Download Name: Improper Integrals Done Properly

Math 2414 Activity 15(86 parts)(3 per person)(Due by August 18)
1. Let f be a positive, continuous, decreasing function for x  1 , and suppose that an  f  n  .

If the series

an converges to s, and we call
n 1
a ,
n
the nth partial sum of the series sn 
i
i 1

then the remainder rn 
a
satisfies the
i
i  n 1

double inequality 0  rn 
 f  x  dx .
f  x
Since
n
rn  s  sn ,
we
also
get
that
0  s  sn 
 f  x dx
which
implies
n

sn  s  sn 
n n 1
1 2

that


 f  x dx . This is equivalent to a  a  a   f  x dx .
n
n
i
i 1
n
n
i
n 1
i 1
n
Use this last inequality to approximate the sum of the following convergent series using the
indicated number of terms, and give an estimate of the maximum error.

a)

n 1
1
n4

(six terms)
b)

n 1
1
 
e
n
(four terms)

2. Use the result of problem 1, rn 

f  x  dx , to find the smallest n so that rn  .001 for the
n
following convergent series:

a)

n 1
1
n4

b)

n 1
1
n 1
2
(4 parts)

3. Problem 1 gave us an upper bound for the remainder, rn 
 a , now we’ll find a lower
i
i  n 1
bound.
From the figure, you can see that

rn 

f  x  dx . Use this result to find
n 1
what the minimum error is when you use
the truncated series in the two parts of
problem 2.
4. Use the Integral Test to determine
convergence or divergence for the
following series:


b)

c)

f)

a)
n 0

n 0

n 0

n 1
n
n 3
2
n 1
3
2n 2  1
n2
en
1
n
e
n2


g)


d)
n 1

e)
n 1

1
n 1   ln n 
n 1

1
h)
 n ln n ln n 1
tan 1 n
1  n2
2

 1
ln 1  
 n
2
n 3
5. a) Suppose that f is a positive, continuous decreasing function for x  1 , g is a continuously
differentiable function defined for x  1 and lim g  x    . If g   x  f  x  is positive and
x 
eventually decreasing, then show that the two series


n 1
n 1
 f  n  and  g n f  g  n 
either both converge or both diverge.


{Hint: From the integral test, we have that
 f  n  and  f  x  dx either both converge
n 1
1


or both diverge, and that
 g n f  g  n  and  g x  f  g  x  dx either both
n 1
converge or diverge. But
1


1
g 1
 f  g  x g x  dx   f u  du , so take it away.}
(9 parts)

b) Use part a) to determine convergence or divergence of the series
  ln n  n .
1
2
n2
{Hint: :Let f  x  

c) Use part a) to determine convergence or divergence of the series

n4
{Hint: :Let f  x  
1
and g  x   ln x .}
x2
1
.
n ln  ln n 
1
and g  x   ln x .}
ln x

6. An infinite product
1  a   1  a 1  a 1  a 
n
1
2
converges if only finitely many
3
n 1
of its factors are zero and the sequence of partial products(omitting the zero factors),
n
pn 
1  a  , converges to a non-zero number.
i
Assuming that all of its factors are
i 1
positive, its convergence can also be determined by considering the infinite series




ln 1  an  . If
n 1

ln 1  an  converges to s, then
n 1

1  an   es .
n 1

a) Show that
1   diverges.
1
n
n 1


{Hint: Examine
n 1

n
1  1i  
i 1
ln 1  1n 
 1
ln 1   , with lim
 1 , or use the fact that pn  n  1, since
1
n
 n
n
2 3 4
  
1 2 3


b) Show that
1  
1
n2
n 1
.}
n
 converges, and find its value.
n2

{Hint: ln 1 
1
n2
  ln  n  1  ln  n   ln  n  1  ln n , so
 ln 1     ln i  1  ln i   ln i  1  ln i  , or
n
n
1
i2
i 2
n

i 2
1

1  2  
 i 
i 2
n

n2
 i  1   i  1    3 4
 i   i     2  3 

 

 n  1   1 2

   
 n   2 3
n 1
 n  1 

}
  , pn 
2n
 n 
(4 parts)


  1n 
c) Show that
1 
 converges, and find its value.


n

n2 
n
  1i   1  1  1  1    1n 
pn 
1 
  1  1  1  1   1 


  2  3  4  5  

i
n
i 2 



  1  1    1  1   
1   1 
 1  1    1  1     1 
   1    ; n odd
2
3
4
5
n

1







  n 


 

 1  1 1  1   1  1 1  1    1  1    1  1    1  1  ; n even
  

  
 
 

  2 
 3    4  5    n  2   n  1    n 
 3 2   5 4   n n  1
  2  3   4  5   n  1  n  ; n odd


