Final answer:
The Direct Comparison Test and the Limit Comparison Test can be used to determine the convergence or divergence of a series. By applying the Direct Comparison Test, we can conclude that the given series ∑ (1/n) diverges. Using the Limit Comparison Test, we can determine that the series ∑ (1/n² + 5) converges.
Step-by-step explanation:
The Direct Comparison Test can be used to determine the convergence or divergence of a series. It states that if 0 ≤ an ≤ bn for all n ≥ N, where N is some positive integer, and the series ∑ bn converges, then the series ∑ an also converges.
In the given series ∑ (1/n), we can compare it to the harmonic series. The harmonic series, ∑ (1/n), diverges. So, by direct comparison, the given series also diverges.
The Limit Comparison Test is another method to determine the convergence or divergence of a series. It states that if L = limn→∞ (an/bn) exists and is positive, where an and bn are the terms of the series, and bn is a known series with a known convergence or divergence, then the given series has the same convergence or divergence as the known series.
In the given series ∑ (1/n² + 5), we can compare it to the series ∑ (1/n²). Taking the limit as n approaches infinity, we have limn→∞ {(1/n² + 5)/(1/n²)} which simplifies to limn→∞(n² + 5)/(n²). This limit evaluates to 1. Since the limit is positive and finite, the given series has the same convergence as the series ∑ (1/n²), which is a known converging series.