5 Steps to a 5 AP Calculus BC 2019

352 STEP 4. Review the Knowledge You Need to Score High

Step 2: nlim→∞

an+ 1

an

=nlim→∞

(

(n+1)^3

(ln 2)n+^1

·

(ln 2)n

n^3

)

=

1

ln 2

nlim→∞

(

n+ 1

n

) 3

=

1

ln 2

≈ 1. 443

Since this limit is greater than 1, the series diverges.

Comparison Test

Suppose

∑∞

n= 1

anand

∑∞

n= 1

bnare series with non-negative terms, and

∑∞

n= 1

bnis known to con-

verge. If a term-by-term comparison shows that for alln,an≤bn, then

∑∞

n= 1

anconverges. If

∑∞

n= 1

bndiverges, and if for all n,an≥bn, then

∑∞

n= 1

andiverges. Common series that may be

used for comparison include the geometric series, which converges forr<1 and diverges

forr≥1, and thep-series, which converges forp>1 and diverges forp≤1.

Example 1

Determine whether the series

∑∞

n= 1

1

n^2 + 5

converges or diverges.

Step 1: Choose a series for comparison. The series

∑∞

n= 1

1

n^2 + 5

can be compared to

∑∞

n= 1

1

n^2

,

thep-series withp=2. Both series have non-negative terms.

Step 2: A term-by-term comparison shows that

1

n^2 + 5

<

1

n^2

for all values ofn.

Step 3:

∑∞

n= 1

1

n^2

converges, so

∑∞

n= 1

1

n^2 + 5

converges.

Example 2

Determine whether the series 2+

3

5

+

4

10

+

5

17

+···converges or diverges.

Step 1: The series 2+

3

5

+

4

10

+

5

17

+···=

∑∞

n= 1

n+ 1

n^2 + 1

can be compared to

∑∞

n= 1

1

n

.

Step 2:

n+ 1

n^2 + 1

=

n^2 +n

n^3 +n

≥

n^2 + 1

n^3 +n

=

1

n

,so

n+ 1

n^2 + 1

≥

1

n

for alln≥1.

Step 3: Since

∑∞

n= 1

1

n

diverges, 2+

3

5

+

4

10

+

5

17

+···also diverges.

Limit Comparison Test

If

∑∞

n= 1

anand

∑∞

n= 1

bnare series with positive terms, and if limn→∞

an

bn

=Lwhere 0<L<∞,

then either both series converge or both diverge. By choosing, for one of these, a series that

is known to converge, or known to diverge, you can determine whether the other series

converges or diverges. Choose a series of a similar form so that the limit expression can be

simplified.