Math 208 Worksheet 4-6-11
The comparison test: first “guess” whether the series is convergent or divergent. If your guess is
convergent, construct a larger series (by making the denominator smaller or the numerator larger)
that is known to converge. If your guess is divergent, construct a smaller series(by making the
denominator larger or the numerator smaller) that is known to diverge.
The limit comparison test: extract the dominant term from the numerator and the denominator to
construct bn . (you must be able to tell whether
b
is convergent or divergent.
n
Determine convergence/divergence using a) the comparison test b) the limit comparison test
n 1

2
n 1 n  1

1) a)
n 4
c)
n2  1

b)

n 1
n2  1

3
n 1 n  4


d)
1
e)

2
n 1 n  4
2k  3k

k
n 1 5  k

2) Determine convergence/divergence for each of the following (any method):

A)
1
sin  

n
n 1

B)

n 1
n ln n
n3  1
 n  1



n 1  n 

C)
n

D)
 cos(n)
n 1
a) Comparison: guess divergent, so construct a smaller divergent series by dropping plus 1 from the
numerator and minus 1 from the denominator:
n 1
n 1
 2  .
2
n 1 n
n
The smaller series is divergent,
so the original is divergent by the comparison test.
Limit Comparison:

Since
n 1
an  2
n 1

1
n 1
2
1
 n is divergent,  n
n 1
n 1
(from the series),
n 1
2
n 1
an
n2  n
bn  2  . lim
 lim n  1  lim 2
 1.
n  n  1
n
n n  bn n  1
n
is divergent.
b) Comparison: guess convergent, so construct a larger convergent series by replacing
n 4
by
n  4 n  5 n in the numerator and dropping plus 1 from the denominator to produce a
n 4 5 n
5
 2  3
2
n 1
n
n2
monomial in the denominator and the numerator :
. The larger series is
convergent, so the original is convergent by the comparison test.
Limit Comparison:
an 
n
n 4
2
(from the series),
n 4
3
2
2
2
an
n

4
n
lim
 lim n  1  lim
 1.
2
n b
n
n n  1
1
n
bn 

Since
3

n 1
1
n
3
2
n
n
2

1
n
3
2
.

is convergent,

n 1
n 4
n2  1
is convergent.
n2
c) Comparison: guess divergent, so construct a smaller divergent series as follows: key here is to
replace terms so that you obtain a monomial in the numerator and the denominator:
n2
n2
n 
n2  1
2  2  1

3
2
n  4 n  4n3 5n3 10n
2
(this is valid after n = 2). The smaller series is divergent, so the
original is divergent by the comparison test.
Limit Comparison:
an 

Since
1
is divergent,

n 1 n
n2  1
n3  4
bn 
(from the series),
n2  1

3
n 1 n  4
n2 1
 .
n3 n
n2  1
3
a
n3  n
lim n  lim n  4  lim 3
1.
n b
n 
n  n  4
1
n
n

is divergent.
d) Comparison: guess convergent, so construct a larger convergent series as follows: key here is to
replace terms so that you obtain a monomial in the denominator:
1

n 4
2
1
2
n
n 
4

2
4
3n 2
(the
inequality is valid after n = 5). The larger series is convergent, so the original is convergent by the
comparison test.
Limit Comparison:

Since
1
n
n 1
2
1
an  2
n 4
(from the series),

is convergent,
n
n 1
2
1
4
1
bn  2
n
1
2
an
n2
lim
 lim n  4  lim 2
 1.
n  b
n 
n  n  4
1
n
n2
.
is convergent.
e) Comparison: guess convergent, so construct a larger convergent series by replacing
2 k  3k
by
3  3  2(3 ) in the numerator and dropping minus from the denominator : key is to produce
k
k
k
monomial in the denominator and the numerator.
2 k  3k 2(3k )
3
 k  2 
k
5 k
5
5
k
. The larger series is
convergent, so the original is convergent by the comparison test.
Limit Comparison:
an 
2k  3k
5k  k
(from the series),
bn 
3k
5k
.
2k  3k
k

k
a
5k 2k  3k 5k
5k 2k / 5k 3k  3k 5k / 5k 3k
3
.
Since
lim n  lim 5 k k  lim k k

lim

1
  is

n  b
n 
n   5 3  k 3k
n   5k 3k / 5k 3k  k 3k / 5k 3k
3
n 1  5 
n
5k

2k  3k
convergent,  k
is convergent.
n 1 5  k
1
1
2)a) Limit Comparison: an  sin   (from the series), bn  .(the choice of bn will become clear
n
n
1
sin  


1
a
1
 n   1 . Since
after studying the MacLaurin series) lim n  lim
is divergent,  sin  

n  b
n 
1
n
n 1 n
n 1
n
n
is divergent.
b) Comparison: guess convergent, so construct a larger convergent series as follows: key here is
to observe
ln n  n
(this can be shown using the mean value theorem) .

larger series is convergent, so

n 1
c) Test of divergent:
The
n ln n
is convergent by the comparison test.
n3  1
 n 1  1 

  1  
 n   n
n
n ln n
nn 1
 3  3
3
n 1
n
n2
n
and take the limit using L’hopital to show the limit is not
0. Thus this is divergent
d) Divergent by the test of divergence. The limit does not exist since the function oscillates.