How to determine if $\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}$ is absolutely or conditionally convergent?
1. Take absolute value: $\sum_{n=1}^{\infty} \frac{1}{n^2}$. 2. This is a convergent p-series (p=2). 3. Therefore, the series is absolutely convergent.
How to determine the convergence of $\sum_{n=1}^{\infty} \frac{\cos(n)}{n^2}$?
1. Take the absolute value: $\sum_{n=1}^{\infty} \frac{|\cos(n)|}{n^2}$. 2. Since $|\cos(n)| \leq 1$, compare to $\sum_{n=1}^{\infty} \frac{1}{n^2}$. 3. The p-series converges, so the original series is absolutely convergent.
Steps to check for conditional convergence.
1. Verify the series converges using Alternating Series Test. 2. Take the absolute value of the terms. 3. Show the absolute value series diverges. 4. Conclude it's conditionally convergent.
How to test $\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}}$ for absolute/conditional convergence?
1. Alternating Series Test shows convergence. 2. Absolute value gives $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$, a divergent p-series (p=1/2). 3. Conditionally convergent.
How to test $\sum_{n=1}^{\infty} \frac{(-1)^n n}{n^2 + 1}$ for absolute/conditional convergence?
1. Alternating Series Test shows convergence. 2. Absolute value gives $\sum_{n=1}^{\infty} \frac{n}{n^2 + 1}$, which diverges by Limit Comparison Test with $\frac{1}{n}$. 3. Conditionally convergent.
How to test $\sum_{n=1}^{\infty} \frac{\sin(n)}{n!}$ for absolute/conditional convergence?
1. Take absolute value: $\sum_{n=1}^{\infty} \frac{|\sin(n)|}{n!}$. 2. Since $|\sin(n)| \leq 1$, compare to $\sum_{n=1}^{\infty} \frac{1}{n!}$. 3. Ratio Test shows $\sum_{n=1}^{\infty} \frac{1}{n!}$ converges. 4. Absolutely convergent.
How to test $\sum_{n=2}^{\infty} \frac{(-1)^n}{\ln(n)}$ for absolute/conditional convergence?
1. Alternating Series Test shows convergence. 2. Absolute value gives $\sum_{n=2}^{\infty} \frac{1}{\ln(n)}$, which diverges by Comparison Test with $\frac{1}{n}$. 3. Conditionally convergent.
How to test $\sum_{n=1}^{\infty} \frac{(-1)^n n^2}{2^n}$ for absolute/conditional convergence?
1. Take absolute value: $\sum_{n=1}^{\infty} \frac{n^2}{2^n}$. 2. Apply Ratio Test. 3. The series converges absolutely.
How to test $\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n} + n}$ for absolute/conditional convergence?
1. Alternating Series Test shows convergence. 2. Absolute value gives $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n} + n}$, which converges by Limit Comparison Test with $\frac{1}{n^{3/2}}$. 3. Absolutely convergent.
How to test $\sum_{n=1}^{\infty} \frac{(-1)^n (n+1)}{n}$ for absolute/conditional convergence?
1. Alternating Series Test fails since $\lim_{n \to \infty} \frac{n+1}{n} = 1 \neq 0$, so the series diverges. 2. No need to check absolute convergence.
Explain the first step in determining absolute or conditional convergence.
First, take the absolute value of the terms in the series, i.e., consider $\sum |a_n|$.
What does it mean if $\sum |a_n|$ converges?
If $\sum |a_n|$ converges, then $\sum a_n$ is absolutely convergent.
What does it mean if $\sum |a_n|$ diverges, but $\sum a_n$ converges?
If $\sum |a_n|$ diverges, but $\sum a_n$ converges, then $\sum a_n$ is conditionally convergent.
Why test for absolute convergence first?
It's often easier to determine absolute convergence first. If a series is absolutely convergent, you don't need to check for conditional convergence.
How does the Alternating Series Test relate to conditional convergence?
The Alternating Series Test can be used to show that an alternating series converges. If the absolute value of that series diverges, then the original series is conditionally convergent.
Explain the role of comparison tests in determining absolute convergence.
Comparison tests, like the Direct Comparison Test, can be used to determine if $\sum |a_n|$ converges or diverges, thus helping to establish absolute convergence.
How does the behavior of $\sin(n)$ affect convergence?
Since $-1 \leq \sin(n) \leq 1$, taking the absolute value means $|sin(n)| \leq 1$. This is useful for comparison tests.
What is a harmonic series, and why is it important in convergence tests?
A harmonic series is $\sum_{n=1}^{\infty} \frac{1}{n}$, and it's a classic example of a divergent series. It's often used for comparison.
Explain the difference between absolute and conditional convergence in terms of error estimation.
Absolutely convergent series have better error estimation properties than conditionally convergent series, as rearranging terms in a conditionally convergent series can change its sum.
How do you handle series that are not alternating but also not strictly positive?
Take the absolute value of the terms and then apply convergence tests. If the absolute value converges, the series is absolutely convergent.
What is the key difference between absolute and conditional convergence?
Absolute: $\sum |a_n|$ converges. Conditional: $\sum a_n$ converges, but $\sum |a_n|$ diverges.
Compare the convergence of $\sum \frac{1}{n}$ and $\sum \frac{(-1)^n}{n}$.
$\sum \frac{1}{n}$: Diverges (Harmonic). $\sum \frac{(-1)^n}{n}$: Conditionally Converges (Alternating Harmonic).
Compare the convergence of $\sum \frac{1}{n^2}$ and $\sum \frac{(-1)^n}{n^2}$.
$\sum \frac{1}{n^2}$: Converges (p-series, p=2). $\sum \frac{(-1)^n}{n^2}$: Absolutely Converges.
Contrast the tests used for absolute vs. conditional convergence.
Absolute: Ratio, Root, Comparison. Conditional: Alternating Series Test, then check absolute value for divergence.
Compare the impact of rearranging terms in absolutely vs. conditionally convergent series.
Absolutely: Rearranging doesn't change the sum. Conditional: Rearranging can change the sum.
What is the difference between using the Direct Comparison Test and the Limit Comparison Test?
Direct Comparison: Directly compare terms. Limit Comparison: Compare the limit of the ratio of terms.
Compare the convergence of $\sum_{n=1}^{\infty} \frac{1}{n^p}$ for $p>1$ and $p \le 1$.
For $p > 1$, the series converges. For $p \le 1$, the series diverges.
Compare absolute convergence to divergence.
Absolute convergence: Series converges even with absolute values. Divergence: Series does not approach a finite limit.
Contrast the behavior of $\frac{1}{n}$ and $\frac{1}{n!}$ as $n$ approaches infinity.
$\frac{1}{n}$ approaches 0 slower than $\frac{1}{n!}$. $\sum \frac{1}{n}$ diverges, while $\sum \frac{1}{n!}$ converges.
Compare the Alternating Series Test with the p-series test.
Alternating Series Test: Tests convergence of alternating series. p-series test: Tests convergence of series of the form $\sum \frac{1}{n^p}$.
