Calculus Lesson 9.4 Comparison Tests Theorem 9.4: Direct Comparison Test (p. 481) Similar to Direct Comparison Test from lesson 7.8. Let

€

an∑ be a series with nonnegative terms. a)

€

an∑ converges if there is a convergent series

€

bn∑ such that

€

bn ≥ an for all n > N. In simple terms, if the larger series converges, then the smaller must converge.

b)

€

an∑ diverges if there is a divergent series

€

bn∑ such that

€

bn∑ has nonnegative terms and

€

bn ≤ an for all n > N. In simple terms, if the smaller series diverges, then the larger diverges.

Example: Does

€

1n3 + 5nn=1

∞

∑ converge or diverge?

Solution: Start with

€

5n ≥ 0 (for

€

n ≥1), then

€

n3 + 5n ≥ n3, giving

€

1n3 + 5n

≤1n3

.

€

1n3n=1

∞

∑ is a p-series with p = 3 > 1, therefore it converges.

Since

€

1n3

≥1

n3 + 5n and

€

1n3n=1

∞

∑ converges, then by the Direct Comparison Test

€

1n3 + 5nn=1

∞

∑

also converges.

Example: Does

€

5n +12n −1n=1

∞

∑ converge or diverge?

Solution: Start with

€

1≥ 0, then

€

5n +1≥ 5n , which would give

€

5n +12n −1

≥5n

2n −1, which would then give

€

5n +12n −1

≥5n

2n.

€

5n

2nn=1

∞

∑ is a geometric series with

€

r =52

>1, therefore it diverges.

Since

€

5n

2n≤5n +12n −1

and

€

5n

2nn=1

∞

∑ diverges, then by the Direct Comparison Test

€

5n +12n −1n=1

∞

∑ also

diverges.

Theorem 9.5: Limit Comparison Test (p. 483) Suppose an > 0 and bn > 0 for all n > N.

(a) If

€

limn→∞

anbn

≥ 0 (except ∞) and

€

bn∑ converges, then

€

an∑ converges. This means if the limit is 0

or any number but ∞, and the comparison series converges, then the original series will converge.

(b) If

€

limn→∞

anbn

> 0 (including ∞) and

€

bn∑ diverges, then

€

an∑ diverges. This means if the limit

equals any number but 0 and the comparison series diverges, then the original series will diverge.

Example: Does

€

1n3 − 5nn=1

∞

∑ converge or diverge?

Solution: Compare

€

1n3 + 5n

to

€

1n3

.

€

1n3n=1

∞

∑ is a p-series with p = 3 > 1, therefore it converges.

€

limn→∞

1

n3−5n1n3

€

= limn→∞

n3

n3 − 5n (divide all terms by n3)

€

= limn→∞

11− 5

n2

€

=1

1− 5∞2

.

Since the 0 < limit < ∞ and

€

1n3n=1

∞

∑ converges, then by the Limit Comparison Test

€

1n3 − 5nn=1

∞

∑ also converges.

Example: Does

€

ln n( )( )3

n3n=1

∞

∑ converge or diverge?

Solution: Compare

€

ln n( )( )3

n3 to .

€

1n2n=1

∞

∑ is a p-series with p = 2 > 1, therefore it converges.

€

limn→∞

ln n( )( )3

n3

1n2

€

= limn→∞

ln n( )( )3

n=∞∞

, this is indeterminate, therefore use L'Hôpital's Rule,

€

limn→∞

3 ln n( )( )2

n=∞∞

, still indeterminate, therefore use L'Hôpital's Rule again,

€

limn→∞

6ln n( )n

=∞∞

, still indeterminate, therefore use L'Hôpital's Rule again,

€

limn→∞

6n1

=6∞

= 0 .

Since the limit = 0 and

€

1n2n=1

∞

∑ converges, then by the Limit Comparison Test

€

ln n( )( )3

n3n=1

∞

∑

also converges.

Example: Does

€

1n ln n( )n=2

∞

∑ converge or diverge?

Solution: Compare

€

1n ln n( )

to .

€

1nn=2

∞

∑ is the harmonic series, therefore it diverges.

€

limn→∞

1

n ln n( )1n

€

= limn→∞

nln n( )

=∞∞

, this is indeterminate, therefore use L'Hôpital's Rule,

€

limn→∞

1

2 n−1

2

1n

€

= limn→∞

n2 n

€

= limn→∞

n2

.

Since the limit = ∞ and

€

1nn=2

∞

∑ diverges, then by the Limit Comparison Test

€

1n ln n( )n=2

∞

∑

also diverges.