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  2. AP Calculus
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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.

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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​.

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.

What does the Alternating Series Test state?

If ana_nan​ is decreasing and lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0limn→∞​an​=0, then ∑(−1)nan\sum (-1)^n a_n∑(−1)nan​ converges.

What does the Direct Comparison Test state?

If 0≤an≤bn0 \leq a_n \leq b_n0≤an​≤bn​ and ∑bn\sum b_n∑bn​ converges, then ∑an\sum a_n∑an​ converges. If an≥bn≥0a_n \geq b_n \geq 0an​≥bn​≥0 and ∑bn\sum b_n∑bn​ diverges, then ∑an\sum a_n∑an​ diverges.

What does the Limit Comparison Test state?

If lim⁡n→∞anbn=c\lim_{n \to \infty} \frac{a_n}{b_n} = climn→∞​bn​an​​=c, where 0<c<∞0 < c < \infty0<c<∞, then ∑an\sum a_n∑an​ and ∑bn\sum b_n∑bn​ either both converge or both diverge.

What does the p-series test state?

The series ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p}∑n=1∞​np1​ converges if p>1p > 1p>1 and diverges if p≤1p \leq 1p≤1.

State the Ratio Test.

Let L=lim⁡n→∞∣an+1an∣L = \lim_{n \to \infty} |\frac{a_{n+1}}{a_n}|L=limn→∞​∣an​an+1​​∣. If L<1L < 1L<1, the series converges absolutely. If L>1L > 1L>1, the series diverges. If L=1L = 1L=1, the test is inconclusive.

State the Root Test.

Let L=lim⁡n→∞∣an∣nL = \lim_{n \to \infty} \sqrt[n]{|a_n|}L=limn→∞​n∣an​∣​. If L<1L < 1L<1, the series converges absolutely. If L>1L > 1L>1, the series diverges. If L=1L = 1L=1, the test is inconclusive.

What is the absolute convergence theorem?

If ∑∣an∣\sum |a_n|∑∣an​∣ converges, then ∑an\sum a_n∑an​ converges.

What is the nth-term test for divergence?

If lim⁡n→∞an≠0\lim_{n \to \infty} a_n \neq 0limn→∞​an​=0, then the series ∑an\sum a_n∑an​ diverges.

State the integral test.

If f(x)f(x)f(x) is continuous, positive, and decreasing on [1,∞)[1, \infty)[1,∞), then ∑n=1∞f(n)\sum_{n=1}^{\infty} f(n)∑n=1∞​f(n) and ∫1∞f(x)dx\int_{1}^{\infty} f(x) dx∫1∞​f(x)dx either both converge or both diverge.

State the theorem on rearrangement of absolutely convergent series.

If a series is absolutely convergent, then any rearrangement of the series converges to the same sum.