 


  3  2   5  4   n  1  n  2   1  1  ; n even
  2 3   4 5   n  2 n  1   n 



d) Show that

n 1
n
pn 

i 1

1 
1 

i
i

2




n



1
1 
 converges, and find its value.
n
n

2





i 1
 i 1


 i 
  n  1 
n

n
i 1
 i i  2
1 



i
i

2
i
i

2






i 1
 i  2    1  2  3 
2 3 4

i 1
e) For the infinite product

n
n 1  2 3 4
   
n  3 4 5


i 1
  i  12 


i
i

2




n 1 

n2

1  x   1  x  1  x 1  x  1  x 
n
2
i 1
n
1  x 2 , use induction to show that the partial product
n 0
pn 

 i 2  2i  1 


i
i

2




2
 n  2

i
n
2
4
2
i 0
n
n 1
1  x2

, for x  1.
1 x
1  x2
{Hint: For n  0 , the left side is 1  x , and the right side is
 1  x . Assuming that the
1 x
formula is true for n, pn1 
2n 1


n 1
1 x
 1  x2 
1 x
 
1  x2
n 1
1 x
2

}
(3 parts)
f) If x  1 , show that the partial product
 1  x   1  x  1  x 1  x  1  x   2
n
pn 
2i
2
2n
4
n1
i 0

g) Under what conditions on x will the infinite product
converge to? {Watch out for x  1.}
1  x  converge, and what will it
2n
n 0

h) Use the previous part to find the value of the infinite product
 1  2  .
2n
n 0


4
1

converges, and find its value.

2
n 1 
  2n  1 
3
5
4
7
 4
 4  4
 4  4 
{Hint: p1  1     , p2  1    1     , p3  1    1    1     , so
1
3
5
 1
 1  9
 1   9   25 
it looks like the numerators are odd numbers starting at 3, and the denominators are
odd numbers starting at 1. As a formula, it looks like
n

4 
2n  1
pn 


. To prove it by induction, it’s true for n  1 ,
1 

2
2
n

1
2
i

1




i 1 

assuming it’s true for n, then


2n  1 
4
2n  1 
4


pn1  
 1 



1




2n  1   2  n  1  1 2 
2n  1   2n  12 
 
 
2

2n  1   2n  1
4
 2n  1 2n  3 2n  1  .}




2
2
2
2n  1   2n  1  2n  1 
 2n  1 2n  1

i) Show that


(4 parts)
j) The procedure begins with a circle of radius 1. An equilateral triangle is inscribed, then a
circle is inscribed, then a square is inscribed, then another circle, then a regular pentagon,
another circle, then a regular hexagon, another circle, …. The question is what happens? Do
the closed figures continue to diminish until they vanish into a point? Do they converge to a
closed curve, and if so what is it?
Let’s look at the sequence of the polygonal radii:
cos 3

3

1
cos 3
(0 parts)
cos 3 ,  cos 3  cos 4 
 cos 3  cos 4 



4
cos 3
cos 3 ,  cos 3  cos 4  ,  cos 3  cos 4  cos 5 
 cos 3  cos 4  cos 5 
 cos 3  cos 4 

5
(0 parts)
So the sequence of polygonal radii is given by
cos 3 ,  cos 3  cos 4  ,  cos 3  cos 4  cos 5  ,  cos 3  cos 4  cos 5   cos n  , , where n is the
number of sides of the regular polygon.
If lim  cos 3  cos 4  cos 5   cos n  exists, then the limiting curve will be a circle with this
n

limit as its radius.
So we need to investigate the infinite product

n 3

equivalently,

n 3
 
cos   or
n
 
   
1  1  cos     . This infinite product converges if the associated
 n  
 
 
ln 1  x 
   
ln 1  1  cos     converges. Since lim
 1, we only need to check the
x0
n
x






n 3

1  cos  n 

1  cos  n   . Compute lim
convergence of the series
to determine if the
1
n 

series


n 3

infinite product

n 3
n2
 
cos   converges to the radius of a limiting circle.
n




2 
2 
ln 1 
k) For the infinite product
1 
 , the related series is
 . Use the
n
n

1
n
n

1







n2
n2 



2

n  n  2
  n  2  n  1 
2 
ln 1 

ln

ln
fact that




 to find the sum of
n
n

1
n
n

1
n
n

1







 n2 
 n2 

n2
the series, and exponentiate to find the value of the infinite product. Or find it directly.






