BC Calc III
Sample 9.1-9.4 Solutions
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#1.
Complete (no work necessary):

a
a. An example of a divergent series
n 1
n
with an  0 is given by an 
1
n

b. Look at the series
a
k 1
k
. The ratio test states that if
a 
ak  0 for all n and lim  n1   L  1 then the series converges.
n 
 an 

c. An expression of the form
a
k 1
k
is called an infinite series.
Corresponding to this, we have the sequence {Sn}, which is called the sequence
n
of partial sums . The nth term of this sequence is given by Sn =  ak
k 1
The sequence {an} is called the sequence of general terms.
If lim ak  0 then
k 

a
k 1
k
diverges.
k 2
.
k 
k 1
Use this to determine whether the series converges or diverges. If the series converges,
find the value.

#2. Find an expression for the sequence of partial sums, S n , for the series
n
k 2
3
4
5
 n 1
 n2
Sn   ln 
  ln    ln    ln     ln 
  ln 

 k 
1
2
 3
 n 1 
 n 
k 1
 ln 3  ln1  ln 4  ln 2  ln 5  ln 3  ln( n  2)  ln n
  ln1  ln 2  ln(n  2)  ln(n  1)
Therefore, lim Sn  lim   ln1  ln 2  ln(n  2)  ln(n  1)    .
n
n

Hence, the series
k 1
BC CALC III
k 2
 diverges.
k 
 ln 
 ln 
#3.
Determine whether each series converges or diverges. Justify your answer
carefully and completely.

1
n3
a.  2
We use limit comparison with an = . Then
n
n + n
n2
an
1 n2  n
n2  n
 lim 
 lim 2
 1 . Since 0  L   and
n  b
n  n
(n  3) n (n  3n)
n
diverges,

n3
diverges by the limit comparison test.

n2 + n
n2
L  lim

1
n
n 1
 n2  1 
  n  We use the ratio test.
n  1 3

 (k  1) 2  1 


k 1
 (k  1) 2  1 3k 
 (k 2  2k  2) 1  1
a
L  lim k 1  lim  32
 lim 


lim
 

k 1
k  a
k 
k 2  1  k  
k 2 1
3 3
k
 k  1  k   3
k
3


Then L  1 .

 n2  1 
So,   n  converges by the ratio test.
n  1 3


b.
BC CALC III
#3. (continued)Determine whether each series converges or diverges. Justify your answer
carefully and completely


c.
 n! n!
Using the Ratio Test:
(2n)!
 (n  1)!(n  1)! (2n)!
a
L  lim n 1  lim 

n  a
n 
n !n ! 
 (2n  2)!
n
n 1
 (n  1)(n  1) 
 lim 
n  (2n  2)(2n  1) 


1

4

 n! n! converges by the ratio test.
Since L <1, 
n 1 (2n)!

d.
k
e
k 1
k2
We use the integral test.
b k

dk

lim
1 ek 2 b  1 ek 2 dk 
 1 2 b 
 lim   e k

b 
k  1
 2

k
1 
 1
 lim   b2 
b 
2e. 
 2e
1

2e.
k
Since k 2 is positive and decreasing,
e
BC CALC III

k
e
k 1
k2
converges by the integral test.
 n2  1 
Determine whether the sequence an  tan 1 
 converges or diverges. Explain.
 n 
If it converges, find the limit.
#4.
 n2  1 

1
Since 
   as n   and y  tan x has a horizontal asymptote at y  ,
2
 n 

 n2  1  
lim  tan 1 
 
n 
 n  2


Find the value of n such that S n approximates the value of the series
#5.
an error of at most .001. Explain carefully.

If
1
 .001 then

2
k  n 1 k  1
n
1
is within .001 of

2
k 1 k  1

k
k 1
1
.
1
2


 1
1
1
1



dk


2
2
2
n 1 k 2
(n  1)
k  n 1 k  1
k  n 1 k
 1 b 
1
1
.
dk  lim 


2
b 
k
 k n  1 n  1
If n = 9999, then

 1
1
1
1
1
1


dk 



2
2
2
2
n 1 k
(n  1)
(10000) 10000 1000
k  n 1 k  1
Now,


n 1
BC CALC III
k
k 1
1
with
1
2

#6. Look at the series
(1) n
 n 3
n 1
n
.
a. Show that this series converges. This converges by AST since
1
1
ak 
is decreasing and
0.
k
k 3
k  3k
b. Find a value for n such that Sn is within .001 of the actual sum.
Since ak 
1
1
1
is decreasing and
 0 , S  Sn  an 1 
. To get
k
k
k 3
k 3
(n  1)  3n 1
1
 .001 , we need n + 1 > 5, or n = 5.
(n  1)  3n 1
#7. Determine whether the following series converges conditionally, converges
absolutely, or diverges? Show all steps/explain.


n 1
2n
(1)n  2n
  as n   , this series diverges by the nth term test.
Since
n3
n3
BC CALC III