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