(a)

#### To determine

**To describe:** If an>bn for all *n*, then series ∑an convergent or divergent.

#### Answer

If an>bn for all *n*, then the series ∑an is divergent.

#### Explanation

**Given:**

The series ∑bn divergent.

**Result used:** Comparison test

“Suppose that ∑an and ∑bn are the series with positive terms,

(1) If ∑bn is convergent and an≤bn for all n, then ∑an is also convergent.

(2) If ∑bn is divergent and an≥bn for all n, then ∑an is also divergent.”

**Conclusion:**

Consider an>bn for all n,∑an>∑bn.

Since the series ∑bn divergent and use the result state above, it can be concluded that the given series is divergent.

Therefore, the series ∑an is divergent.

(b)

#### To determine

**To describe:** If an<bn for all *n*, then series ∑an convergent or divergent.

#### Answer

The series ∑an may convergent or divergent.

#### Explanation

**Given:**

The series ∑bn diverges.

**Result used:** Comparison test

“Suppose that ∑an and ∑bn are the series with positive terms,

(1) If ∑bn is convergent and an≤bn for all n, then ∑an is also convergent.

(2) If ∑bn is divergent and an≥bn for all n, then ∑an is also divergent.”

**Conclusion:**

Consider an<bn for all n, ∑an<∑bn.

Since ∑bn is divergent and by using the Result stated above, it can be concluded that the

series ∑an may convergent or divergent.

Therefore, we cannot say anything about the series ∑an.

**Examples:**

(1)

Consider the series ∑1n2 and ∑1n.

1n2<1n

Here, it is observed that the series ∑1n is divergent but ∑1n2 is convergent.

(2)

Consider the series ∑1n+1 and ∑1n.

1n+1<1n

Here, it is observed that the series ∑1n is divergent but ∑1n+1 is divergent.