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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.

Use the Direct Comparison Test to determine whether the series converges or diverges: ∑ from n=3 to infinity ln(n)/√n

Determine the convergence or divergence of the series ∑ ln(n)/√n using the Direct Comparison Test. Compare it with the p-series ∑ 1/n^(3/2) to conclude whether the original series converges.

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