Question

∑n=1 to [infinity] (1/n^p) converges if and only if:

A. p>1

B. p≥1

C. p<1

D. p≤1

Answer 1

Step-by-step explanation:

The given series, ∑n=1 to [infinity] (1/n^p), converges if and only if p > 1.

Here's the explanation:

1. To determine if the series converges, we need to analyze the convergence of the harmonic series, which is the special case of the given series when p = 1.

2. When p = 1, the series becomes ∑n=1 to [infinity] (1/n^1), which is the harmonic series.

3. The harmonic series is known to diverge, meaning it does not have a finite sum. This is proven using the integral test or other methods.

4. Therefore, any value of p less than or equal to 1 would cause the series to diverge.

5. On the other hand, when p is greater than 1, the terms in the series approach zero as n approaches infinity. In this case, the series converges.

6. The proof of convergence when p > 1 can be shown using the p-series test, which states that if p > 1, then the series ∑n=1 to [infinity] (1/n^p) converges.

Based on these explanations, the correct inference is:

A) p > 1

Therefore, the series ∑n=1 to [infinity] (1/n^p) converges if and only if p > 1.

Answer 2

The series ∑n=1 to ∞ (1/np) converges if and only if p > 1, as per the comparison and p-series tests.

Step-by-step explanation:

The question discusses the convergence of the series ∑n=1 to ∞ (1/np). This series converges if and only if p > 1. This is a result of the comparison test and the p-series test. For p ≤ 1, the series does not converge. For example, when p = 1, it yields the harmonic series, which is known to diverge. When p > 1, each term in the series becomes smaller at a rate that is sufficient for the sum of the series to remain finite, thus achieving convergence.