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More Challenging Problems
Determine if the series, a1 + a2 + a3 + a4 + … converges or diverges, where:
1. an= 1/(3n + 2)
Answer
1. Converges
Solution
1. First, observe that 0 < 1/(3n + 2) < 1/3n The series with bn = 1/3n is a geometric series with r= 1/3 and so converges. Then the series with an= 1/(3n + 2) converges by the comparison test.
2. an= 1/(n – 1)
Answer
1. Diverges
Solution
1. First, observe that 1/(n – 1) > 1/n The series with bn= 1/n is the harmonic series and so diverges. Then the series with an = 1/(n – 1) diverges by the comparison test.
3. an = 1/(n2 + n)
Answer
1. Converges
Solution
1. This is a little trickier, because we could compare this an with 1/n (the harmonic series, so diverges) or with 1/n2 (a p-series with p = 2 > 1, so converges). Which should be used?
Note: 1/(n2 + n) < 1/n and 1/(n2 + n) < 1/n2 Being less than a diverging series tells us nothing, but being less than a converging series tells us that the series of an converges.
4. an = 1/(3n – 2)
Answer
1. Converges
Solution
1. The Comparison Test cannot be applied, because 1/(3n – 2) > 1/3n and although the geometric series ∑ 1/3n converges, being greater than a converging series tells us nothing. We can use the Limit Comparison Test, because as n → ∞, (1/(3n – 2)/(1/3n) = (3n)/(3n – 2) = 1/(1 – 2/3n) → 1 Then by the Limit Comparison Test, ∑ 1/(3n – 2) converges because ∑ 1/3n converges.
5. an = (√(n3 + 4))/(n2 + 2n + 3)
Answer
1. Diverges
Solution
1. The first issue is to find what series to use for comparison. For large n the numerator √(n3 + 4) is close to n3/2. For large n the denominator n2 + 2n + 3 is close to n2. Then the fraction an is close to n3/2/n2 = 1/n1/2. This is the general term of the series we use for comparison. As n → ∞, (√(n3 + 4))/(n2 + 2n + 3) / (1/n1/2) = (n3/2 √(1 + 4/n3)) / (n2(1 + 2/n + 3/n2)) ⋅ n1/2 = (√(1 + 4/n3)) / (1 + 2/n + 3/n2) → 1 Being a p-series with p = 1/2 < 1, ∑1/n1/2 diverges, so the original series diverges by the Limit Comparison Test.
6. an = (3n + 2)/(5n + 4)
Answer
1. Converges
Solution
1. For large n the ratio (3n + 2)/(5n + 4) is very close to 3n/5n, the general term of a geometric series with ratio 3/5, hence converging. As n → ∞, ((3n + 2)/(5n + 4))/(3n/5n) = ((3n(1 + 2/3n))/(5n(1 + 4/5n))⋅(5n/3n) = (1 + 2/3n)/(1 + 4/5n) → 1 Then by the Limit Comparison Test, the original series converges because the geometric series converges.