The series comparison test or the integral test, we can conclude that the series [infinity]∑ₖ₌₁ = 1/9k² + 3k - 2
The series you provided can be analyzed for convergence using different methods. Here are two approaches:
1. Series Comparison Test:
Find a comparison series. We can compare the given series with the harmonic series, which is known to diverge:
∑[infinity]ₖ₌₁ 1/k
Compare terms. For each term of the given series, we can find a corresponding term in the harmonic series that is greater or equal. In this case:
1/9k² + 3k - 2 ≥ 1/k
This is because 1/9k² and -2 are negligible for large values of k, and 3k dominates the inequality.
Conclusion. Since the given series has terms that are greater or equal to the terms of a divergent series, it must also diverge.
2. Integral Test:
Find the corresponding indefinite integral. For the given series, the term is:
f(k) = 1/9k² + 3k - 2
Taking the indefinite integral:
∫f(k) dk = -k/9 + 3k²/2 - 2k + C
Evaluate the limit as k approaches infinity.
lim_[k→∞] ∫f(k) dk = ∞
Conclusion. Since the integral diverges, the series must also diverge.
Therefore, using either the series comparison test or the integral test, we can conclude that the series:
[infinity]∑ₖ₌₁ = 1/9k² + 3k - 2