Math 2414 Activity 16 (Due by August 14)
Determine convergence or divergence of the following alternating series:

1.

n 1

3.

n 1

5.

n 1
 1
n 1
7.

n 1

8.

n1

2.
n2
 1
n 1
4.
n
n 1

 1
n 1
 n 1  n 


n 1
1  3    2n  1
1

 1
2  4    2n  2n  2

n
 1
n 1
2n   1
n

1
 1 ln 1  
 n
n
{Hint: Look at


n
an 1
.}
an
6.
 1
n
n 1
2 4
{Hint: an   
3 5
2  4    2n  1

1  3    2n  1 n 2

2n  2 2 n 2 n  1 2 n  1
.}



2n  1 2n  1 n 2
n2
n 1
{Hint: Check out the partial sums.}
 1
n1
6n  3  1
n1
{Hint: Check out the partial sums.}

9. The alternating series

 1
n 1
n 1
converges by the Alternating Series Test, but what does it
n
converge to?
Let’s look at the even partial sums:
s2 n  1 
 1 1
1   
 2 3

1 1 1
  
2 3 4
1   1 1 1
  1    
2n   2 3 4
s2 n
Solving the previous equation for s2n , we get


1
2n
1   1 1
  1   
2n   2 3
1
 
n
1   1 1
 1 1
s2 n   1    
  1   
2n   2 3
 2 3
1
1
1


 
n 1 n  2
2n

So

n 1
 1
n 1
n
1
 
n
1
 1
 lim s2 n  lim 


n
n n  1
n2

1
 1
Find lim 


n n  1
n2



1 
.
2n 
1 
 , and you’ll know the sum of the series.
2n 
1
1 
 1
Activity 2, problem #11. Calculate lim 
by rewriting it as

 

n n  1
n

2
n

n


1 1
1
1 
lim  1 


 and identifying it as a definite integral.
2
n n 1 
1  nn 
 n 1 n
Activity 13, problem # 4.From the two graphs, you can conclude that
2 n 1

2n
1
1
1
dx 


x
n 1 n  2

1

nn
n 1

1
dx .
x
n
1
 1
Use this fact to find lim 


n n  1
n2


1 
.
n  n 
10. Show that in the Alternating Series Test, it is not enough to have lim an  0 . Consider the
n
alternating
series
11
1 1 1 1 1 1
     
2 4 3 9 4 16



 1
n 1
an ,
n 1
 2
 n  1 ; n is odd
an  
.
4

; n is even
 n 2
a) What’s lim an ?
n 
 1 1
Consider the even partial sums, s2 n  1   
 2 3
1  1 1
   1   
n  4 9

1
.
n2 
where
b) What’s lim s2 n ?
n
c) What does this imply about the convergence of

1 1 1 1 1 1
11      
2 4 3 9 4 16


 1
n 1
an ?
n 1
n
 1
n 1
n
1  
 1
 1  n
11. a) Show that 1    1   
.
n
n
n




b) Show that if an  is a sequence of positive numbers, then if ln  an  is decreasing, then
an  is decreasing.
In other words, show that if ln  an1   ln  an  , then an1  an .
x
c) For x  0 , show that ln 1  x   x .
{Hint: ln 1  x  

1
1
dt , and
 1.}
1 t
1 t
0
  1 n 
 1   
n 
d) Show that an  ln  
is a decreasing


n




 1
f  x   x ln 1    ln x has a negative derivative.
x


e) Determine whether the alternating series

n 1
divergent using the previous results.
sequence
by
showing
that
{Hint:Use part c).}
 1 n1  1 n 
 1 1    1    is convergent or
 n  
 n 
n
12. A person starts walking from home (at x  0 ) toward a friend’s house (at x  1 ). Threefourths of the way there, he changes his mind and starts walking back home. Three-fourths
of the way back home, he changes his mind again and starts walking back to his friend’s
house. Three-fourths of the way to where he first turned around, he turns around again and
starts walking toward home. Three-fourths of the way to where he turned around the second
time, he turns around again and starts walking toward his friend’s house. If he continues this
pattern indefinitely, where will he end up?
Here are his movements:
3 3 3 3 3 3
     
