Answer:
0 ≤ an ≤ bn
The series ∑₁°° bn converges
The series ∑₁°° an converges by comparison to ∑₁°° bn.
0 ≤ an ≤ bn
The series ∑₁°° bn diverges
The comparison test is inconclusive for our choice of bn.
Step-by-step explanation:
an = 1 / (n² + n + 3) and bn = 1 / n²
The numerators are the same, and the denominator of an is greater than the denominator of bn. So 0 ≤ an ≤ bn.
bn is a p series with p > 1, so it converges.
Since the larger function converges, the smaller function also converges.
an = (3n − 1) / (6n² + 2n + 1) and bn = 1 / (2n)
If we rewrite bn as bn = (3n − 1) / (6n² − 2n), we can tell that when the numerators are equal, the denominator of an is greater than the denominator of bn. So 0 ≤ an ≤ bn.
bn is a p series with p = 1, so it diverges.
The larger function diverges. We cannot conclude whether the smaller function converges or diverges.
