Calculus BC - 2022 AP Live Review Session 4: Convergent and Divergent Series - Solutions
Multiple Choice Practice

an 5n
1. Given the series 
, which of the following could be the correct set-up and conclusion for the
n =1 n !
ratio test?
an+15n+1 n!
5a
5a
1 5an+1
1
lim
×
= lim
×
= lim
× lim n+1 = (0) lim n+1 = 0 < 1Þ converges
n
n®¥ n +1 ! a 5
n®¥ a
( ) n n®¥ n +1 an n®¥ n +1 n®¥ an
n
an 5n (n +1)!
(A) lim
×
< 1; the series converges.
n®¥ n!
an+15n+1
It is the reciprocal so wrong conclusion.
(B) lim
an +1 5n +1 n !

 1; the series diverges.
( n + 1)! an 5n
Correct set-up. Wrong conclusion.
(C) lim
an +1 5n +1 n !

 1; the series converges.
( n + 1)! an 5n
Correct set-up. Wrong conclusion.
(D) lim
an +1 5n +1 n !

 1; the series diverges.
( n + 1)! an 5n
Correct set-up and conclusion.
n →
n →
n →
2. Level: AP2

Let
a
n =1
n
be an infinite series with an = f ( n ) . Which of the following conditions must be met to apply the
integral test?
(A)
f ( x) is differentiable
(B)
f ( x) is positive
Does not have to be differentiable, just continuous.
(C) an is continuous
This a set of discrete values, cannot be continuous.
(D) an is decreasing
This a set of discrete values need to be decreasing and going to zero.
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3. Level: AP2

 (−1)
Given the series
n =1
Series Test?
n
an , which of the following conditions must be met in order to use the Alternating
II. an  0
I. an +1  an
III. lim an = 0
n →
(A) I only
¥
å (-1) a
(B) I and III only
n
n
assumes that an is positive.
n=1
(C) III only
(D) I, II, and III
4. Level: AP2
n 2 + 3n − 2
Given the series  3
, which of the following series can be used as a comparison for the Limit
2
n =1 n − n + 1
Comparison test.


(A)
1
n
n =1

(B)
3
1
n
n =1
2

(C)
n
n =1

(D)
1

n =1 n
For large n,
n2 + 3n - 2
n2 1
behaves
like
=
n3 - n2 +1
n3 n
5. Level: AP2
Which of the following infinite series diverges by the n-th term test?

(A)
 ln(n)
n =1
lim
ln ( n) = ¥
n®¥

(B)
ln(n)
n
n =1

lim
n®¥

(C)
n3

n
n =1 2
lim
(D)
3n 2 + n − 4

n 3 + 2n
n =1
lim

n®¥
ln ( n)
n
= 0 comparing magnitudes. n dominates ln ( n)
n3
= 0 comparing magnitudes. Exponential dominates power.
2n
3n2 + n - 4
= 0 comparing magnitudes. n3 dominates n2
n®¥
n3 + 2n
AP Calculus BC: Convergent and Divergent Series
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6. Level: AP3
Which of the following converge?

(A)
1
n

n =1

(B)
−1
divergent p-series with p =
1
<1
2
n
divergent Harmonic series p-series with p = 1
n −2

3
n
n =1
n-2
n =1

(C)

(D)

n =1
3
3
n4
6n 2
3
n
=
1
n2 3 n
=
1
7
n3
convergent p-series with p > 1
4
n4 n 3 1
= 2 = 2 divergent p-series with p < 1
n2
n
n3
7. Level: AP3

2
4 

n =1  3 
n −1
is
(A) 6
(B) 8
convergent geometric series with 0 < r < 1.
(C) 12
(D) divergent
8. Level: AP3
n2 - 2
å n3 + 4n2 +1 is true?
¥
n=3
1
¥
(A) å is a divergent
1
(A) The series converges by the limit comparison test with å .
n=3 n
n=3 n
Harmonic series, shows

divergence.
1
(B) The series converges by the limit comparison test with  3 .
n2 - 2
n =3 n
3
2

(B) & (D) lim n + 4n3 + 1
1
n®¥
1n
(C) The series diverges by the limit comparison test with  .
n =3 n
2
3
( n - 2) n
= lim 3

n®¥
1
n + 4n2 + 1
(D) The series diverges by the limit comparison test with  3 .
n5 - 2n3
n =3 n
=
lim
=¥
n®¥ n 3 + 4n 2 + 1
n2 - 2
3
2
n2 - 2 ) n
(
n3 - 2n
Þ Cannot apply test.
n
+
4n
+
1
(C) lim
= lim 3
=
lim
=1
2
3
2
n®¥
n®¥
n®¥
1n
n + 4n + 1
n + 4n +1
See (A) Þ Both diverge.
¥
Which of the following statements about the series
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9. Level: AP3

