IDNLearn.com is your go-to platform for finding accurate and reliable answers. Ask your questions and receive prompt, detailed answers from our experienced and knowledgeable community members.
Sagot :
To determine whether the given infinite series converge or diverge, we will use the integral test, which states that for a positive, continuous, and decreasing function [tex]\( f(x) \)[/tex] on the interval [tex]\([a, \infty)\)[/tex], the infinite series [tex]\(\sum_{n=a}^{\infty} f(n)\)[/tex] converges if and only if the improper integral [tex]\(\int_{a}^{\infty} f(x) \, dx\)[/tex] converges.
### Part (A): [tex]\(\sum_{n=1}^{\infty} \frac{\ln n}{n^2}\)[/tex]
#### Step 1: Define the Function
Let [tex]\( f(x) = \frac{\ln x}{x^2} \)[/tex]. We'll check if this function is positive, continuous, and decreasing for [tex]\( x \geq 1 \)[/tex].
- Positivity: For [tex]\( x \geq 1 \)[/tex], both the numerator [tex]\( \ln x \)[/tex] and the denominator [tex]\( x^2 \)[/tex] are positive, so [tex]\( f(x) > 0 \)[/tex].
- Continuity: The function [tex]\( f(x) = \frac{\ln x}{x^2} \)[/tex] is continuous for [tex]\( x > 0 \)[/tex].
- Decreasing: To check if [tex]\( f(x) \)[/tex] is decreasing, we can find its derivative:
[tex]\[ f'(x) = \frac{d}{dx} \left( \frac{\ln x}{x^2} \right) = \frac{(1/x) \cdot x^2 - \ln x \cdot 2x}{x^4} = \frac{x - 2x \ln x}{x^3} = \frac{1 - 2 \ln x}{x^3} \][/tex]
For [tex]\( x \geq 3 \)[/tex], [tex]\( 2 \ln x \geq 1 \)[/tex], making [tex]\( 1 - 2 \ln x \leq 0 \)[/tex]. Hence, [tex]\( f'(x) \leq 0 \)[/tex] for [tex]\( x \geq 3 \)[/tex], indicating that [tex]\( f(x) \)[/tex] is decreasing for large [tex]\( x \)[/tex].
#### Step 2: Evaluate the Improper Integral
Evaluate the improper integral [tex]\(\int_{1}^{\infty} \frac{\ln x}{x^2} \, dx\)[/tex]:
[tex]\[ \int_{1}^{\infty} \frac{\ln x}{x^2} \, dx \][/tex]
Using integration by parts, let [tex]\( u = \ln x \)[/tex] and [tex]\( dv = \frac{1}{x^2} dx \)[/tex]. Then [tex]\( du = \frac{1}{x} dx \)[/tex] and [tex]\( v = -\frac{1}{x} \)[/tex].
[tex]\[ \int \ln x \cdot \frac{1}{x^2} \, dx = -\frac{\ln x}{x} - \int -\frac{1}{x^2} \cdot \frac{1}{x} \, dx = -\frac{\ln x}{x} + \int \frac{1}{x^2} \, dx = -\frac{\ln x}{x} - \frac{1}{x} = -\frac{\ln x + 1}{x} \][/tex]
So,
[tex]\[ \left[ -\frac{\ln x + 1}{x} \right]_{1}^{\infty} \][/tex]
Evaluating this from 1 to infinity,
[tex]\[ \lim_{b \to \infty} \left( -\frac{\ln b + 1}{b} \right) - \left( -\frac{\ln 1 + 1}{1} \right) = 0 + 1 = 1 \][/tex]
Thus, the integral converges to 1, and therefore, the series:
[tex]\[ \sum_{n=1}^{\infty} \frac{\ln n}{n^2} \][/tex]
converges.
### Part (B): [tex]\(\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{1 / 2}}\)[/tex]
#### Step 1: Define the Function
Let [tex]\( f(x) = \frac{1}{x (\ln x)^{1/2}} \)[/tex]. Check if this function is positive, continuous, and decreasing for [tex]\( x \geq 2 \)[/tex].
- Positivity: For [tex]\( x \geq 2 \)[/tex], both [tex]\( x \)[/tex] and [tex]\( \ln x \)[/tex] are positive, so [tex]\( f(x) > 0 \)[/tex].
- Continuity: The function [tex]\( f(x) = \frac{1}{x (\ln x)^{1/2}} \)[/tex] is continuous for [tex]\( x > 1 \)[/tex].
- Decreasing: To check if [tex]\( f(x) \)[/tex] is decreasing, we find its derivative.
