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The given problem asks 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-by-Step Solution:

1. **Simplify the given expression** inside the square root:
   
   \[
   \sqrt{\frac{k^2 - k}{k^4}} = \sqrt{\frac{k^2(1 - \frac{1}{k})}{k^4}} = \frac{\sqrt{k^2(1 - \frac{1}{k})}}{k^2}
   \]
   Simplifying further:
   \[
   = \frac{k\sqrt{1 - \frac{1}{k}}}{k^2} = \frac{1}{k}\sqrt{1 - \frac{1}{k}}.
   \]

2. **Analyze the behavior for large \( k \):**

   As \( k \to \infty \), the term \( \frac{1}{k} \) becomes small, and thus:
   \[
   \sqrt{1 - \frac{1}{k}} \to 1.
   \]
   Hence, for large \( k \), the term approximates:
   \[
   \frac{1}{k}\sqrt{1 - \frac{1}{k}} \sim \frac{1}{k}.
   \]

3. **Choose a comparison series**:

   Since the given series behaves like \( \frac{1}{k} \) for large \( k \), we compare it with the harmonic series:
   \[
   \sum_{k=2}^{\infty} \frac{1}{k}.
   \]
   The harmonic series is known to **diverge**.

4. **Apply the Limit Comparison Test**:

   The Limit Comparison Test states that for two series \( \sum a_k \) and \( \sum b_k \), if:
   \[
   \lim_{k \to \infty} \frac{a_k}{b_k} = L,
   \]
   where \( L \) is a positive, finite number, then both series either converge or both diverge.

   Here, we set:
   \[
   a_k = \frac{1}{k}\sqrt{1 - \frac{1}{k}} \quad \text{and} \quad b_k = \frac{1}{k}.
   \]
   We compute the limit:
   \[
   \lim_{k \to \infty} \frac{\frac{1}{k}\sqrt{1 - \frac{1}{k}}}{\frac{1}{k}} = \lim_{k \to \infty} \sqrt{1 - \frac{1}{k}} = 1.
   \]
   Since the limit is a positive finite number (specifically, 1), the Limit Comparison Test tells us that the given series behaves like the harmonic series. 

5. **Conclusion**:

   Since the harmonic series diverges, the original series:
   \[
   \sum_{k=2}^{\infty} \sqrt{\frac{k^2 - k}{k^4}}
   \]
   also **diverges**.

Would you like further clarification on any step, or any additional details?

Here are 5 related questions for further exploration:
1. What is the harmonic series, and why does it diverge?
2. Can the Limit Comparison Test be used for all types of series?
3. How would the result change if the numerator of the original series were different, say \( k^3 \)?
4. What are other methods to test the convergence of series?
5. How does the Limit Comparison Test compare to the Ratio Test?

**Tip:** Always simplify the series terms first before applying comparison tests to understand their behavior better at infinity.

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

This problem uses the Limit Comparison Test to determine if the series ∑(sqrt(k^2 - k)/k^4) converges or diverges. Key concepts include the comparison to the harmonic series and simplifying complex expressions.

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