4 4 4 4 4 4
.
His final position will be the sum of an alternating series which meets the conditions of the
Alternating Series Theorem, so he must end up at some point. Find it.
13. Suppose that f , f  , and f  are all continuous on 0,1 , with f  0 . Consider the series


f
 1n  .
n 1

a) If the series

f
 1n 
is convergent, then what must be the values of f  0  and f   0  ?
n 1
{Hint: f  0   lim f  1n  , so use the nth term test. f   0   lim
n
n
that f   0   0 and use The Limit Comparison Test.}
f  1n   f  0 
1
n
, so suppose
b) If f  0   0 and f   0   0 , can you conclude anything about the convergence or

divergence of

f
 1n  ?
n 1
 1n   lim f  x  ,
f  x
exists. Apply L’Hopital’s Rule
2
1
n
x0
x0
x
x
2
n
twice to get a result in the (extended)Limit Comparison Test.}
{Hint: lim
f
2
provided that lim

14. Determine if the improper integral

sin  x  sin  x 2  dx is convergent or divergent.
0


Hint: Let u  x 2 and convert the integral into
the book, sin
 u  sin u  



sin
 u  sin u  du .
2 u
0
cos u  u  cos u  u
By a trig identity from
 , so we get that
2
 cos u  u  cos u  u 
sin u sin  u 
 du . But
du  


2 u
4 u

0
0 

1
 cos u  u  cos u  u 
 cos u  u  cos u  u 

 du  
 du 




4 u
4 u


0 
0 

 cos u  u 
 du ,
 


4 u

1 

 






















1


 cos u  u 

 du


4 u


provided that each integral converges. The first integral does, so let’s look at the integrand
of the second one: f  x  
the interval 1,  .

cos x  x
4 x




on
 cos u  u 

 du is
It appears that


4 u

1 
equivalent to an alternating series to which
the Alternating Series Test can be applied.
What about the second integral:



1


 cos u  u 

 du ?


4 u




15. It can be shown that the alternating series
 1
n 1
1
converges, but what does it
n n
2
n 1
converge to?

{Hint:

 1
n 1
n 1
1

n2  n


 1
n 1
n 1
1 
1
 

 n n 1


 1
n 1
n 1
1

n


n 1
 1
n 1
1
,
n 1
so
use the result of problem #2.}
16. The sequence  xn  is defined recursively by x0  0 , x1  1 , and xn1 
xn  xn1
; n  1 . Find
2
lim xn .
n 
xn  xn1
1
1 x  x

 xn    xn  xn1     n1 n2  xn1 
2
2
2
2

{Hint: xn1  xn 
So xn   xn  xn1    xn1  xn2    xn2  xn3  
n   .}
n
17. The sequence  xn  is defined recursively by x0  0 , x1  1 , and xn1 

xn  nxn1
; n  1 . Find
n 1
lim xn .
n 
{Hint: xn1  xn 

xn  nxn1
n
n  xn1   n  1 xn2
 xn  

x
 xn  xn1   

n 1 
n 1
n 1
n  1
n

n 1
 1 x  x   1 .

 xn1  xn2   
 1 0
n 1
n 1
n 1
So xn   xn  xn1    xn1  xn2    xn2  xn3     x1  x0 
n
n 1

 1
k
 k  1 . Now let n   .}
k 0
n
n
 1
 1
     x1  x0     
 2
 2
k
n 1
 1
  x1  x0  
   Now let
2
k 0 
1
  xn1  xn2  
4
18. Determine absolute convergence, conditional convergence, or divergence for the following
series:

a)


 1
n 1
 n 1  n 


b)
n 1

 1
n 1
n 1

19. The alternating series

 1
n 1
n 1
1 
 1
 n  n 1


1
1 1 1 1 1 1 1
1       
n
2 3 4 5 6 7 8
is not absolutely
convergent, but it is conditionally convergent. Show that the following rearrangement of
1 1 1 1 1 1 1 1
this series: 1         
has a different sum. Multiply the original
3 2 5 7 4 9 11 6
1
series by , insert a term of 0 before each of its terms, and then add it to the original series
2
term by term.
1 1 1 1
1    
2 2 3 4
1
1
1
1
 1 1 1 1
 