[tex]\[ f'(x) = \frac{d}{dx} \left( \frac{1}{x (\ln x)^{1/2}} \right) = \frac{(\ln x)^{1/2} \cdot (-1/x^2) - 1 \cdot \frac{1}{2} (\ln x)^{-1/2} \cdot \frac{1}{x}}{(\ln x)^1} \][/tex]
Simplified:
[tex]\[ f'(x) = \frac{- (\ln x)^{1/2} - \frac{1}{2x (\ln x)^{1/2}} }{x^2 (\ln x)} \][/tex]
For [tex]\( x \geq 2 \)[/tex], the numerator is negative, making [tex]\( f'(x) \leq 0 \)[/tex], indicating that [tex]\( f(x) \)[/tex] is decreasing for [tex]\( x \geq 2 \)[/tex].
#### Step 2: Evaluate the Improper Integral
Evaluate the improper integral [tex]\(\int_{2}^{\infty} \frac{1}{x (\ln x)^{1/2}} \, dx\)[/tex]:
[tex]\[ \int_{2}^{\infty} \frac{1}{x (\ln x)^{1/2}} \, dx \][/tex]
Let [tex]\( u = \ln x \)[/tex], thus [tex]\( du = \frac{1}{x} dx \)[/tex]. Changing the limits, for [tex]\(x = 2\)[/tex], [tex]\( u = \ln 2 \)[/tex], and as [tex]\( x \to \infty \)[/tex], [tex]\( u \to \infty \)[/tex].
[tex]\[ \int \frac{1}{x (\ln x)^{1/2}} dx = \int \frac{1}{(\ln x)^{1/2}} \frac{1}{x} dx = \int \frac{1}{u^{1/2}} du = 2 u^{1/2} = 2 (\ln x)^{1/2} \][/tex]
Evaluate this from [tex]\( 2 \)[/tex] to [tex]\( \infty \)[/tex]:
[tex]\[ \left[ 2 (\ln x)^{1/2} \right]_{2}^{\infty} \][/tex]
[tex]\[ \lim_{b \to \infty} \left( 2 (\ln b)^{1/2} \right) - 2 (\ln 2)^{1/2} \][/tex]
Since [tex]\( \ln b \to \infty \)[/tex] as [tex]\( b \to \infty \)[/tex],
[tex]\[ 2 (\ln b)^{1/2} \to \infty \][/tex]
Therefore, the integral diverges to infinity, and thus, the series:
[tex]\[ \sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{1/2}} \][/tex]
diverges.
### Part (A): [tex]\(\sum_{n=1}^{\infty} \frac{\ln n}{n^2}\)[/tex]
#### Step 1: Define the Function
Let [tex]\( f(x) = \frac{\ln x}{x^2} \)[/tex]. We'll check if this function is positive, continuous, and decreasing for [tex]\( x \geq 1 \)[/tex].
- Positivity: For [tex]\( x \geq 1 \)[/tex], both the numerator [tex]\( \ln x \)[/tex] and the denominator [tex]\( x^2 \)[/tex] are positive, so [tex]\( f(x) > 0 \)[/tex].
- Continuity: The function [tex]\( f(x) = \frac{\ln x}{x^2} \)[/tex] is continuous for [tex]\( x > 0 \)[/tex].
- Decreasing: To check if [tex]\( f(x) \)[/tex] is decreasing, we can find its derivative:
[tex]\[ f'(x) = \frac{d}{dx} \left( \frac{\ln x}{x^2} \right) = \frac{(1/x) \cdot x^2 - \ln x \cdot 2x}{x^4} = \frac{x - 2x \ln x}{x^3} = \frac{1 - 2 \ln x}{x^3} \][/tex]
For [tex]\( x \geq 3 \)[/tex], [tex]\( 2 \ln x \geq 1 \)[/tex], making [tex]\( 1 - 2 \ln x \leq 0 \)[/tex]. Hence, [tex]\( f'(x) \leq 0 \)[/tex] for [tex]\( x \geq 3 \)[/tex], indicating that [tex]\( f(x) \)[/tex] is decreasing for large [tex]\( x \)[/tex].
#### Step 2: Evaluate the Improper Integral
Evaluate the improper integral [tex]\(\int_{1}^{\infty} \frac{\ln x}{x^2} \, dx\)[/tex]:
[tex]\[ \int_{1}^{\infty} \frac{\ln x}{x^2} \, dx \][/tex]
Using integration by parts, let [tex]\( u = \ln x \)[/tex] and [tex]\( dv = \frac{1}{x^2} dx \)[/tex]. Then [tex]\( du = \frac{1}{x} dx \)[/tex] and [tex]\( v = -\frac{1}{x} \)[/tex].
