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How to determine if ∑n=1∞(−1)nn2\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}∑n=1∞​n2(−1)n​ is absolutely or conditionally convergent?

  1. Take absolute value: ∑n=1∞1n2\sum_{n=1}^{\infty} \frac{1}{n^2}∑n=1∞​n21​. 2. This is a convergent p-series (p=2). 3. Therefore, the series is absolutely convergent.
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How to determine if ∑n=1∞(−1)nn2\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}∑n=1∞​n2(−1)n​ is absolutely or conditionally convergent?

  1. Take absolute value: ∑n=1∞1n2\sum_{n=1}^{\infty} \frac{1}{n^2}∑n=1∞​n21​. 2. This is a convergent p-series (p=2). 3. Therefore, the series is absolutely convergent.

How to determine the convergence of ∑n=1∞cos⁡(n)n2\sum_{n=1}^{\infty} \frac{\cos(n)}{n^2}∑n=1∞​n2cos(n)​?

  1. Take the absolute value: ∑n=1∞∣cos⁡(n)∣n2\sum_{n=1}^{\infty} \frac{|\cos(n)|}{n^2}∑n=1∞​n2∣cos(n)∣​. 2. Since ∣cos⁡(n)∣≤1|\cos(n)| \leq 1∣cos(n)∣≤1, compare to ∑n=1∞1n2\sum_{n=1}^{\infty} \frac{1}{n^2}∑n=1∞​n21​. 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 ∑n=1∞(−1)nn\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}}∑n=1∞​n​(−1)n​ for absolute/conditional convergence?

  1. Alternating Series Test shows convergence. 2. Absolute value gives ∑n=1∞1n\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}∑n=1∞​n​1​, a divergent p-series (p=1/2). 3. Conditionally convergent.

How to test ∑n=1∞(−1)nnn2+1\sum_{n=1}^{\infty} \frac{(-1)^n n}{n^2 + 1}∑n=1∞​n2+1(−1)nn​ for absolute/conditional convergence?

  1. Alternating Series Test shows convergence. 2. Absolute value gives ∑n=1∞nn2+1\sum_{n=1}^{\infty} \frac{n}{n^2 + 1}∑n=1∞​n2+1n​, which diverges by Limit Comparison Test with 1n\frac{1}{n}n1​. 3. Conditionally convergent.

How to test ∑n=1∞sin⁡(n)n!\sum_{n=1}^{\infty} \frac{\sin(n)}{n!}∑n=1∞​n!sin(n)​ for absolute/conditional convergence?

  1. Take absolute value: ∑n=1∞∣sin⁡(n)∣n!\sum_{n=1}^{\infty} \frac{|\sin(n)|}{n!}∑n=1∞​n!∣sin(n)∣​. 2. Since ∣sin⁡(n)∣≤1|\sin(n)| \leq 1∣sin(n)∣≤1, compare to ∑n=1∞1n!\sum_{n=1}^{\infty} \frac{1}{n!}∑n=1∞​n!1​. 3. Ratio Test shows ∑n=1∞1n!\sum_{n=1}^{\infty} \frac{1}{n!}∑n=1∞​n!1​ converges. 4. Absolutely convergent.

How to test ∑n=2∞(−1)nln⁡(n)\sum_{n=2}^{\infty} \frac{(-1)^n}{\ln(n)}∑n=2∞​ln(n)(−1)n​ for absolute/conditional convergence?

  1. Alternating Series Test shows convergence. 2. Absolute value gives ∑n=2∞1ln⁡(n)\sum_{n=2}^{\infty} \frac{1}{\ln(n)}∑n=2∞​ln(n)1​, which diverges by Comparison Test with 1n\frac{1}{n}n1​. 3. Conditionally convergent.

How to test ∑n=1∞(−1)nn22n\sum_{n=1}^{\infty} \frac{(-1)^n n^2}{2^n}∑n=1∞​2n(−1)nn2​ for absolute/conditional convergence?

  1. Take absolute value: ∑n=1∞n22n\sum_{n=1}^{\infty} \frac{n^2}{2^n}∑n=1∞​2nn2​. 2. Apply Ratio Test. 3. The series converges absolutely.