Which of the following statements describes the series
 (−1)
n =1
(A) converges absolutely
n
n
using the alternating series test?
n+3
n
= 0 by comparing magnitudes,
n®¥ n + 3
might converge.
lim
(B) converges conditionally
(C) diverges
(D) alternating series test is inconclusive
n
= 0 by comparing magnitudes, might converge.
n®¥ n + 3
lim
¥
å (-1)n
n=1
¥
1
n
n
n
=å
does not converge because for large n it behaves like
= n2 ,
n + 3 n=1 n + 3
n
1
n
1
because for large n it behaves like
= n 2 , the p-series with Þ p = Þ Not absolutely convergent
n
2
æ 1 ö
2
n + 3) ç
- (1) n
(
÷
n
+
3
2
n
æ
ö
(
)
è 2 nø
( n + 3) - 2n
n
d
n
>0
=
=
ç
÷=
2
2
2
n+3
dx è n + 3ø
2 n ( n + 3)
2 n ( n + 3)
( n + 3)
( )
=
¥
2 n ( n + 3)
å (-1)
n=1
3- n
n
2
< 0 when n > 3 Þ
n
n +1
n
is decreasing or
<
.
n+3
n+4 n+3
n
converges by the AST.
n+3
n
is decreasing for n > 3 Look at the graph.
n+3
10. Level: AP3
Which of the following series converge?


I.
3

n
n =0 2
(A) I only
II.
I.
(D) I, II, and III
¥
II.
1
ån
n=1
¥
III.
n
III.
2
 ( −1)
n
n =1
n
æ 1ö
1
= 3å ç ÷ Þ convergent geometric series r = < 1
2
n=0 è 2 ø
¥
3
å2
n=0
(B) II only
(C) I and II only
n
n =1
¥

1
2
Þ convergent p-series p = 2 > 1
å( -1)
n
Þ divergent by nth term test.
n=1
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11. Level: AP3

Let
a
n =1
n
be a series whose n th partial sum is Sn =

a
(A) The series
n
diverges by the nth term test. We need an to do the nth term test.
n
diverges by the p -series test. We need an to do the p-series test.
n
converges because lim Sn  0 . We need an to do the nth term test, not S n .
n =1

a
(B) The series
n =1

a
(C) The series
n =1
n →

a
(D) The series
2n + 1
. Which of the following statements is true?
5n - 3
n =1
n
converges to
2n +1 2
=
n®¥
5n - 3 5
2
.
5
S = lim Sn
lim
n®¥
12. Level: AP4
 x−2
For which of the following values of x does the series  

3 
n =1 

æ x - 2ö
å çè 3 ÷ø
n=1
¥
(A) x = −1
(B) x = 1
n-1
is geometric with r =
n −1
converge?
x-2
3
1
x - 2 < 1Þ x - 2 < 3 Þ -1 < x < 5
3
x = 1 is in this interval
r =
(C) x = −5
(D) x = 5
13. Level: AP4

When applied to the series
a
n =1
n
, the ratio test yields the result lim
n →
an +1
= L . Which of the following could
an
be true?
(A) L =
1
and
2
2
(B) L = and
3
(C) L = 1 and

n =1
3
and
2
converges conditionally.
L=
1
< 1 converges by the Ratio Test
2
n
diverges.
L=
2
< 1 converges by the Ratio Test
3

a
n =1

a
n =1
(D) L =
n
a
n
converges conditionally.
L = 1 Ratio Test inconclusive
converges absolutely.
L=

a
n =1
n
AP Calculus BC: Convergent and Divergent Series
3
> 1 diverges by the Ratio Test
2
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14. Level: AP4
Which of the following series converge conditionally?

2
I.  (−1)  
3
n =0
¥
n
n
n=1
n
I.
(B) I and III only
¥
(C) III only
æ 1ö
çè n ÷ø
3/2
¥
III.
å (-1)
n=1
n+1
æ 2ö
çè n ÷ø
n
¥
n æ 2ö
æ 2ö
2
-1
=
å ( ) çè 3 ÷ø å çè 3 ÷ø Þ geometric series r = 3 < 1 absolutely convergent
n=0
n=0
¥
(A) I and II only
å (-1)
II.
n+1
II.
å (-1)
n+1
n=1
(D) I, II, and III
¥
III.
å ( -1)
æ 1ö
çè n ÷ø
n+1
n=1
¥
å( -1)
n+1
n=1
3/2
æ 1 ö
3
= å ç 3/2 ÷ Þ p-series p = > 1Þ absolutely convergent
2
ø
n=1 è n
¥
æ 2ö ¥ æ 2ö
çè n ÷ø = å çè n ÷ø Þ divergent Harmonic series
n=1
æ 2ö
çè n ÷ø Þ convergent by AST Þ conditionally convergent
15. Level: AP3
Which of the following geometric series converge?
 
