that $g(1)$ converges, so we can use the alternating series error bound to find an upper bound on the error of the approximation.
If the first two terms of $g(1)$ are used to approximate $g(1)$, then the alternating series error bound would be the absolute value of the third term, which is when $n = 2$.
The alternating series error bound is:
\(
\frac{1}{2e^2 + 3}.
\)

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