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We are asked to use the **Limit Comparison Test** to determine whether the series

\[
\sum_{k=2}^{\infty} \sqrt{\frac{k^2 - k}{k^4}}
\]

converges or diverges.

### Step 1: Simplify the given term
First, simplify the general term of the series:

\[
\sqrt{\frac{k^2 - k}{k^4}} = \frac{\sqrt{k^2 - k}}{k^2}.
\]

As \( k \) becomes large, \( k^2 - k \) behaves like \( k^2 \) since the \( -k \) becomes negligible. So, for large \( k \):

\[
\sqrt{k^2 - k} \approx \sqrt{k^2} = k.
\]

Thus, for large \( k \), the general term behaves like:

\[
\frac{\sqrt{k^2 - k}}{k^2} \approx \frac{k}{k^2} = \frac{1}{k}.
\]

### Step 2: Choose a comparison series
The simplified term behaves like \( \frac{1}{k} \), which suggests we compare the given series with the harmonic series:

\[
\sum_{k=2}^{\infty} \frac{1}{k}.
\]

We know that the harmonic series \( \sum_{k=2}^{\infty} \frac{1}{k} \) diverges.

### Step 3: Apply the Limit Comparison Test
Let \( a_k = \frac{\sqrt{k^2 - k}}{k^2} \) and \( b_k = \frac{1}{k} \). The Limit Comparison Test says that if

\[
\lim_{k \to \infty} \frac{a_k}{b_k} = L,
\]

where \( 0 < L < \infty \), then both series either converge or diverge together.

Now, compute the limit:

\[
\lim_{k \to \infty} \frac{\frac{\sqrt{k^2 - k}}{k^2}}{\frac{1}{k}} = \lim_{k \to \infty} \frac{k \sqrt{1 - \frac{1}{k}}}{k^2} \cdot k = \lim_{k \to \infty} \frac{k}{k^2} = 1.
\]

Since the limit is a finite positive number (1), the Limit Comparison Test tells us that the given series behaves like the harmonic series, which **diverges**.

### Conclusion:
The series \( \sum_{k=2}^{\infty} \sqrt{\frac{k^2 - k}{k^4}} \) **diverges**.

---

Would you like further details or have any questions about the solution? 

Here are five related questions to explore further:
1. What are the conditions for the Limit Comparison Test to be applicable?
2. How do you handle cases where the Limit Comparison Test gives \( L = 0 \) or \( L = \infty \)?
3. How can we determine the behavior of a series if the comparison is inconclusive?
4. What are other common methods to test for series convergence besides the Limit Comparison Test?
5. What are the differences between the Limit Comparison Test and the Direct Comparison Test?

**Tip:** Always simplify the general term in a series as much as possible before deciding on the appropriate test for convergence.

Use the Limit Comparison Test to determine whether the series ∑ (k=2 to ∞) sqrt((k^2 - k)/k^4) converges or diverges.

Explore the application of the Limit Comparison Test to determine the convergence or divergence of the series ∑ sqrt((k^2 - k)/k^4). Learn how the series behaves similarly to the harmonic series and why it diverges.

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