I.   
n=0  3 


n
 ( 0.99 )
II.
  
III.  2 

n =1  e + 1 

n
n =0
n
(B) I and II only
æpö
p
I. å ç ÷ Þ r = > 1 divergent
3
n=0 è 3 ø
(C) II and III only
II.
¥
(A) II only
¥
(D) I, II, and III
n
å(0.99)
n
Þ r = 0.99 < 1 convergent
n=0
n
æ p ö
p
3.14
III. å 2 ç
Þr =
»
< 1 convergent
÷
e +1 2.7 +1
n=1 è e +1ø
¥
16. Level: AP3
Which of the following series converges?
¥
(A)
å
(-1)
2n-1
n
n=1
¥
(B)
å( -1)
n
n=1
¥
(C)
å
n=1
¥
(-1)
3
1
3
1
×
n +1
3n - 2
n-1
n
2n
(D) å ( -1) × 99
n
n=1
n
(-1) + ( -1) + ( -1)
=
2
5
3
Þ divergent Harmonic series
n +1 1
= divergent by nth term test
n®¥ 3n - 2
3
lim
¥
=å
n=1
( -1)
n-1
n1 3
convergent by AST
2n
lim 99 ® ¥ The exponential dominates the power, divergent by nth term test.
n®¥ n
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17. Level: AP3
Which of the following series converge?

I.

k =4
k2 + k
k −3
¥
(A) I and III only
(B) I and II only
(C) III only
k +3
å (k - 2)2 + 1
k=1
¥
II.
å
I.
k2 + k
lim
k -3
k=4
k2 + k
k -3
k®¥
6k + 3
3
+k
k =1

k
III.
k2
= ¥ divergent
k®¥ k 1 2
= lim
¥
¥
k +3
k
1
»
»
II. å
like Harmonic, divergent
å
å
2
2
k=1 (k - 2) +1
k=1 ( k )
k=1 k
¥
(D) II and III only
k +3
( k + 3) k = lim ( k ) k = 1 Both diverge by Limit Comparison Test
(k - 2)2 + 1
lim
=
lim
k®¥
k®¥
1k
(k - 2)2 +1 k®¥ (k)2
¥
6k + 3 ¥ 6k ¥ 6
III. å 3
» å 3 » å 2 like convergent p-series p = 2 > 1
k=1 k + k
k=1 k
k=1 k
6k + 3
6k 3
3
6k + 3) k 2
(
k
+
k
lim
= lim 3
= lim
= 1 Both converge by Limit Comparison Test
k®¥ 6 k 2
k®¥ k + k 6
k®¥ 6k 3
(
( )
( )
)
18. Level: AP4
Let a n = n . Which of the following series converges?
¥
(A)
å
n=1
¥
(B)
(5nan + 3)
2n 6 +3n5
4
(
lim
n®¥
n!
×an
lim
5n - 1
å n×a
n=1
n
lim
å2
n=1
n
n®¥
¥
(C)
n®¥
)
5n n + 3
4
2n 6 +3n5
n!
2 × n
n
5n -1
n× n
(5n )
= lim
32
=¥
= lim
n®¥
2n 6
n®¥
4
(5 n ) = 5
= lim
4
n®¥
6
4
2n 6
2
¹ 0 Divergent by nth term test.
because factorial dominates. Divergent by nth term test.
5n
n× n
= lim
n®¥
5
n
=0
¥
å
n=1
5
n
is a divergent p-series.
5n -1
32
(5n) n = lim 5n3 2 = 5 Both converge by Limit Comparison Test
lim n
= lim
32
n®¥
n®¥ n3 2
1 n n®¥ n

(D)

n =1
( −1)
n
n
an
¥
å
n=1
( -1)
n
n
× n
¥
=å
n=1
AP Calculus BC: Convergent and Divergent Series
( -1)
n
n
Convergent by AST
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19. Level: AP4
Which of the following series can be shown to converge by the alternating series test?

I.