         0   0   0   0   0 
2
4
6
8
 2 4 6 8
 
 1 1 1 1 1 1 1

1         
 2 3 4 5 6 7 8

1
1
1
1


0   0   0   0   0  
2
4
6
8





1 1 1 1 1
1     
3 2 5 7 4

From Problem #2, we know that

n 1
 1
n 1
 ln 2 . What’s the value of the rearranged series?
n

20. Find the value of the convergent alternating series
 1
n 1
n 1
2n  1 1
1
 
{Hint:
, so
n  n  1 n n  1

 1
n 1
the result of Problem #2.}
n 1
2n  1

n  n  1

 1
n 1
2n  1
.
n  n  1
n 1
1

n

 1
n 1
n 1
1
, and use
n 1

21. a) If


an is absolutely convergent, then what can be said about

n 1
n 1
an2
?
1  an2

b) What if


an is only conditionally convergent?
{Hint: Think about

 1
n 1
n 1
Determine convergence or divergence for the following series:

22.

n 1
 n!
 3n !

24.

n 1

26.

n 1

{ratio test}
1 3  5 
 n  2 !
2
2n  n!
r
 n  1n 1 


23.
n!
{ratio test}
 2n  1
30.

n 0

31.

n 1
25.

n 1

{ratio test}
27.

n 1

; 0  r 1
29.

n 0
  n  1n

 n !
nn
n!
 2  4  6   2n  
2  5  8   3n  1

n 1
n 1


 1
n 1
 1

28.

2
7
 
 19 
n



n 1
1
n
 5   1n 
 2 


n3  3n 
n
en1  n  1
n
{ratio test}
 n!  125 n 

 
n 
46

 
n

1



{ratio test}

32.
{ratio test}

n 1
2n n !
nn
{ratio test}
n
n
.}

Determine the convergence or divergence of the series

an whose terms are defined
n 1
recursively by the following:
1  sin n
33. a1  2 ; an1 
an
n
35. a1  1; an1 
37. a1  5 ; an1 
tan 1 n
an
2
n
n
an
2
1
1
; an1   an  n1
3
{Hint: Find a formula for an or…}
39. a1 
1
3n  1
34. a1  ; an1 
an
3
2n  5
n
an
n 1
{Hint: Find a formula for an }
36. a1  3 ; an1 
1
n  ln n
; an1 
an
2
n  10
{Hint: Is an increasing or decreasing?}
38. a1 
4
n 1
; an1   an 
5
{Hint: Find a formula for an or…}
40. a1 
a  5  an 
3
; an1  n
2
2
{Hint: Determine lim an by cob-web analysis.}
41. a1 
n 
42. Let a1  2 and an1  2  4  an , n  1.
{Hint: Determine lim an and apply the Ratio Test.}
n 

43. For which positive integers k is the series

n 1


44. If

n 1
an converges, then what about

n 1
 n! convergent?
 kn !
2
n
 1  sin  an  

 ? {Root Test}
2


45. In this problem, you’ll extend the Root Test to series containing factorials.
 2k !  2k   2k  1   2k  2  
 2k   2k  1   2k  2  
 k k k 
 k   k  1 
1
k
k
k 1
 k k 1
Which implies that  2k !  k k 1 and  2k  1!   2k  1 k k 1
2k
 2k !  2 k k k 1
k 1
 k 2k
So you get
 k
Which implies that
2 k 1
2k
 2 k ! 
k.
 2k  1!  2 k 1  2k  1 k k 1

And you also get
2 k 1
2k  1 k
k 1
2 k 1
 k
Which implies that
2 k 1
 2k  1! 
k.
a) Determine lim n n! by considering what happens for odd and even integers in the previous
n
inequalities.

b) Apply the Root Test to the series
2n
.

