Test each of the following series for convergence by the Integral Test. (Don't use any other tests.) If the Integral Test, by itself, can be applied to the series to determine whether it converges, enter CONV if it converges or DIV if it diverges. If the integral test cannot be applied to the series, enter NA. (Note: this means that even if you know a given series converges by using some other test, or some other test combined with the integral test, but the integral test, all by itself, does not determine the answer then you must enter NA rather than CONV.) 1. \( \sum_{n=1}^{\infty} \sum_{n=1}^{\infty} n e^{3 n} \) 3. \( \sum_{n=1}^{\infty} \frac{n+8}{n e^{-3 n}} \) 4. \( \sum_{n=1}^{\infty} \frac{-3)^{n}}{n(\ln (4 n))^{5}} \) 5. \( \sum_{n=1}^{\infty} \frac{\ln (4 n)}{n} \) Note: To get full credit, all answers must be correct. Having all but one correct is worth \( 50 \% \). Two or more incorrect answers gives a score of

To apply the Integral Test, we need to ensure that the series meets the necessary conditions: the function must be positive, continuous, and decreasing for \( n \geq N \) for some integer \( N \). We will analyze each series one by one. ### 1. \( \sum_{n=1}^{\infty} n e^{3n} \) **Analysis:** - The term \( n e^{3n} \) is positive for all \( n \). - The function \( f(x) = x e^{3x} \) is continuous for \( x \geq 1 \). - The function \( f(x) \) is increasing because its derivative \( f'(x) = e^{3x}(3x + 1) \) is positive for \( x \geq 1 \). Since the function is not decreasing, the Integral Test cannot be applied. **Conclusion:** NA ### 2. \( \sum_{n=1}^{\infty} \frac{n+8}{n e^{-3n}} \) **Analysis:** - Simplifying the term: \( \frac{n+8}{n e^{-3n}} = \frac{1 + \frac{8}{n}}{e^{-3n}} \). - As \( n \to \infty \), \( \frac{8}{n} \to 0 \), so the term behaves like \( \frac{1}{e^{-3n}} \), which is positive. - The function \( f(x) = \frac{x + 8}{x e^{-3x}} \) is continuous for \( x \geq 1 \). - The function is decreasing for large \( n \) since the exponential decay dominates. We can apply the Integral Test. **Integral:** \[ \int_{1}^{\infty} \frac{x + 8}{x e^{-3x}} \, dx \] This integral diverges because the exponential term in the denominator grows faster than the polynomial in the numerator. **Conclusion:** DIV ### 3. \( \sum_{n=1}^{\infty} \frac{-3^n}{n(\ln(4n))^5} \) **Analysis:** - The term \( \frac{-3^n}{n(\ln(4n))^5} \) is negative for all \( n \). - Since the terms are not positive, the Integral Test cannot be applied. **Conclusion:** NA ### 4. \( \sum_{n=1}^{\infty} \frac{\ln(4n)}{n} \) **Analysis:** - The term \( \frac{\ln(4n)}{n} \) is positive for all \( n \). - The function \( f(x) = \frac{\ln(4x)}{x} \) is continuous for \( x \geq 1 \). - To check if it is decreasing, we can find the derivative: \[ f'(x) = \frac{1 - \ln(4x)}{x^2} \] This derivative is negative for sufficiently large \( x \) (specifically, when \( x > e^{1/4} \)), indicating that the function is decreasing. We can apply the Integral Test. **Integral:** \[ \int_{1}^{\infty} \frac{\ln(4x)}{x} \, dx \] This integral diverges (it behaves like \( \int \frac{\ln(x)}{x} \, dx \), which diverges). **Conclusion:** DIV ### 5. \( \sum_{n=1}^{\infty} \frac{\ln(4n)}{n} \) **Analysis:** - The term \( \frac{\ln(4n)}{n} \) is positive for all \( n \). - The function \( f(x) = \frac{\ln(4x)}{x} \) is continuous for \( x \geq 1 \). - As in the previous case, the function is decreasing for sufficiently large \( x \). We can apply the Integral Test. **Integral:** \[ \int_{1}^{\infty} \frac{\ln(4x)}{x} \, dx \] This integral diverges. **Conclusion:** DIV ### Final Answers: 1. NA 2. DIV 3. NA 4. DIV 5. DIV