This problem involves using the Direct Comparison Test to determine whether the given series converges or diverges. Let's break it down step by step.

The given series is:

$\sum_{n=3}^{\infty} a_n = \sum_{n=3}^{\infty} \frac{\ln n}{\sqrt{n}}$

The problem suggests comparing this series with a simpler series, denoted as $b_n$:

$b_n = c \left(\frac{1}{n^p}\right)$

where $c$ is a constant, and $p$ is the exponent of $n$ that determines the type of the comparison series.

Step 1: Find the comparison series

To compare the series $a_n = \frac{\ln n}{\sqrt{n}}$, we will approximate its behavior. Notice that for large $n$, the term $\ln n$ grows very slowly compared to powers of $n$, so we can compare $a_n$ with the series $\frac{1}{n^{3/2}}$ (since $\sqrt{n} = n^{1/2}$).

Thus, the comparison series can be chosen as:

$b_n = \frac{1}{n^{3/2}}$

Step 2: Check if the comparison series converges or diverges

This comparison series $\sum_{n=3}^{\infty} \frac{1}{n^{3/2}}$ is a p-series with $p = 3/2$.

• A p-series $\sum_{n=1}^{\infty} \frac{1}{n^p}$ converges if $p > 1$, and diverges if $p \leq 1$.

Since $p = 3/2 > 1$, the series $\sum_{n=3}^{\infty} b_n = \sum_{n=3}^{\infty} \frac{1}{n^{3/2}}$ converges.

Step 3: Apply the Direct Comparison Test

Now that we know the comparison series converges, we need to check if the terms $a_n = \frac{\ln n}{\sqrt{n}}$ are smaller than or comparable to the terms $b_n = \frac{1}{n^{3/2}}$ for large $n$.

Since $\ln n$ grows much more slowly than $n^{1/2}$ for large $n$, we have:

$\frac{\ln n}{\sqrt{n}} < \frac{1}{n^{3/2}} \text{ for large } n$

Thus, we can use the Direct Comparison Test, which tells us that if $0 \leq a_n \leq b_n$ for all $n \geq 3$ and the series $\sum b_n$ converges, then the series $\sum a_n$ also converges.

Conclusion:

Since $\sum b_n$ converges and $a_n \leq b_n$ for large $n$, the series $\sum a_n$ also converges by the Direct Comparison Test.

Final Answer:

$\sum_{n=3}^{\infty} \frac{\ln n}{\sqrt{n}} \text{ converges.}$

Would you like any further details or explanations?

Relative Questions:

1. What is the general form of the Direct Comparison Test?
2. Why does a p-series converge when $p > 1$?
3. How does $\ln n$ grow in comparison to powers of $n$?
4. Can the Limit Comparison Test also be used in this scenario? If so, how?
5. What other types of comparison tests can be used to determine convergence or divergence?

Tip:

When comparing series, always ensure the chosen comparison series behaves similarly for large $n$ and satisfies the conditions of the comparison test.