n
!
n 1
46. Consider the series 1 
1 1 1 1 1 1 1 1
       
2 2 4 4 8 8 16 16
1
 n2 ; n is even
2
.
an  
1
 n 1 ; n is odd
 2 2
 an 
, where
a) Show that the ratio test fails.
b) What happens in the root test?
47.
Remember
from
Activity
13,
Problem
Fn   1,1,2,3,5,8,13,21,34,55,89,  ,
#15,
the
Fibonacci

determine if the series

n 1
sequence
Fn  ,
1
converges or
Fn
diverges.
48. Show that the series 1 
1 1 1 1 1 1 1 1
       
2 3 4 5 6 7 8 9
 1 1 1 1 1 1 1 1
S3 n   1              
 2 3  4 5 6  7 8 9
  1
n 1
 1 1 1 1 1 1 1 1
 1              
 2 3  4 5 6  7 8 9
  1
n 1
converges.
1
1 
 1

 

 3n  2 3n  1 3n 
 27 n 2  18n  2 


 3n  3n  1 3n  2  
which is a partial sum of an alternating series. Since any partial sum of the original series
differs from a multiple of 3 partial sum by at most 2 terms which go to zero as n tends to
infinity, if S 3n is convergent then so is any partial sum.

See if you can show that
 1
n 1
n 1
27n 2  18n  2
converges.
3n  3n  1 3n  2 
49. Show that the series 1 
1 2 1 1 2 1 1 2 1
        
2 3 4 5 6 7 8 9 10
converges and find its sum.
1 2 1 1 2
1
1
2
     

 .
2 3 4 5 6
3n  2 3n  1 3n
3 
1 1
1
3 3
So S3n        1     , but this means that
3n 
2 3
3n
3 6
S3n  1 
 1
S3 n   1  
 2
1
1 1
  1  
n
2 3

1
1
, and so S3n 

3n
n 1

1
. Find lim S3n .
n
3n


1
1 
 1



.
4
n

1
4
n

3
2
n

2


n 0
a) Determine if the series is convergent or divergent using comparison.
1
1
1
 4n  3 2n  2    4n  1 2n  2    4n  1 4n  3
{Hint:



4n  1 4n  3 2n  2
 4n  1 4n  3 2n  2 
50. Consider the series
8n2  14n  6  8n 2  10n  2  16n2  16n  3
}

 4n  1 4n  3 2n  2
b) Determine if the series is convergent or divergent using the limit of its partial sums.

N
{
n 0
1
1 
 1




 4n  1 4n  3 2n  2 

N

n 0

k 1
1
1
1
1
1
1
n 0
1
1
1  1
 1





 4n  1 4 n  2 4 n  3 4 n  4  2
4 N 4

 4n  1  4n  2  4n  3  4n  4  4n  2  4n  4 
N
 1
k
k 1
1

2
2 N 2

k 1
 1
k
k 1
.}

N
1 
 1



2n  1 2n  2 
n 0 
51. Is the alternating series
1 1 1 1 1 1 1 1 1 1 1 1
           
3 3 2 2 5 5 4 4 7 7 6 6
convergent or
divergent?
{Notice that you can’t use the Alternating Series Test since the an ’s aren’t decreasing.}

52. Determine convergence or divergence of


 sin  2n   sin  2n  1  .
1
1
n 1
cos  xn* 
 1 
 1 
{Hint: The Mean Value Theorem says that sin    sin 
for some

 2n 
 2n  1  2n  2n  1
1
1
 xn*  .}
2n  1
2n
53. Consider the series
1
1   1
1
1   1
1
1 






1 



3
2
5
7
4
9
11
6

 
 

1
1 
 1




4
n

3
4
n

1
2
n



n 1 1
This series is a rearrangement of the convergent alternating series
.
 1
n
n 1
1
1
1
1
1
1





4n  3
4n  1
2n
3
1
2n
2 n 1
2 n 1
4n
4n
.




1  1
1
1



 1 
2 n
3
1
 2 n
1

1



4n
4n
2
n
This means that Sn for n a multiple of 3, satisfies Sn 
i 1

rearrangement of the convergent series
 1
n 1
n 1
1
do?
n
1
n
.
So what does the
54. Adding grouping symbols to a convergent series doesn’t change its convergence or its sum.

