23. The alternating series test can be used to show convergence for which of the following series? I. $$ 1-\frac{1}{4}+\frac{1}{9}-\frac{1}{16}+\frac{1}{25}-\frac{1}{36}+\cdots+a_{n}+\cdots $$, where $$ a_{n}=(-1)^{n+1} \frac{1}{n^{2}} $$ II. $$ \sin 1-\frac{\sin 2}{4}+\frac{\sin 3}{9}-\frac{\sin 4}{16}+\frac{\sin 5}{25}-\frac{\sin 6}{36}+\cdots+b_{n}+\cdots $$, where $$ b_{n}=(-1)^{n+1} \frac{\sin n}{n^{2}} $$ III. $$ \frac{1}{\sqrt{2}+1}-\frac{1}{\sqrt{2}-1}+\frac{1}{\sqrt{3}+1}-\frac{1}{\sqrt{3}-1}+\frac{1}{\sqrt{4}+1}-\frac{1}{\sqrt{4}-1}+\cdots+c_{n}+\cdots $$, where $$ c_{n}=\left\{\begin{array}{ll}\frac{1}{\sqrt{k+1}+1} & ext { if } n=2 k-1 \frac{1}{\sqrt{k+1}+1}-\frac{1}{\sqrt{k+1}-1} \ -\frac{1}{\sqrt{k+1}-1} & ext { if } n=2 k \ ext { (A) I only } \ ext { (B) II only } \ ext { (C) I and II only } \ ext { (D) I and III only }\end{array} ight. $$

Question

23. The alternating series test can be used to show convergence for which of the following series? I. $$ 1-\frac{1}{4}+\frac{1}{9}-\frac{1}{16}+\frac{1}{25}-\frac{1}{36}+\cdots+a_{n}+\cdots $$, where $$ a_{n}=(-1)^{n+1} \frac{1}{n^{2}} $$ II. $$ \sin 1-\frac{\sin 2}{4}+\frac{\sin 3}{9}-\frac{\sin 4}{16}+\frac{\sin 5}{25}-\frac{\sin 6}{36}+\cdots+b_{n}+\cdots $$, where $$ b_{n}=(-1)^{n+1} \frac{\sin n}{n^{2}} $$ III. $$ \frac{1}{\sqrt{2}+1}-\frac{1}{\sqrt{2}-1}+\frac{1}{\sqrt{3}+1}-\frac{1}{\sqrt{3}-1}+\frac{1}{\sqrt{4}+1}-\frac{1}{\sqrt{4}-1}+\cdots+c_{n}+\cdots $$, where $$ c_{n}=\left\{\begin{array}{ll}\frac{1}{\sqrt{k+1}+1} & 	ext { if } n=2 k-1 \frac{1}{\sqrt{k+1}+1}-\frac{1}{\sqrt{k+1}-1} \ -\frac{1}{\sqrt{k+1}-1} & 	ext { if } n=2 k \ 	ext { (A) I only } \ 	ext { (B) II only } \ 	ext { (C) I and II only } \ 	ext { (D) I and III only }\end{array}ight. $$

UpStudy AI · Accepted Answer

To determine which of the given series can be shown to converge using the alternating series test, we need to analyze each series individually. The alternating series test states that a series of the form $$ \sum (-1)^{n} a_n $$ converges if the following two conditions are met:

1. The sequence $$ a_n $$ is positive.
2. The sequence $$ a_n $$ is monotonically decreasing (i.e., $$ a_{n+1} \leq a_n $$ for all $$ n $$).
3. The limit of $$ a_n $$ as $$ n $$ approaches infinity is zero: $$ \lim_{n \to \infty} a_n = 0 $$.

Let's analyze each series:

### Series I:
$$
a_n = (-1)^{n+1} \frac{1}{n^2}
$$
- **Positivity**: $$ a_n = \frac{1}{n^2} > 0 $$ for all $$ n \geq 1 $$.
- **Monotonicity**: $$ a_n $$ is decreasing since $$ \frac{1}{n^2} > \frac{1}{(n+1)^2} $$.
- **Limit**: $$ \lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{1}{n^2} = 0 $$.

Since all conditions are satisfied, Series I converges.

### Series II:
$$
b_n = (-1)^{n+1} \frac{\sin n}{n^2}
$$
- **Positivity**: The term $$ b_n $$ is not guaranteed to be positive since $$ \sin n $$ oscillates between -1 and 1. Thus, $$ b_n $$ can be negative for odd $$ n $$.
- **Monotonicity**: We cannot determine monotonicity without knowing the behavior of $$ \sin n $$.
- **Limit**: $$ \lim_{n \to \infty} b_n $$ does not exist because $$ \sin n $$ oscillates.

Since the first condition fails (positivity), Series II does not converge by the alternating series test.

### Series III:
$$
c_n = \begin{cases}
\frac{1}{\sqrt{k+1}+1} & \text{if } n=2k-1 \
\frac{1}{\sqrt{k+1}+1}-\frac{1}{\sqrt{k+1}-1} & \text{if } n=2k \
-\frac{1}{\sqrt{k+1}-1} & \text{if } n=2k
\end{cases}
$$
To analyze this series, we need to check the positivity and behavior of $$ c_n $$:
- **Positivity**: The terms $$ c_n $$ need to be checked for positivity. The first term is positive, but the second term's positivity depends on $$ k $$.
- **Monotonicity**: The behavior of $$ c_n $$ is complex due to the piecewise definition, making it difficult to determine if it is monotonically decreasing.
- **Limit**: The limit as $$ n \to \infty $$ needs to be evaluated, but the oscillation and complexity of the terms make it unclear.

Due to the complexity and lack of clear positivity and monotonicity, we cannot definitively conclude that Series III converges using the alternating series test.

### Conclusion:
Based on the analysis:
- Series I converges.
- Series II does not converge.
- Series III cannot be confirmed to converge.

Thus, the correct answer is:
$$
\text{(A) I only}
$$
Series I converges using the alternating series test.

The alternating series test can be used to show convergence for which of the following series?
I. , where
II. , where
III. ,
where

Solución de inteligencia artificial de Upstudy

Responder

Solución

Bonus Knowledge

preguntas relacionadas

Latest Calculus Questions

The alternating series test can be used to show convergence for which of the following series? I. , where II. , where III. , where