[tex]\[ \int \ln x \cdot \frac{1}{x^2} \, dx = -\frac{\ln x}{x} - \int -\frac{1}{x^2} \cdot \frac{1}{x} \, dx = -\frac{\ln x}{x} + \int \frac{1}{x^2} \, dx = -\frac{\ln x}{x} - \frac{1}{x} = -\frac{\ln x + 1}{x} \][/tex]
So,
[tex]\[ \left[ -\frac{\ln x + 1}{x} \right]_{1}^{\infty} \][/tex]
Evaluating this from 1 to infinity,
[tex]\[ \lim_{b \to \infty} \left( -\frac{\ln b + 1}{b} \right) - \left( -\frac{\ln 1 + 1}{1} \right) = 0 + 1 = 1 \][/tex]
Thus, the integral converges to 1, and therefore, the series:
[tex]\[ \sum_{n=1}^{\infty} \frac{\ln n}{n^2} \][/tex]
converges.
### Part (B): [tex]\(\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{1 / 2}}\)[/tex]
#### Step 1: Define the Function
Let [tex]\( f(x) = \frac{1}{x (\ln x)^{1/2}} \)[/tex]. Check if this function is positive, continuous, and decreasing for [tex]\( x \geq 2 \)[/tex].
- Positivity: For [tex]\( x \geq 2 \)[/tex], both [tex]\( x \)[/tex] and [tex]\( \ln x \)[/tex] are positive, so [tex]\( f(x) > 0 \)[/tex].
- Continuity: The function [tex]\( f(x) = \frac{1}{x (\ln x)^{1/2}} \)[/tex] is continuous for [tex]\( x > 1 \)[/tex].
- Decreasing: To check if [tex]\( f(x) \)[/tex] is decreasing, we find its derivative.
[tex]\[ f'(x) = \frac{d}{dx} \left( \frac{1}{x (\ln x)^{1/2}} \right) = \frac{(\ln x)^{1/2} \cdot (-1/x^2) - 1 \cdot \frac{1}{2} (\ln x)^{-1/2} \cdot \frac{1}{x}}{(\ln x)^1} \][/tex]
Simplified:
[tex]\[ f'(x) = \frac{- (\ln x)^{1/2} - \frac{1}{2x (\ln x)^{1/2}} }{x^2 (\ln x)} \][/tex]
For [tex]\( x \geq 2 \)[/tex], the numerator is negative, making [tex]\( f'(x) \leq 0 \)[/tex], indicating that [tex]\( f(x) \)[/tex] is decreasing for [tex]\( x \geq 2 \)[/tex].
#### Step 2: Evaluate the Improper Integral
Evaluate the improper integral [tex]\(\int_{2}^{\infty} \frac{1}{x (\ln x)^{1/2}} \, dx\)[/tex]:
[tex]\[ \int_{2}^{\infty} \frac{1}{x (\ln x)^{1/2}} \, dx \][/tex]
Let [tex]\( u = \ln x \)[/tex], thus [tex]\( du = \frac{1}{x} dx \)[/tex]. Changing the limits, for [tex]\(x = 2\)[/tex], [tex]\( u = \ln 2 \)[/tex], and as [tex]\( x \to \infty \)[/tex], [tex]\( u \to \infty \)[/tex].
[tex]\[ \int \frac{1}{x (\ln x)^{1/2}} dx = \int \frac{1}{(\ln x)^{1/2}} \frac{1}{x} dx = \int \frac{1}{u^{1/2}} du = 2 u^{1/2} = 2 (\ln x)^{1/2} \][/tex]
Evaluate this from [tex]\( 2 \)[/tex] to [tex]\( \infty \)[/tex]:
[tex]\[ \left[ 2 (\ln x)^{1/2} \right]_{2}^{\infty} \][/tex]
[tex]\[ \lim_{b \to \infty} \left( 2 (\ln b)^{1/2} \right) - 2 (\ln 2)^{1/2} \][/tex]
Since [tex]\( \ln b \to \infty \)[/tex] as [tex]\( b \to \infty \)[/tex],
[tex]\[ 2 (\ln b)^{1/2} \to \infty \][/tex]
Therefore, the integral diverges to infinity, and thus, the series:
[tex]\[ \sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{1/2}} \][/tex]
diverges.
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Find the answers you need at IDNLearn.com. Thanks for stopping by, and come back soon for more valuable insights.