We know from Problem #2 that ln 2  1  12  13  14  15 


 1
n 1
a) By inserting grouping symbols into the series 1  12  13  14  15 
series
1
1
1



1 2 3  4 5  6



n 1
1
1
1
b) and also the sum of the series 1 



23 45 67
n 1
succinctly
 a b  S b   S b
i i
n n
i
i 1
i 1
1
.
n
, find the sum of the
1
2n  2n  1

1
55. Given a1 , , an and b1 , , bn , let Sk  a1  a2   ak .
Show that a1b1  a2b2  anbn  S1  b1  b2   S2 b2  b3  
n
n 1

n 1
1
2n  2n  1
 Sn1 bn1  bn   Snbn or more
 bi  , which is commonly referred to as the
i 1
summation by parts formula analogous to the integration by parts formula:
n 1
 a b  S b   S b
n
i
i
i 1 dv u
n
v
n
u
i
i 1
v
i 1
 bi  .
du
For n  1 , the left side of the formula is a1b1 and the right side of the formula is a1b1 .
Now suppose the formula is true for n, so we have
a1b1  a2b2  anbn  S1  b1  b2   S2 b2  b3    Sn1 bn1  bn   Snbn ,
an 1bn 1 on both sides to get
a1b1  a2b2 
 anbn  an1bn1  S1  b1  b2   S2 b2  b3  
 S1  b1  b2   S2  b2  b3  
and
add
 Sn1 bn1  bn   Snbn  an1bn1
 Sn1  bn1  bn   Sn  bn  bn1   an1bn1  Snbn1
See if you can show that an1bn1  Snbn1  Sn1bn1 .
let’s
56. Suppose that bn  is a non-negative decreasing sequence and m 
Show that b1m  a1b1  a2b2 
m  b1  b2  
i
for all n.
i 1
 anbn  b1M for all n. From problem #55, we know that
 anbn  S1  b1  b2   S2  b2  b3  
a1b1  a2b2 
a  M
n
 Sn1 bn1  bn   Snbn
 m  bn1  bn   mbn  a1b1  a2b2 
 anbn  M b1  b2  
 M bn1  bn   Mbn
  bn1  bn   bn   a1b1  a2b2 
 anbn  M  b1  b2  
  bn1  bn   bn  .
m  b1  b2  
So simplify the contents of the brackets to get the result.
57. Suppose that bn  is a non-negative decreasing sequence and m 
Show that for each k, bk m  ak bk  ak 1bk 1 
#48, we know that
a  M
n
i
for all n.
i 1
 anbn  bk M for all n  k . From problem
ak bk  ak 1bk 1   anbn  Sk  bk  bk 1   Sk 1 bk 1  bk 2  
Finish it off like problem #49.
 Sn1 bn1  bn   Snbn

58. (Dirichlet’s Convergence Test) Suppose that the partial sums of the series
a
n
are
n 1
a  M
n
bounded, i.e. m 
k
for all n, and bn  is a decreasing sequence with lim bn  0 .
n
k 1

Show that the series
 a b converges.
n n
n 1
To show that a series is convergent, it is enough to show that the partial sums of the series
bunch up, i.e.  Sn  Sm  can be made arbitrarily small for n  m sufficiently large.
 Sn  Sm    aibi ,
n
and
from
the
previous
problem,
we
know
that
i  m 1
ab  b
n
bm1m 
i i
M , and since lim bn  0 , we can make
m 1
i  m 1
choosing n and m sufficiently large.
n
 Sn  Sm 
arbitrarily small by

Apply Dirichlet’s Test to the series

n 1
{Hint: Let an  sin n , bn 
From
you
know
cosk  isin k ,
n
e
 eni 
0i


1
, and consider the sum
n
trigonometry,
1  e i  e 2 i  e 3i 
sin n
and
n
n 1
k
cos n
n
 sin n .
n 1
that
ei  cos  i sin  ,
so
n
k 0
but also from the geometric series
k 0
1  e 
formula that 1  e  e  e   e 
,
1  ei
1  cos  n  1  i sin  n  1 1  cos1  i sin1
so 1  ei  e2i  e3i   eni  