( −1)
n =1
n
2n

II.
2

cos ( n )
n
n =0
(A) I only

III.
 ( −1)
n −1
n =1

sin ( n )
n
I. Not an Alternating Series
¥
(B) II only
II.
(C) I and III only
å
cos (p n )
n=0
(D) II and III only
III. lim
n
sin ( n)
n®¥
n
¥
=å
( -1)
n=0
n
n
and convergent by AST
= 0 because n dominates sin(n), but the terms do not strictly decrease.
d æ sin ( n) ö ncos ( n ) - sin ( n)
Þ n > 1 then ncos ( n ) dominates but -1 £ cos ( n ) £ 1
ç
÷=
dx è n ø
n2
so the derivative is not always negative.
20. Level: AP4
Which of the following series can be shown to converge by the alternating series test?
3
1
27
125
n −1 n
I.
−2+
−4+
+ ... + an + ... , where an = ( −1)  n .
2
8
32
2
 2 n 
cos 

1 1 1 1
1
1
1
1
3 

II. 1 − − + − − + −
.
−
+ ... + an + ... , where an =
2
8 18 16 50 72 49 128 162
( n + 1)
3
1 32 33 34 35 36
n
n +1
−
+
−
+
−
+ ... + an + ... , where an = ( −1)
III.
.
2 5
8
11 14 17
3n − 1
(A) II only
(B) III only
(C) I and III only
(D) II and III only
n3
I. lim an = lim n = 0 because the exponential will eventually dominate the power.
n®¥
n®¥ 2
Therefore this series does converge by AST.
( n +1)
3
2 n+1
n3
< n Þn³4
2
æ 2p n ö
II. This is not an alternating series because cos ç
¹ ±1
è 3 ÷ø
n
= 0 because n eventually dominates n1 3 .
n®¥
n®¥ 3n -1
3
n
n1 3
1
1
»
= 2 3 Þ Eventually an behaves like 2 3 which is strictly decreasing. Therefore this
3n -1 3n 3n
3n
series does converge by AST.
III. lim an = lim
3
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21. Level: AP
Which of the following series converge?

I.
II.
(A) I only
(B) III only
3n+1 ( n +1)
100
I.
lim
n®¥
3n − 1

n =1 7 n

3n n100

n!
n =1
( n +1)!
(C) I and III only
3( n +1)
( n +1)
n!
× n 100 = lim
=
3lim
n®¥ n +1 n100
n®¥
3n
n100
( )
100
  −1 



e 
n =1 

III.
n
(D) I, II, and III
99
= 0 < 1Þ converges by ratio test
3n -1 3
= ¹ 0 Þ diverges by nth term test.
n®¥ 7n
7
II. lim
III. Geometric with r =
p -1 3.14 -1 2.14
e
»
2.72
=
2.72
< 1Þ converges by geometric test.
22. Level: AP4
3
3
3
 −1  3
What is the sum of the series −
+
−
+ ... +  
+ ... ?
 2 4 8
 2  
n
3
(A)
(B)
(C)
2

Geometric with a =
1
3p
3 p 2p
6
2
,r = - Þ S =
=
×
=
=
p
2
1- ( -1 2 ) 3 2p 3p p
2
3
6

6
2 + 1
(D) The series diverges.
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23. Level: AP4

Consider the series
 an where f (n) = an . Which of the following statements about f ( x) and
n=2

a
n=2
n
could
be true?
1
and
x
(A) f ( x) =
¥
ò
2

a
n=2
converges by the integral test.
n
b
f ( x ) dx = lim ò f ( x ) dx = lim éë ln x ùû 2 = lim éë ln b - ln 2 ùû = ¥ Þ diverges
b
b®¥
b®¥
b®¥
2
(B) f ( x) =
1
and
2x
¥

a
n=2
n
b
ò f ( x ) dx = lim ò 2
-x
b®¥
2
2
(C) f ( x) = xe x and
converges by the integral test.
(
) (
)
b
ln 2
dx = lim éë -2- x ln 2 ùû2 = lim éë -2- b ln2 - -2-2 ln 2 ùû = 0 +
Þ converges
b®¥
b®¥
4

a
n=2
n
diverges by the integral test.
f ( x ) is not decreasing so the integral test cannot be applied. lim an ¹ 0
n®¥
(D) f ( x) =

1
x ( ln ( x ) )
2
and
a
n=2
n
diverges by the integral test.
24. Level: AP4
If 0 < an < b n where an and bn are both decreasing sequences for n ³ 0, which of the following must be
true?

