To determine if the series

$\sum_{n=1}^{\infty} \frac{\sqrt{n} + 4}{n^2}$

converges or diverges, we can analyze the behavior of the terms as $n \to \infty$ and potentially use a comparison test.

Step-by-Step Solution

1. Rewrite the Terms: Separate the terms in the numerator: $\frac{\sqrt{n} + 4}{n^2} = \frac{\sqrt{n}}{n^2} + \frac{4}{n^2}$ So, the series becomes $\sum_{n=1}^{\infty} \left(\frac{\sqrt{n}}{n^2} + \frac{4}{n^2}\right).$

2. Simplify Each Term:

   • For $\frac{\sqrt{n}}{n^2}$: Rewrite $\sqrt{n}$ as $n^{1/2}$, so $\frac{\sqrt{n}}{n^2} = \frac{n^{1/2}}{n^2} = \frac{1}{n^{3/2}}$.
   • For $\frac{4}{n^2}$: This is already in a suitable form.

   The series now is: $\sum_{n=1}^{\infty} \left(\frac{1}{n^{3/2}} + \frac{4}{n^2}\right).$

3. Analyze Each Component:

   • The term $\frac{1}{n^{3/2}}$ represents a $p$-series with $p = 3/2 > 1$, which converges.
   • The term $\frac{4}{n^2}$ is also a $p$-series with $p = 2 > 1$, which also converges.

4. Conclusion: Since both series $\sum \frac{1}{n^{3/2}}$ and $\sum \frac{4}{n^2}$ converge, their sum also converges by the comparison test.

Final Answer

The series $\sum_{n=1}^{\infty} \frac{\sqrt{n} + 4}{n^2}$ converges.

Would you like a detailed breakdown of any step or have further questions?

Further Related Questions:

1. What is a $p$-series, and how do we determine its convergence?
2. How does the comparison test work, and when can it be applied to series?
3. Could we use the limit comparison test in this problem instead?
4. What are alternative tests for convergence, like the ratio or root tests?
5. How would we determine if an alternating series converges?

Tip:

For series convergence, identifying the dominant term (largest power in the denominator) often helps in applying comparison tests efficiently.