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The given series is:

\[
\sum_{n=1}^{\infty} \frac{n^p}{n^q} = \sum_{n=1}^{\infty} n^{p-q}
\]

where \(p\) and \(q\) are positive numbers.

### Step 1: Simplifying the Series
The general term of the series simplifies to:

\[
n^{p-q}
\]

So the series becomes:

\[
\sum_{n=1}^{\infty} n^{p-q}
\]

### Step 2: Applying the p-series Test
This series resembles a p-series, which is of the form:

\[
\sum_{n=1}^{\infty} \frac{1}{n^r}
\]

for some value of \(r\). In our case, the term is \(n^{p-q} = n^{-(q-p)}\). Thus, we can compare this to a p-series with exponent \(-(q-p)\).

The p-series convergence criterion states:
- If \(r > 1\), the series converges.
- If \(r \leq 1\), the series diverges.

### Step 3: Testing Convergence
For our series to converge, we need the exponent \(q - p > 1\), meaning that \(q > p + 1\). If this condition holds, the series converges. Otherwise, the series diverges.

### Conclusion:
- **If \(q > p + 1\)**, the series **converges**.
- **If \(q \leq p + 1\)**, the series **diverges**.

Would you like a deeper explanation on any step? Here are five related questions to consider:

1. How does the p-series test work for other series forms?
2. Can this series be tested using the ratio test for more complex cases?
3. What happens if \(p = q\) or \(p > q\)?
4. How would the behavior of the series change if we altered the powers of \(n\)?
5. Can we express the sum of the series in closed form for specific values of \(p\) and \(q\)?

**Tip:** Understanding convergence tests (like p-series and ratio tests) is essential for working with infinite series.

Test for convergence or divergence of the series: (2^p / 1^q) + (3^p / 2^q) + (4^p / 3^q) + ... ∞ where p and q are positive numbers.

Analyze the convergence or divergence of the series ∑ (n^p / n^q) using the P-Series test. This detailed solution covers the simplification process and the application of the P-Series convergence criterion for various values of p and q. Suitable for advanced calculus students.

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