7. Cauchy Condensation Test. Consider the series

an where an  is a decreasing sequence
n 1
of positive numbers that converges to zero. If sn  a1  a2 
s2n  a1  a2  a3  a4   a2n
 a1   a2  a3    a4  a5  a6  a7    a8  a9 
 a1  2a2  4a4 
 2n1 a2n 1  a2n
 a1  2a2  4a4 
 2n1 a2n 1  2n a2n

So if the series
2 a
  a2n 1  a2n 1 1 
 a2n 1   a2n

n
n 1
 a15  
 an , then
2n
converges then so does

an .
n 1
On the other hand,
(2 parts)
2s2n  a1  a1  a2  a2 
 a2n  a2n
 a1   a1  a2    a2  a3  a3  a4    a4  a5 
 a1  2a2  4a4 
 2n a2n  a2n
 a1  2a2  4a4 
 2n a2n
This means s2n 
1
2
 a  2a
1
2
 4a4 
  a2n 1  a2n 1 1 
 a8  
 a2 n   a2 n

2 a
 2 a2n  , so if
n
n
diverges then so does
2n
n 1


an .
n 1

For the series

 
n 1
1
,
n
n 1

2 a2n 
n

n 1
1
2  n
2
n

1, which diverges.
n 1
Try the Cauchy Condensation Test on the following series:

a)

n2

1
n ln n
b)

n 3
1
n  ln n   ln  ln n  
8. Find the values of c for which the following series are convergent:

a)
 n ln n

1
b)
c
n2
 nln nln ln n
1
c
n 3
9. This problem shows that you have to be careful in applying the integral test:


b) Determine the convergence or divergence of the infinite series

a) Determine the convergence or divergence of the improper integral
sin 2  x  dx .
1

sin 2  n  .
n 1
c) Explain the result in terms of the Integral Test.

10. In Kummer’s Acceleration Method, one writes



 a  b   a  b  , where  a
n
n 1
n
n
n 1
n
n
n 1
n 1

is the series whose sum we want to approximate and where
b
n
is a series whose sum
n 1
can be computed exactly and whose terms are so close to the an ’s that the series
(7 parts)


 a  b  converges much faster than  a
n
n
n
and so can be approximated with fewer terms.
n 1
n 1

Apply this method to find the sum of each of the following series
 a with an
n
n 1

error of at most .01 by using the suggested series
b .
Compare the number of terms
n
n 1
required with Kummer to the number of terms required with the integral estimate.
a) an 
1
1
, bn 
2
n
n  n  1
c) an 
1
1
, bn 
3
n
n  n  1 n  2 
b) an 
n
1
, bn 
n 1
n  n  1
3

11. Find all the nonnegative values of b for which the series
b
ln n
converges.
n2
{Hint: For b  0 , bln n   eln b 
ln n
 nln b 
1
.
ln  b1 
n
12. Although the Harmonic Series diverges to infinity, it does so very slowly. You can show
n
1
 1  ln n .
that ln  n  1 
i
i 1
1  ln n


n
i 1
1
i
ln  n  1
(4 parts)
Scientists believe that the universe is 13 billion years old. Suppose that you began adding
the terms of the Harmonic Series 13 billion years ago and that you added one term per
second.
Assuming a 365 day year, you would have added about
60  60  24  365  13 billion   5,329,584,000,000,000 terms. Determine the approximate
total of these 5,329,584,000,000,000 terms using the average of the upper and lower
2
n
1 1  ln  n  n 
bounds
.

i
2
i 1

13. The problem is to cross a desert by jeep. There are no sources of fuel in the desert, and you
cannot carry enough fuel in a jeep in order to make the crossing in one go. You do not have
time to establish fuel dumps, but you do have a large supply of jeeps and drivers, none of
which you want to lose. How can you get across the desert using the minimum amount of
fuel?
Here’s a solution to the problem:
We will measure the distance a jeep can travel in terms of a tankful of fuel; one jeep by
itself can travel a distance of one tankful. If two jeeps set out together, they should travel
for 13 of a tankful, then Jeep 2 transfers 13 of its tankful to Jeep 1, and returns to base on the
remaining 13 tankful. Jeep 1 is then able to travel a total of 1  13 tankfuls. With three jeeps,
they should stop after traveling 15 of a tankful, then transfer 15 of a tankful from Jeep 3 into
each of Jeeps 1 and 2, which are now full. Jeep 3 now has 52 of a tankful, Jeeps 1 and 2
now proceed as before with Jeep 2 returning with an empty tank to Jeep 3. Between them,
they have enough fuel to get back to the base. Meanwhile, Jeep 1 has traveled a total of
1 13  15 tankfuls. The same reasoning shows that with four jeeps you can achieve a
distance of 1  13  15  71 tankfuls, and with n jeeps you can get a jeep across a desert that is
1
1  13  15  71   2 n
1 tankfuls wide.
1  13  15  17 
 2 n11  1  12  13  14 
 1  12  13  14 
 2 n11  21n    12  14  16  81 
 21n 
 2 n11  21n   12 1  12  13  14 
 1n 
But we have from the inequality in the previous problem that
ln  n  1  1  12  13  14   1n  1  ln n and ln  2n  1  1  12  13  14   21n  1  ln  2n  .
Putting them together, we get that
ln  2n  1  12 ln n  12  1  13  15  71   2 n11  1  ln  2n   12 ln  n  1
Or more simply
 2n  1  1
 2n 
1
1
1
1
ln 


