sigma(1, infinity) (n^2 - 5n)/(n^3 + n + 1) Determine whether the series converges or diverges.

sigma(1, infinity) (n^2 - 5n)/(n^3 + n + 1) Determine whether the series converges or diverges. We are asked to determine whether the series: Σ (n=1, ∞) (n^2 - 5n) / (n^3 + n + 1) converges or diverges. Step 1: Analyze the behavior of the terms for large n. We need to examine the general term: a_n = (n^2 - 5n) / (n^3 + n + 1) As n becomes large, the highest degree terms in both the numerator and the denominator dominate the behavior of the sequence. So, for large n, we can approximate: a_n ≈ n^2 / n^3 = 1 / n Step 2: Compare with a known series. The general term of the series behaves like 1 / n for large n. We know that the series Σ (1/n) (the harmonic series) diverges. Step 3: Apply the Limit Comparison Test. To rigorously confirm the behavior, we apply the Limit Comparison Test. Compare the given series with the harmonic series Σ (1/n), which is known to diverge. Let b_n = 1/n. We compute the limit of the ratio of a_n to b_n as n approaches infinity: lim (n → ∞) [a_n / b_n] = lim (n → ∞) [(n^2 - 5n) / (n^3 + n + 1)] / (1/n) Simplify the ratio: = lim (n → ∞) (n^3 - 5n^2) / (n^3 + n + 1) = lim (n → ∞) (1 - 5/n) / (1 + 1/n^2 + 1/n^3) As n approaches infinity, the ratio approaches 1: lim (n → ∞) (1 - 5/n) / (1 + 1/n^2 + 1/n^3) = 1 Step 4: Conclusion from the Limit Comparison Test. Since the limit is a finite positive number (1), and we know that the harmonic series Σ (1/n) diverges, the Limit Comparison Test tells us that the given series also **diverges**. Therefore, the series Σ (n=1, ∞) (n^2 - 5n) / (n^3 + n + 1) **diverges**.