Section: AP Calc AB

49)
Determine convergence or divergence Σ n/2n3+1
n=1

Explanation

Using the comparison test,  Σ n/2n3+1  < Σ n/2n3
n=1 n=1

Since  Σ n/2n3  = Σ 1/2n2
n=1 n=1

Thus  Σ n/2n3+1 is convergence
n=1


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