Use the Limit Comparison Test with bn=1n2b_n = \frac{1}{n^2}bn=n21 to determine the convergence of ∑n=1∞n+1n3+2\sum_{n=1}^{\infty} \frac{n+1}{n^3+2}∑n=1∞n3+2n+1.
Converges because limn→∞anbn=1>0\lim_{n \to \infty} \frac{a_n}{b_n} = 1 > 0limn→∞bnan=1>0 and ∑bn\sum b_n∑bn converges
Diverges because limn→∞anbn=1>0\lim_{n \to \infty} \frac{a_n}{b_n} = 1 > 0limn→∞bnan=1>0 and ∑bn\sum b_n∑bn diverges
Inconclusive
Converges because limn→∞anbn=0\lim_{n \to \infty} \frac{a_n}{b_n} = 0limn→∞bnan=0