Consider the two seriesS =∞Xn=1n2 + 4n + sin(n)n3.01 , T =∞Xn=1n + 4 + cos(n)n1.99 .One of these is convergent, and one is divergent. Which is which? Justify youranswer, clearly stating any results used from lectures or the SupplementaryMaterials.

To determine whether the series S and T are convergent or divergent, we can use the comparison test or limit comparison test.

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

The limit comparison test states that given two series ∑an and ∑bn with positive terms, if the limit as n approaches infinity of an/bn is a positive finite number, then either both series converge or both diverge.

Let's start with series S:

S = ∑ (n^2 + 4n + sin(n)) / n^3.01

We can compare this series with ∑ 1/n. Since n^2 + 4n + sin(n) ≤ n^2 + 4n + 1 ≤ n^2 + 4n^2 + n^2 = 6n^2 for all n, we have 0 ≤ (n^2 + 4n + sin(n)) / n^3.01 ≤ 6n^2 / n^3.01 = 6/n^1.01.

The series ∑ 1/n^p is known to be convergent if p > 1 and divergent if p ≤ 1. Since 1.01 > 1, the series ∑ 6/n^1.01 is convergent. Therefore, by the comparison test, the series S is also convergent.

Now let's consider series T:

T = ∑ (n + 4 + cos(n)) / n^1.99

We can compare this series with ∑ 1/n. Since n + 4 + cos(n) ≥ n for all n, we have 0 ≤ 1/n ≤ (n + 4 + cos(n)) / n^1.99 = 1/n^0.99 + 4/n^1.99 + cos(n)/n^1.99.

The series ∑ 1/n^p is divergent if p ≤ 1. Since 0.99 ≤ 1, the series ∑ 1/n^0.99 is divergent. Therefore, by the comparison test, the series T is also divergent.

So, the series S is convergent and the series T is divergent.