Answer :
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.