1  cos1  i sin1
1  cos1  i sin1
n 1 i
i
2i
3i
ni
1  cos1  cos  n  1  cos1cos  n  1  sin1sin  n  1 

2  2cos1
sin1  sin1cos  n  1  sin  n  1  cos1sin  n  1 
i 
.
2  2cos1
So putting things together, we get that

n
k 0
n
1  cos1  cos  n  1  cos1cos  n  1  sin1sin  n  1 
cosk  
2  2cos1
sin1  sin1cos  n  1  sin  n  1  cos1sin  n  1 
.
sin k  
2

2cos1
k 0


n
k 0
5
and
cosk 
2  2cos1
sink  1  cos1 .}
n
k 0
2
and
So
we
that
have
that

59. (Abel’s Convergence Test) Suppose that the series
a
converges and the sequence bn 
n
n 1
is either monotone decreasing or monotone increasing and is bounded, i.e. m  bn  M for

all n. Show that the series
a b
n n
converges.
n 1
{Hint: If bn  is increasing, then it must have a limit, b, and the new sequence b  bn  is a

decreasing sequence whose limit is 0. Since
a
n
converges, its partial sums must
n 1

be bounded, so Dirichlet’s Test applied to the series
 a b  b  implies that it
n
n
n 1

converges,
and
the
series
 ba is
convergent,
n
so
n 1



 a b  ba   a b  b  .}
n n
n 1
n
n 1
n
n
n 1

Use Abel’s Test so show that if

n 1
cn
converges, then so does
np


n 1
cn
for q  p .
nq
{Hint: Let an 
Let’s establish a connection between the Ratio and Root tests.
a
Suppose that an  0 for all n and lim n1  L
n a
n
for
cn
1
and bn  q  p .}
p
n
n
0  L  .
Then
lim ln  an1   ln  an   ln L , and for   0 , ln L    ln  an1   ln  an   ln L   for n  N . So
n
we get that
ln L    ln  aN 1   ln  aN   ln L  
ln L    ln  aN  2   ln  aN 1   ln L  
ln L    ln  aN  m   ln  aN  m1   ln L  
for m  1 .
So adding the inequalities, we get that m  ln L     ln  aN m   ln  aN   m  ln L    , which
1
1
leads to ln L    ln  aN m   ln  aN   ln L   ,
m
m
1
1
1
ln  aN     ln  aN m   ln L  ln  aN    . Since   0 is arbitrary, it must be
m
m
m
1
1
1
m N
the case that lim ln  aN m   ln L . Since ln  aN m  
, we also
ln  aN m  
m m
m
m N
m
1
get that lim
ln  aN m   ln L , and hence that lim n an  L . If L  0 , then for   0 ,
m m  N
n
a
0  n1   for n  N . So we get that
an
a
0  N 1  
aN
and to
0
aN 2

aN 1
, and
aN m

a N  m1
a
a
a
Multiplying the inequalities, we get 0  N 1  N 2   N m   m , which implies that
aN aN 1
aN m1
a
0  N  m   m . So 0  m aN m  m aN  , and again since   0 is arbitrary, and that
aN
0
m N
1

 m
n a  0  L . In the case of L   , then we have
m

N
a

a
 N m   N m   , we get lim
n
n


a
a
a
a
that for any B  0 , n1  B for n  N . As before, N 1  N 2   N m  B m , and hence,
an
aN aN 1
aN m1
1
m
m N
m


it
aN m  m aN B . Since B  0 is arbitrary, and  aN m    aN m  


a
must be the case that lim n an    L . This means that if lim n1  L , then lim n an  L
n
n
n a
n
and therefore, a conclusion about a series based on the ratio test will always yield the same
conclusion from the root test. However, the root test may yield a result while the ratio test
fails to yield a result.
aN  m
 B m . So
aN
m
1
m
1
m N
Use the previous result to find the following limits:
n
 n 1 
n  1
an 1

