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Math  /  Calculus

QuestionWhich of the following series converges conditionally? (A) 1−32+(32)2−(32)3+⋯+(−32)n−1+⋯1-\frac{3}{2}+\left(\frac{3}{2}\right)^{2}-\left(\frac{3}{2}\right)^{3}+\cdots+\left(-\frac{3}{2}\right)^{n-1}+\cdots (B) 1−21!+222!−233!+⋯+(−2)−1(n−1)!+⋯1-\frac{2}{1!}+\frac{2^{2}}{2!}-\frac{2^{3}}{3!}+\cdots+\frac{(-2)^{-1}}{(n-1)!}+\cdots (C) 1−12+13−14+⋯+(−1)−13+⋯1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+\cdots+\frac{(-1)^{-1}}{\sqrt{3}}+\cdots (D) 1−15+(15)2−(15)3+⋯+(−15)n−1+⋯1-\frac{1}{5}+\left(\frac{1}{5}\right)^{2}-\left(\frac{1}{5}\right)^{3}+\cdots+\left(-\frac{1}{5}\right)^{n-1}+\cdots

Studdy Solution

STEP 1

1. We are given four series and need to determine which one converges conditionally.
2. A series converges conditionally if it converges, but does not converge absolutely.
3. To check for conditional convergence, we will first check for absolute convergence and then for convergence.

STEP 2

1. Analyze each series for absolute convergence.
2. Analyze each series for convergence.
3. Determine which series converges conditionally.

STEP 3

Analyze series (A) for absolute convergence:
Series (A) is:
1−32+(32)2−(32)3+⋯+(−32)n−1+⋯ 1 - \frac{3}{2} + \left(\frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^3 + \cdots + \left(-\frac{3}{2}\right)^{n-1} + \cdots
This is a geometric series with common ratio r=−32 r = -\frac{3}{2} .
A geometric series converges absolutely if ∣r∣<1 |r| < 1 .
Here, ∣r∣=32>1 |r| = \frac{3}{2} > 1 , so series (A) does not converge absolutely.

STEP 4

Analyze series (B) for absolute convergence:
Series (B) is:
1−21!+222!−233!+⋯+(−2)n−1(n−1)!+⋯ 1 - \frac{2}{1!} + \frac{2^2}{2!} - \frac{2^3}{3!} + \cdots + \frac{(-2)^{n-1}}{(n-1)!} + \cdots
This is the exponential series for e−2 e^{-2} , which converges absolutely for all real numbers.
Therefore, series (B) converges absolutely.

STEP 5

Analyze series (C) for absolute convergence:
Series (C) is:
1−12+13−14+⋯+(−1)n−1n+⋯ 1 - \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{4}} + \cdots + \frac{(-1)^{n-1}}{\sqrt{n}} + \cdots
Consider the absolute series:
1+12+13+⋯+1n+⋯ 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}} + \cdots
This is a p p -series with p=12 p = \frac{1}{2} .
A p p -series converges if p>1 p > 1 .
Here, p=12<1 p = \frac{1}{2} < 1 , so the series does not converge absolutely.

STEP 6

Analyze series (D) for absolute convergence:
Series (D) is:
1−15+(15)2−(15)3+⋯+(−15)n−1+⋯ 1 - \frac{1}{5} + \left(\frac{1}{5}\right)^2 - \left(\frac{1}{5}\right)^3 + \cdots + \left(-\frac{1}{5}\right)^{n-1} + \cdots
This is a geometric series with common ratio r=−15 r = -\frac{1}{5} .
A geometric series converges absolutely if ∣r∣<1 |r| < 1 .
Here, ∣r∣=15<1 |r| = \frac{1}{5} < 1 , so series (D) converges absolutely.

STEP 7

Analyze series (A) for convergence:
Since series (A) does not converge absolutely, check for conditional convergence using the Alternating Series Test.
The Alternating Series Test requires: - The terms an=(32)n−1 a_n = \left(\frac{3}{2}\right)^{n-1} decrease in magnitude. - The limit of the terms as n→∞ n \to \infty is zero.
The terms do not decrease in magnitude because (32)n−1 \left(\frac{3}{2}\right)^{n-1} increases.
Therefore, series (A) does not converge.

STEP 8

Analyze series (C) for convergence:
Since series (C) does not converge absolutely, check for conditional convergence using the Alternating Series Test.
The Alternating Series Test requires: - The terms an=1n a_n = \frac{1}{\sqrt{n}} decrease in magnitude. - The limit of the terms as n→∞ n \to \infty is zero.
Both conditions are satisfied for series (C).
Therefore, series (C) converges conditionally.

STEP 9

Determine which series converges conditionally:
Series (C) converges conditionally.
The series that converges conditionally is C \boxed{\text{C}} .

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