Let an = 6n3n + 1.(a)Determine whether {an} is convergent.convergentdivergent    (b)Determine whether ∞n = 1an is convergent.convergentdivergent

(a) To determine whether the sequence {an} = 6n3^n + 1 is convergent or divergent, we need to examine the limit as n approaches infinity.

The limit of a sequence {an} as n approaches infinity is the value that the terms of the sequence get arbitrarily close to as n gets larger and larger. If such a limit exists, the sequence is said to be convergent. If no such limit exists, the sequence is said to be divergent.

In this case, as n approaches infinity, the term 3^n grows much faster than the linear term n. Therefore, the sequence {an} will grow without bound as n approaches infinity. This means that the sequence {an} is divergent.

(b) To determine whether the series ∑ (from n=1 to ∞) 1/an is convergent or divergent, we can use the comparison test.

The comparison test states that if 0 ≤ an ≤ bn for all n and the series ∑ bn converges, then the series ∑ an also converges. Conversely, if the series ∑ bn diverges, then the series ∑ an also diverges.

In this case, we can compare the series ∑ 1/an to the series ∑ 1/(6n3^n). Since 1/(6n3^n) ≤ 1/an for all n, if the series ∑ 1/(6n3^n) converges, then the series ∑ 1/an also converges.

The series ∑ 1/(6n3^n) is a p-series with p = -3, which is less than 1. Therefore, the series ∑ 1/(6n3^n) diverges, and by the comparison test, the series ∑ 1/an also diverges.