60. lim 
{Hint:
Let
and
look
at
the
ratio
.}
a

1 
n
n
n 
an
n!
  n! 
n
61. lim
n
n!
n
{Hint: Let an 
 n  1 n  2 
62. lim 
n
n
{Hint:
Let
an 
 n  n 

 n  1
n 1
(Kummer’s
n
 n  1 n  2  n  n 
an1   n  1  1   n  1  2 

an
2  2n  1 nn
1
a
n!
and look at the ratio n 1 .}
n
n
an
and
look
at
nn
  n  1  n  1   n  1  n    n  1  n  1 
 n  1
n 1
the

ratio
nn
 n  1  n  n 
.}
Test
and
Raabe’s
Test)
Suppose
that
an , bn  0 ,
b   ,
and
n
1 a
1 
lim   n 
an converges, and if L  0 , then
  L . If L  0 , then
n  b
a
b
n 1 
 n n1
diverges.
Case I: L  0 .
1 an
1
L
an an1 Lan1
Then for n  N ,


 , which implies that


bn an1 bn1 2
bn bn1
2

an1 
n
or
2  an an1 
 
 , but this means that
L  bn bn1 

M
2
SM 
ak 1 
L
k N

so is
a
a .
n
n 1
 ak ak 1  2  aN aM 1  2 aN

. So
 
 
 
b
b
L
b
b
L
b
k
k

1
N
M

1
N




k N

M

a
k 1
k N
is convergent, and
Case II: L  0 .
1 an
1
a


 0 , which implies that an1  n bn1 , so we get that
bn an1 bn1
bn
Then for n  N ,
aN 1 
aN
bN 1
bN
aN  2 
aN 1
bN  2
bN 1
aN
bN 1
bN
a

 bN  2  N bN  2
bN 1
bN
aN
bN  2
aN 2
bN
a
a N 3 
bN 3 
 bN 3  N bN  2
bN  2
bN  2
bN
aN
bM 1
aM 1
bN
a
aM 
bM 
 bM  N bM
bM 1
bM 1
bN
M 1
M 1
a 
So
ak 1   N 
bk 1 , therefore
b
 N  k N
k N



a
k 1

and
k N
a
n
are divergent.
n 1
Raabe’s Test is a special case of Kummer’s Test that is easier to apply:
 a

Suppose that an  0 and lim  n n   n  1   L . If L  0 , then
n
 an1


a
n
converges, and if
n 1

L  0 , then
a
n
diverges.
n 1
63. Show that Raabe’s Test follows from Kummer’s Test.
{Hint: Let bn 

64. Use Raabe’s Test on the series
246
35 7
n 1
 2n 
 2n  1
.
1
in Kummer’s Test.}
n

65. Use Raabe’s Test on the series
 2n  .
 2n  3 
246
57
n 1

66. Use Raabe’s Test on the series
 2 n!
n
2
  2n  1! .
n 1
67. a) Let’s estimate the error for a partial sum of a series for which the ratio test applies.

n

a
Consider the series
an 
ak 
ak . If k 1    1 for k  n , then show that
ak
n 1
k 1
k  n 1
  
Sn
Rn  an 

.
1
Rn

Rn 

a   a
k
k  n 1
k
 an1  an 2  an3 
k  n 1
  an   an1   an 2 
{Hint:
  an   2 an   3 an 
.}

 an

i
i 1

th
b) Estimate the error, from part a), in using the 10 partial sum of the series

n 1
{Hint:
n
.
2n
ak 1  1  1 11
 1    
for k  10 .}
ak
 k  2 20
68. a) Let’s estimate the error for a partial sum of a series for which the root test applies.

Consider the series

n
a  a   a .
n
n 1
k
k 1
k

k
ak    1 for k  n , then show that
k  n 1
Sn
Rn 
If
Rn
n1
1
.

