AP Calculus BC 10.6 Comparison Tests for Convergence - FRQs - Exam Style Questions

(a) Integral test
Conditions: for \(x\ge 0\), choose \(f(x)\) positive, continuous, and decreasing.
Take \(f(x)=e^{-x}\).
\(\displaystyle \int_{0}^{\infty} e^{-x}\,dx=\lim_{b\to\infty}\int_{0}^{b} e^{-x}\,dx=\lim_{b\to\infty}\big[-e^{-x}\big]_{0}^{b}=1.\)
Since the improper integral converges, \(\displaystyle \sum_{n=0}^{\infty}\frac{1}{e^{\,n}}\) converges by the integral test.

(b) Limit comparison for absolute convergence
Let \(a_n=\dfrac{1}{e^{\,n}}\) and \(b_n=\left|\dfrac{(-1)^n}{2e^{\,n}+3}\right|=\dfrac{1}{2e^{\,n}+3}\).
\(\displaystyle \lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{1/e^{\,n}}{1/(2e^{\,n}+3)}=\lim_{n\to\infty}\!\left(2+\frac{3}{e^{\,n}}\right)=2>0.\)
Because \(\sum a_n\) converges and the limit is positive and finite, \(\sum b_n\) converges by the limit comparison test.
Hence \(\displaystyle g(1)=\sum_{n=0}^{\infty}\frac{(-1)^n}{2e^{\,n}+3}\) converges absolutely.

(c) Radius of convergence
For \(c_n=\dfrac{(-1)^n}{2e^{\,n}+3}\), apply the ratio test:
\(\displaystyle \lim_{n\to\infty}\left|\frac{c_{n+1}}{c_n}\right| =\lim_{n\to\infty}\frac{2e^{\,n}+3}{2e^{\,n+1}+3} =\frac{1}{e}.\)
Convergence requires \(\dfrac{1}{e}|x|<1\Rightarrow |x|<e\). Therefore the radius of convergence is \(\boxed{R=e}\).

(d) Alternating series error bound
At \(x=1\), the series has terms \(a_n=\dfrac{(-1)^n}{2e^{\,n}+3}\) with \(|a_{n+1}|<|a_n|\) and \(a_n\to 0\).
Using the first two terms (\(n=0,1\)), the remainder satisfies \(\displaystyle \text{Error}\le |a_2|=\frac{1}{2e^{\,2}+3}\).
Thus an upper bound on the error is \(\boxed{\dfrac{1}{2e^{\,2}+3}}\).