Let's analyze the given series from the image and determine their convergence or divergence.

Problem 4:

The series is: $\sum_{n=1}^\infty \frac{\sqrt{n} + 7}{n^2}$

Steps:

1. Asymptotic Behavior: The dominant term in the numerator as $n \to \infty$ is $\sqrt{n}$, and the dominant term in the denominator is $n^2$. Thus, the series behaves asymptotically like: $\frac{\sqrt{n}}{n^2} = \frac{1}{n^{3/2}}$

2. Comparison Test: Compare this series with the p-series $\sum \frac{1}{n^p}$, where $p = \frac{3}{2} > 1$. Since the p-series converges for $p > 1$, the given series converges by the Comparison Test.

Conclusion:

The series is convergent.

Problem 5:

The series is: $\sum_{n=2}^\infty \frac{5}{n \ln(n)}$

Steps:

1. Integral Test: Consider the function $f(x) = \frac{5}{x \ln(x)}$ for $x > 1$. This function is positive, continuous, and decreasing for $x > 1$. Use the integral test: $\int_2^\infty \frac{5}{x \ln(x)} dx$

   Substitution: Let $u = \ln(x)$, so $du = \frac{1}{x} dx$. The integral becomes: $\int \frac{5}{x \ln(x)} dx = 5 \int \frac{1}{u} du = 5 \ln(u) = 5 \ln(\ln(x))$

   Evaluating at the bounds, as $x \to \infty$, $\ln(\ln(x)) \to \infty$. Thus, the integral diverges.

2. Conclusion from the Integral Test: Since the integral diverges, the series also diverges.

Final Answers:

1. Problem 4: Convergent
2. Problem 5: Divergent

Let me know if you would like further clarification or details on the steps!

Related Questions:

1. What is the Comparison Test and when can it be used effectively?
2. How does the p-series test determine convergence for different values of $p$?
3. Why is the Integral Test a valid method for determining series convergence?
4. What are other tests (e.g., Ratio Test) that can be used to analyze these series?
5. How does the behavior of logarithmic functions ($\ln(n)$) affect series convergence?

Tip: Always simplify the dominant terms for large $n$ to better understand the series' behavior.