How to test ∑n=1∞(−1)nn+n\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n} + n}∑n=1∞​n​+n(−1)n​ for absolute/conditional convergence?

  1. Alternating Series Test shows convergence. 2. Absolute value gives ∑n=1∞1n+n\sum_{n=1}^{\infty} \frac{1}{\sqrt{n} + n}∑n=1∞​n​+n1​, which converges by Limit Comparison Test with 1n3/2\frac{1}{n^{3/2}}n3/21​. 3. Absolutely convergent.

How to test ∑n=1∞(−1)n(n+1)n\sum_{n=1}^{\infty} \frac{(-1)^n (n+1)}{n}∑n=1∞​n(−1)n(n+1)​ for absolute/conditional convergence?

  1. Alternating Series Test fails since lim⁡n→∞n+1n=1≠0\lim_{n \to \infty} \frac{n+1}{n} = 1 \neq 0limn→∞​nn+1​=1=0, so the series diverges. 2. No need to check absolute convergence.

Define absolute convergence.

A series ∑an\sum a_n∑an​ is absolutely convergent if ∑∣an∣\sum |a_n|∑∣an​∣ converges.

Define conditional convergence.

A series ∑an\sum a_n∑an​ is conditionally convergent if ∑an\sum a_n∑an​ converges but ∑∣an∣\sum |a_n|∑∣an​∣ diverges.

What does convergence mean?

A series converges if the sequence of its partial sums approaches a finite limit.

What does divergence mean?

A series diverges if the sequence of its partial sums does not approach a finite limit.

Define harmonic series.

A harmonic series is a series of the form ∑n=1∞1n\sum_{n=1}^{\infty} \frac{1}{n}∑n=1∞​n1​.

What is an alternating series?

A series where the terms alternate in sign, often involving (−1)n(-1)^n(−1)n or (−1)n+1(-1)^{n+1}(−1)n+1.

Define p-series.

A p-series is a series of the form ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p}∑n=1∞​np1​, where p is a constant.

What is the Direct Comparison Test?

A test to determine convergence or divergence by comparing a given series to a known convergent or divergent series.

What is the Alternating Series Test?

A test used to prove convergence of an alternating series if the absolute value of the terms decreases monotonically to zero.

Define sequence.

An ordered list of numbers.

What is the key difference between absolute and conditional convergence?

Absolute: ∑∣an∣\sum |a_n|∑∣an​∣ converges. Conditional: ∑an\sum a_n∑an​ converges, but ∑∣an∣\sum |a_n|∑∣an​∣ diverges.

Compare the convergence of ∑1n\sum \frac{1}{n}∑n1​ and ∑(−1)nn\sum \frac{(-1)^n}{n}∑n(−1)n​.

∑1n\sum \frac{1}{n}∑n1​: Diverges (Harmonic). ∑(−1)nn\sum \frac{(-1)^n}{n}∑n(−1)n​: Conditionally Converges (Alternating Harmonic).

Compare the convergence of ∑1n2\sum \frac{1}{n^2}∑n21​ and ∑(−1)nn2\sum \frac{(-1)^n}{n^2}∑n2(−1)n​.

∑1n2\sum \frac{1}{n^2}∑n21​: Converges (p-series, p=2). ∑(−1)nn2\sum \frac{(-1)^n}{n^2}∑n2(−1)n​: 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 ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p}∑n=1∞​np1​ for p>1p>1p>1 and p≤1p \le 1p≤1.

For p>1p > 1p>1, the series converges. For p≤1p \le 1p≤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 1n\frac{1}{n}n1​ and 1n!\frac{1}{n!}n!1​ as nnn approaches infinity.

1n\frac{1}{n}n1​ approaches 0 slower than 1n!\frac{1}{n!}n!1​. ∑1n\sum \frac{1}{n}∑n1​ diverges, while ∑1n!\sum \frac{1}{n!}∑n!1​ 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 ∑1np\sum \frac{1}{n^p}∑np1​.