6. the function g has derivatives of all orde...
6. the function g has derivatives of all orders for all real numbers. the maclaurin series for g is given by g(x)=∑n = 0∞(−1)nxn2en+3 on its interval of convergence. (a) state the conditions necessary to use the integral test to determine convergence of the series ∑n = 0∞1en. use the integral test to show that ∑n = 0∞1en converges. (b) use the limit - comparison test with the series ∑n = 0∞1en to show that the series g(1)=∑n = 0∞(−1)n2en+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)=∑n = 0∞(−1)n2en+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
# Answer:
## (a)
### Conditions for integral - test:
Let \(a_n = f(n)\), where \(f(x)\) is a continuous, positive, and decreasing function for \(x\geq N\) (\(N\) is a non - negative integer).
### Proof of convergence of \(\sum_{n = 0}^{\infty}\frac{1}{e^n}\):
Let \(f(x)=\frac{1}{e^x}=e^{-x}\).
1. **Continuity**: The function \(y = e^{-x}\) is continuous for all real \(x\).
2. **Positivity**: For \(x\geq0\), \(e^{-x}>0\) since \(e^x>0\) for all real \(x\).
3. **Decreasing**: Take the derivative \(f^\prime(x)=-e^{-x}<0\) for \(x\geq0\).
Now, calculate the improper integral \(\int_{0}^{\infty}e^{-x}dx=\lim_{t\rightarrow\infty}\int_{0}^{t}e^{-x}dx\).
\[
\begin{align*}
\lim_{t\rightarrow\infty}\int_{0}^{t}e^{-x}dx&=\lim_{t\rightarrow\infty}\left[-e^{-x}\right]_{0}^{t}\\
&=\lim_{t\rightarrow\infty}\left(-e^{-t}+e^{0}\right)\\
&= 1
\end{align*}
\]
Since the improper integral \(\int_{0}^{\infty}e^{-x}dx\) converges, by the integral test, \(\sum_{n = 0}^{\infty}\frac{1}{e^n}\) converges.
## (b)
Let \(a_n=\frac{1}{2e^n + 3}\) and \(b_n=\frac{1}{e^n}\).
Calculate the limit \(\lim_{n\rightarrow\infty}\frac{a_n}{b_n}=\lim_{n\rightarrow\infty}\frac{\frac{1}{2e^n+3}}{\frac{1}{e^n}}=\lim_{n\rightarrow\infty}\frac{e^n}{2e^n + 3}\).
Divide numerator and denominator by \(e^n\): \(\lim_{n\rightarrow\infty}\frac{e^n}{2e^n + 3}=\lim_{n\rightarrow\infty}\frac{1}{2+\frac{3}{e^n}}=\frac{1}{2}\).
Since \(0<\lim_{n\rightarrow\infty}\frac{a_n}{b_n}=\frac{1}{2}<\infty\) and \(\sum_{n = 0}^{\infty}b_n=\sum_{n = 0}^{\infty}\frac{1}{e^n}\) converges (from part (a)), by the limit - comparison test, \(\sum_{n = 0}^{\infty}\left|\frac{(- 1)^n}{2e^n+3}\right|=\sum_{n = 0}^{\infty}\frac{1}{2e^n + 3}\) converges. So, \(g(1)=\sum_{n = 0}^{\infty}\frac{(-1)^n}{2e^n+3}\) converges absolutely.
## (c)
Use the ratio test. Let \(a_n=\frac{(-1)^n x^n}{2e^n+3}\). Then \(a_{n + 1}=\frac{(-1)^{n+1}x^{n + 1}}{2e^{n+1}+3}\).
Calculate the ratio \(\left|\frac{a_{n + 1}}{a_n}\right|=\left|\frac{\frac{(-1)^{n + 1}x^{n+1}}{2e^{n + 1}+3}}{\frac{(-1)^n x^n}{2e^n+3}}\right|=\left|x\right|\left|\frac{2e^n + 3}{2e^{n+1}+3}\right|\).
Divide numerator and denominator by \(e^n\): \(\left|\frac{2e^n + 3}{2e^{n+1}+3}\right|=\left|\frac{2+\frac{3}{e^n}}{2e+\frac{3}{e^n}}\right|\).
Taking the limit as \(n\rightarrow\infty\), \(\lim_{n\rightarrow\infty}\left|\frac{2+\frac{3}{e^n}}{2e+\frac{3}{e^n}}\right|=\frac{1}{e}\).
By the ratio test, the series converges when \(\left|x\right|\frac{1}{e}<1\), i.e., \(\left|x\right|<e\). So the radius of convergence \(R = e\).
## (d)
The alternating series is \(S=\sum_{n = 0}^{\infty}\frac{(-1)^n}{2e^n+3}\), and the approximation uses the first two terms \(S\approx\frac{(-1)^0}{2e^0+3}+\frac{(-1)^1}{2e^1+3}=\frac{1}{5}-\frac{1}{2e + 3}\).
For an alternating series \(\sum_{n = 0}^{\infty}(-1)^n a_n\) (\(a_n>0\), \(a_{n+1}\leq a_n\) and \(\lim_{n\rightarrow\infty}a_n = 0\)), the error bound \(E_N\leq a_{N + 1}\). Here \(N = 1\), \(a_n=\frac{1}{2e^n+3}\), so \(a_2=\frac{1}{2e^2+3}\).
# Explanation:
## (a)
### Step1: State integral - test conditions
For \(a_n = f(n)\), \(f(x)\) must be continuous, positive, and decreasing.
### Step2: Check conditions for \(f(x)=e^{-x}\)
Show continuity, positivity, and decreasing nature.
### Step3: Evaluate integral
\(\lim_{t\rightarrow\infty}\int_{0}^{t}e^{-x}dx=\lim_{t\rightarrow\infty}\left[-e^{-x}\right]_{0}^{t}=1\).
## (b)
### Step1: Define \(a_n\) and \(b_n\)
Let \(a_n=\frac{1}{2e^n + 3}\), \(b_n=\frac{1}{e^n}\).
### Step2: Calculate limit
\(\lim_{n\rightarrow\infty}\frac{a_n}{b_n}=\lim_{n\rightarrow\infty}\frac{e^n}{2e^n + 3}=\frac{1}{2}\).
### Step3: Apply limit - comparison test
Since \(\sum b_n\) converges and \(0<\lim\frac{a_n}{b_n}<\infty\), \(\sum a_n\) converges.
## (c)
### Step1: Apply ratio test
Find \(\left|\frac{a_{n + 1}}{a_n}\right|=\left|x\right|\left|\frac{2e^n + 3}{2e^{n+1}+3}\right|\).
### Step2: Simplify ratio
Divide by \(e^n\) to get \(\left|\frac{2+\frac{3}{e^n}}{2e+\frac{3}{e^n}}\right|\).
### Step3: Find limit
\(\lim_{n\rightarrow\infty}\left|\frac{2+\frac{3}{e^n}}{2e+\frac{3}{e^n}}\right|=\frac{1}{e}\), radius \(R = e\).
## (d)
### Step1: Identify alternating - series error formula
\(E_N\leq a_{N + 1}\) for \(\sum_{n = 0}^{\infty}(-1)^n a_n\).
### Step2: Determine \(a_2\)
With \(N = 1\), \(a_2=\frac{1}{2e^2+3}\).