(A) If
a
n =0

n
converges, then
n =0

(B) If
 bn diverges, then
n =0
(C) If lim bn = 0 , then
n →

(D) If
 ( −1)
n =0
b
¥
n
converges.

a
n =0
n
converges.
å ( -1) an converges.
n
n=0
n
a
n =0
n®¥
n®¥
¥
Þ å ( -1) an converges by AST.
n
n=0

bn converges, then
lim bn = 0 and 0 < an < b n then lim an = 0
n
converges.
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25. Level: AP4
¥
Given
å (-1)
np
n
n -3
2
n=1
, for what value(s) of p will the series be conditionally convergent?
(A) p  1
lim
= lim
np
np
= 0 Þ0 < p <1
n®¥ n
= lim
n2 - 3 n®¥ n2
numerator power less than denominator power
n®¥
(B) p = 1
(C)
np
¥
1
3
 p
2
2
å (-1)
n=1
(D) 0  p  1
np
n
n2 - 3
¥
å (-1)n
np
converges by AST.
¥
=å
np
¥
np ¥ 1
= å 1- p divergent p-series
n=1 n
n=1 n
Ȍ
n2 - 3 n=1 n2 - 3
because 1- p < 1 on the interval 0 < p < 1Þ conditionally convergent
n=1
26. Level: AP4

Consider the series
a
n =1
n
where lim an =
n →
1
. Which of the following must be true?
2

(A) The series
a
n
diverges by the nth term test.
n
converges by the nth term test.
n
converges by the ratio test.
n =1

(B) The series
a
n =1

(C) The series
a
n =1

(D) The series
a
n =1
n
could converge or diverge.
27. Level: AP5
 −2 
4   is −1 where k is a constant. What is the value of k ?

n =1  k 

The sum of the geometric series
(A) k = −
2
5
æ -2 ö
n = 1Þ a = 4 ç ÷
è k ø
n
(B) k = 6
1
(C) k = 8
(D) k = 10
æ -2 ö
4ç ÷
è k ø
-8
S=
=
= -1Þ - ( k + 2) = -8 Þ k + 2 = 8 Þ k = 6
æ -2 ö k + 2
1- ç ÷
è k ø
AP Calculus BC: Convergent and Divergent Series
2022 AP Live
Bryan Passwater & Tony Record
28. Level: AP5
¥
Which of the following is true when the integral test is applied to the series
å ne
-n
where f ( x) = xe− x ?
n=1
¥

2
 f ( x ) dx = e
(A)
and
1

(B)

1
-n
n=1

1
converges.
¥
å ne
= ò f ( x ) dx .
¥
-n
1
n=1
å ne

-n
n=1
f ( x ) dx converges and
¥
(C)
å ne
is a Riemann Sum ¹ ò f ( x ) dx is the area under the curve.
¥
1
¥
å ne
f ( x ) dx diverges and the series
-n
diverges. The integral converges.
n=1
(D) The conditions for the integral test are not met and the integral test cannot be applied.
u = x Þ du = dx
dv = e- x dx Þ v = -e- x
ò xe dx = -xe - ò -e dx = -xe - e = -e ( x + 1)
ò xe dx = lim ò xe dx = lim éë-e ( x + 1)ùû = - lim éë e ( b + 1) - ( 2e )ùû
-x
¥
-x
-x
b
-x
b®¥
1
1
-x
-x
-x
-x
b
-x
b®¥
1
-b
-1
b®¥
é æ b + 1ö æ 2 ö ù
é
æ 2ö ù 2
= - lim
=
0
(
)
ê
ú
ê
b
ç
÷
ç
÷
çè e ÷ø ú = e Þ integral converges
b®¥
ëè e ø è e ø û
ë
û
29. Level: AP5

Which of the following could be a valid conclusion of the integral test for the series
a
n =1
n
where f (n) = an
and f (n) is known to be positive and decreasing ?

(A) The series
a
n =1
n
()
diverges because f x is not continuous over the interval 1,  ) .
()
If f x is not continuous then you cannot apply the integral test.

(B) The series
a
n =1
n

(C) The series
a
n =1


1
n
diverges because


1
f ( x)dx  0 . Diverges if the integral is not a finite, positive value.
()
converges to 2 because f x is differentiable over the interval 1,  ) and
f ( x)dx = 2 . The sum is not equal to the integral.

(D) The series
a
n =1
n
()

converges because f x is continuous over the interval 1,  ) and  f ( x)dx = 8 .
1
The integral is a finite, positive value and all conditions of the integral test are met.
AP Calculus BC: Convergent and Divergent Series
2022 AP Live
Bryan Passwater & Tony Record
30. Level: AP5

For which value(s) of k do both
n3
n
n =1
k2
æk ö
and å ç + 1÷
ø
n=1 è 2
¥
-n
converge?
(A) −  k  −2
(B) −4  k  − 3
(C) k < -4 or k > 2
(D) −4  k  0 and k  3
AP Calculus BC: Convergent and Divergent Series
2022 AP Live
Bryan Passwater & Tony Record