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The problem in the image is asking to determine for which values of \( k \) the following series converges and for which values it diverges, where \( k > 0 \) is a real number:

\[
\sum_{n=1}^{\infty} \frac{n^k}{2n + (-1)^n n}
\]

### Step-by-step approach:

1. **Series structure:** The series has the form:
   \[
   \frac{n^k}{2n + (-1)^n n}.
   \]
   We can rewrite the denominator:
   \[
   2n + (-1)^n n = n(2 + (-1)^n),
   \]
   which simplifies the general term of the series to:
   \[
   \frac{n^k}{n(2 + (-1)^n)} = \frac{n^{k-1}}{2 + (-1)^n}.
   \]
   The value of \( 2 + (-1)^n \) oscillates between \( 1 \) (for even \( n \)) and \( 3 \) (for odd \( n \)), so the general term can be approximated by something between \( \frac{n^{k-1}}{1} \) and \( \frac{n^{k-1}}{3} \).

2. **Behavior of the series:** For large \( n \), the term \( \frac{n^{k-1}}{2 + (-1)^n} \) behaves approximately like \( n^{k-1} \). So, we can analyze the behavior of this new series:
   \[
   \sum_{n=1}^{\infty} n^{k-1}.
   \]

3. **Convergence analysis:** The series \( \sum_{n=1}^{\infty} n^p \) converges if and only if \( p < -1 \). In this case, \( p = k-1 \), so we require:
   \[
   k-1 < -1 \implies k < 0.
   \]
   However, since the problem restricts \( k > 0 \), the only possibility for convergence is when \( k = 1 \).

4. **Conclusion:**
   - The series **converges** for \( k = 1 \).
   - The series **diverges** for \( k > 1 \) and \( k < 1 \) within the range \( k > 0 \).

Would you like a more detailed breakdown of any step or further clarifications?

### Related Questions:
1. What is the importance of oscillating terms in determining convergence of a series?
2. How would the convergence behavior change if the denominator had different powers of \( n \)?
3. Can you derive the series behavior for \( k = 1 \) using the comparison test?
4. How does the alternating nature of the series impact convergence testing methods?
5. What are other methods to test series convergence beyond the \( p \)-series test?

### Tip:
When analyzing series, simplifying the general term for large \( n \) often gives the most insight into its convergence behavior.

Låt k > 0 vara ett reellt tal. Avgör för vilka k som summan (from n=1 to ∞) [nk / (2n + (-1)^n * n)] konvergerar och för vilka k den divergerar.

Analyze the convergence of the infinite series Σ (n^k / (2n + (-1)^n n)) for different values of k > 0. Learn how the p-Series test helps determine when the series converges or diverges, focusing on advanced calculus concepts.

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