Rn 

a   a
k
k  n 1
k
 an1  an 2  an3 
k  n 1
  n1   n 2   n3 
{Hint:
.}



i
i  n 1

n
th
b) Estimate the error, from part a), in using the 10 partial sum of the series
n!
n
.
n 1
{Hint:
69. Determine if the non-alternating series
1
1


2 1  3 1
k



  1 n  1 .
1
n
n2
1
1  1k 

by grouping the terms in pairs.
1
1


70. Do the same for the series
2  1 3  1
ak 
  1
n2
k

1
for k  1 .}
2
1
n
n 1
converges
a b a b
    for a, b  0 .
1 2 3 4
a) Determine convergence or divergence if a  b .
a b a b
{Hint:    
1 2 3 4
71. Consider the alternating series

a
 1
n 1
n 1
1
.}
n
b) Determine convergence or divergence if a  b . Suppose that a  b , then


2 a  b n  b
a b a b
b 
a b a b
 a
.
         
 

1 2 3 4
1
2
3
4
2
n

1
2
n
2
n
2
n

1



 

 n1
n 1 
Finish it from here.


72. Try the Ratio Test on the series
2
 1n  n 



. Determine convergence or divergence with
n 1
another method.

73. Try the Root Test on the series

n 1
with another method.
n
 5   1n 

 . Determine convergence or divergence
2


74. There is an extension of the Ratio Test called the Second Ratio Test. Here’s what it says:

a
Second Ratio Test: Let
an be a series with positive terms, and suppose that lim 2 n and
n a
n
n 1

a2 n1
n a
n
lim
both
exist.

a
a 
L  max lim 2 n ,lim 2 n1 
 n an n an 
Let
and

a
a 
l  min lim 2 n ,lim 2 n1  . Then
 n an n an 
1
i) If L  , then
2

a
n
converges.
n 1


1
, then
an diverges.
2
n 1
1
iii) If l   L , then the test fails.
2
ii) If l 

a) If you attempt the Ratio Test on the series
n
1
2
, you don’t get a result. Use the Second
n 1
Ratio Test on it.

b) If you attempt the Ratio Test on the series
n
n 1
1
1
2
, you don’t get a result. Use the
Second Ratio Test on it.

c) If you attempt the Ratio Test on the series

n 1
Use the Second Ratio Test on it.
1  3  5  2n  1
, you don’t get a result.
2n  n  1!

75. Consider the alternating series
  1
1
n
n   1
n2

a)
  1
n
n2
.

1
n   1
b) Show that  1
n
n

  1
n
un . Show that un  0 for all n  2 , and lim un  0 .
n
n2
1
n
n   1
n
 1

1
n   1

1
n
n2

n

 1
n
n   1

n

n2
n
, and hence that
n


n
  1

1

.

n
 n
n   1 n 
 convergent

divergent


c) Is the alternating series convergent or divergent?
76. a) For p  1 , find a simple formula for the value of

f  p 
n

 1
1
1
1



p
p
p
2
3
4

1
1
1
1 p  p  p 
2
3
4
1
p
n1

n1
{Hint: 1 
1
n1
np
1
1
1
 p p
p
2
3
4
b) Find lim f  p  .
p1
c) What’s wrong with f  p  for p  1 ?
1
1
1

 1  p  p  p 
3
4
 2
.
1
1

 1
  2 p  p  p 
4
6

2

 .}

77. Let’s use the Alternating Series Test Estimation to show that a rearrangement of the
Alternating Harmonic Series converges to a different limit. We’ll start with the

1 1 1
n1 1
Alternating Harmonic Series in standard order:
 1    .
 1
n
2 3 4
n1

a) Show that the sum of the Alternating Harmonic Series in standard order is less than
5
.
6
Consider the rearrangement of the Alternating Harmonic Series:
1 1 1 1 1 1 1 1
.
1       
3 2 5 7 4 9 11 6
Adding grouping symbols results in another alternating series:
4 1 12 1 20 1
 1 1 1 1 1 1 1  1
1









   
  .






3 2 35 4 99 6
 3  2  5 7  4  9 11  6
1

 n ; n even
n 1
This new alternating series can be written as
.
 1 an , with an  
4
n

n 1
; n odd
 4n 2  1
b) The new alternating series satisfies the conditions of the Alternating Series Test and
5
hence converges. Show that its sum is greater than , and hence different than the
6
original sum.
