5 Steps to a 5 AP Calculus BC 2019

Series 363

4. Determine whether the series

∑∞

n= 1

n

n^2 + 1

converges or diverges.

Answer:nlim→∞

n

n^2 + 1

1

n

=nlim→∞

n^2

n^2 + 1

=1 and

∑∞

n= 1

1

n

diverges, so

∑∞

n= 1

n

n^2 + 1

diverges by

limit comparison with

∑∞

n= 1

1

n

.

5. Determine whether the series

∑∞

n= 1

50

n(n+1)

converges or diverges.

Answer:Since

∫∞

1

50

x(x+1)

dx=50 limk→∞

∫k

1

(

1

x

+

− 1

x+ 1

)

dx

=50 limk→∞[lnx −ln(x+1)]k 1 =50 limk→∞[lnk−ln(k+1)+ln 2]

=50 limk→∞

[

ln

k

k+ 1

+ln 2

]

=50 ln 2,

∑∞

n= 1

50

n(n+1)

converges by integral test.

6. Determine whether the series

∑∞

n= 1

(− 1 )n

3

2 n

converges absolutely, converges

conditionally, or diverges.

Examine the absolute values of

∑∞

n= 1

∣∣

∣∣(− 1 )n^3

2 n

∣∣

∣∣=

∑∞

n= 1

3

2 n

=

3

2

∑∞

n= 1

1

n

. Since

∑∞

n= 1

1

n

is a

harmonic series that diverges, the series

∑∞

n= 1

∣∣

∣

∣(−^1 )

n^3

2 n

∣∣

∣

∣diverges and

∑∞

n= 1

(− 1 )n

3

2 n

does

not converge absolutely. Now examine the alternating series

∑∞

n= 1

(− 1 )n

3

2 n

. Rewrite

∑∞

n= 1

(− 1 )n

(

3

2 n

)

as

3

2

∑∞

n= 1

(−1)n

n

. Since

∑∞

n= 1

(−1)n

n

is an alternating harmonic series that

converges, therefore

3

2

∑∞

n= 1

(−1)n

n

converges, and the series

∑∞

n= 1

(−1)n

3

2 n

converges

conditionally.

7. Find the interval to convergence for the series

∑∞

n= 1

(x+1)n

√

n

.

Answer:nlim→∞

∣∣

∣∣

∣

(x+1)n+^1

√

n+ 1

·

√

n

(x+1)n

∣∣

∣∣

∣

=nlim→∞

∣∣

∣∣

∣

√

n(x+1)

√

n+ 1

∣∣

∣∣

∣

=|x+ 1 |. Since|x+ 1 |< 1

when− 1 <x+ 1 <1or− 2 ≤x<0, the series converges on (−2, 0). When

x=−2, the series becomes

∑∞

n= 1

(−1)n

√

n

. Since 1>

1

√

2

>

1

√

3

>

1

2

>···and

nlim→∞

1

√

n

=0, this alternating series converges. Whenx=0, the series becomes