5 Steps to a 5 AP Calculus BC 2019

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

Step 1: Consider the series

∑∞

n= 1

∣∣

∣

∣

(−1)n

3 n−^1

∣∣

∣

∣=^1 +

1

3

+

1

9

+

1

27

+···. For this series,s 1 =1,

s 2 = 1 +

1

3

=

4

3

= 1. ̄3,s 3 = 1 +

1

3

+

1

9

=

13

9

= 1 .4, ̄ s 4 = 1 +

1

3

+

1

9

+

1

27

=

40

27

= 1 .481. The

sequence of partial sums,

{

1, 1.3, 1 ̄. ̄4, 1.481,...

}

, converges to 1.5. Or, note

that this is a geometric series witha=1,r=

1

3

; thus, it converges to

3

2

.

Step 2: Since

∑∞

n= 1

∣∣

∣∣(−1)n^1

3 n−^1

∣∣

∣∣converges,

∑∞

n= 1

(−1)n

1

3 n−^1

converges absolutely, and thus

converges.

Example 2

Determine whether the series

∑∞

n= 3

(− 1 )n+^1 lnn

n

converges absolutely, converges conditionally,

or diverges. Begin by examining the series of absolute values

∑∞

n= 3

∣∣

∣∣

∣

(− 1 )n+^1 lnn

n

∣∣

∣∣

∣=

ln 3

3

+

ln 4

4

+

ln 5

5

+....Compare this series with the harmonic series

∑∞

n= 3

1

n

=

1

3

+

1

4

+

1

5

+..., which diverges.

Since the terms of the series

ln 3

3

+

ln 4

4

+

ln 5

5

+...are greater than the terms of the series

1

3

+

1

4

+

1

5

+..., the series

∑∞

n= 3

∣∣

∣

∣∣

(− 1 )n+^1 lnn

n

∣∣

∣

∣∣diverges and thus the series

∑∞

n= 3

(− 1 )n+^1 lnn

n

does

not converge absolutely. Next examine the series

∑∞

n= 3

(−1)n+^1 lnn

n

=

ln 3

3

+

ln 4

4

+

ln 5

5

−

ln 6

6

+

..., which is an alternating series. Now apply the tests for convergence for alternating series.

You must show thatan ≥an+ 1 and limn→∞an=0 for convergence. Let f(x)=

lnx

x

. Note

thatf′(x)=

(

1

x

)

(x)−(1) lnx

x^2

=

1 −lnx

x^2

.Ifx >e, f′(x)<0. Therefore, f(x)isa

strictly decreasing function forx>e. Thus, for the series

∑∞

n= 3

(−1)n+^1 lnn

n

,an≥an+ 1 for

n≥3. Also for the limn→∞an, you have limn→∞an=nlim→∞

lnn

n

with limn→∞lnn=∞and limn→∞n=∞.

ApplyingL’Hoˆpital’sRule, you have limn→∞

1 /n

1

=0. The series

∑∞

n= 3

(−1)n+^1 lnn

n

satisfies the

tests for convergence for an alternating series. Therefore, the series

∑∞

n= 3

(−1)n+^1 lnn

n

converges, but not absolutely, as shown earlier, i.e.,

∑∞

n= 3

(−1)n+^1 lnn

n

converges conditionally.

Example 3

Determine whether the series

∑∞

n= 1

(−1)n

2 k+ 1

5 k− 1

converges absolutely, converges conditionally,

or diverges.