1






1

ln
3
5
7
2 n1
 2

.
 n 
 n 1 
If a jeep can travel 300 miles on one tankful, and the desert is 900 miles wide, estimate the
number of jeeps required to cross the desert using the average of the upper and lower bounds
(1 part)
 4n 2  2n 
 ln 

n2  n 

1
.
 2 n1 
2
1
2
1  13  15  17 
14. For what values of a do the following series converge?


a)
n 1

1 
 a



n2 n4
b)

n2
2a 
 1



 n 1 n 1
15. Determine convergence or divergence for the following series:



1
d)
tan  
 n
a)
n 1

1
sin  
n
b)

n 1

1
sin 2  
n
c)
n 1

e)
n 1


n 1
1
cos 2  
n

 1 
n 1  cos   
 n 

2
x
x2
{Hint: Since you know 1   cos x  1 , you can show 0  1  cos x  . }
2
2

f)

n2
1
g)
2
ln n 

 ln ln n
j)

n2

1
n 3

{Hint: For n  ee ,  ln n 
ln n
h)

n 1
n
k)
  e2 
ln n
.}

1
i)
1 1n
n

n 1

1
ln n 
ln n
ln n

n2
1
ln n 
ln n
Hint:  ln n 
ln n
n
 n  1
n
n n1
ln ln n 

16. We know that the sum of the reciprocals of the counting numbers,  1n  11  12  13 
n 1
diverges to infinity. What if we only add the reciprocals of those counting numbers that
don’t have the digit 6?
There are 8 one-digit counting numbers without a 6:
1,2,3,4,5,7,8,9
 11  12  13  14  15  71  81  91   8 1  8
(14 parts)
There are 72 two-digit counting numbers without a 6:
10,11,12,13,14,15,17,…,99
1
 10  111  121   991   72  101  8  109
There are 648 three-digit counting numbers without a 6:
100,101,102,103,104,105,107,…,999
 1001  1011  1021 

So

n 1
no 6 digit
1
n
1
 999
  648  1001  8   109 
 8  8  109  8   109   8   109  
2
3
2
.
a) Use the previous discussion to find a whole number that bounds the sum of the reciprocals
of whole numbers without a digit of 6.
Here are the missing digit sums:
Missing Digit
0
1
2
3
4
5
6
7
8
9
Sum (to 5 places)
23.10344
16.17696
19.25735
20.56987
21.32746
21.83460
22.20559
22.49347
22.72636
22.92067
b) Find a whole number bound on the sum of the reciprocals of the counting numbers
without any even digits.
There are 5 one-digit counting numbers without any even digits:
1,3,5,7,9
 11  13  15  17  19   2
There are 25 two-digit counting numbers without any even digits:
11,13,15,17,19,31,33,…,99
1
 111  131  151   991   25  111  52 
10
There are 125 three-digit counting numbers without any even digits:
111,113,115,117,119,131,133,…,999
(1 part)
 1111  1131  1151 


So
n 1
no even digit
1
n
1
 999
  125  1111  53 
 2  52 

1
100
1
1
1
 53 
 54 

10
100
1000

n
.
n 1
5
5
 2  5 n1
n 1
n  2 10
n  2 10
2
c) Determine convergence or divergence of the sum of the reciprocals of the counting
numbers that have a ones-digit of 6?
There is 1 one-digit counting number with a one’s digit of 6:
6
1
1
6  6
There are 9 two-digit counting numbers with a one’s digit of 6:
16,26,36,46,56,66,76,86,96
 161  261  361   961   9  961  9  10214
There are 90 three-digit counting numbers with a one’s digit of 6:
106,116,126,136,146,156,166,…,996
1
 1061  1161  1261   9961   90  9961  9 10  3
10  4


So
n 1
one's digit of 6
1
n
  9
1
6
1
96
 9  10 
1
996


   91010n 4 .
1
6
n2
n2
d) Determine convergence or divergence of the sum of the reciprocals of the counting
numbers that don’t have a ones-digit of 6?

17. Suppose that

an converges and an  0 . Determine the convergence or divergence of
n 1


n 1
 sin  an  
1 
 using the Extended Limit Comparison Test with
a
n




an .
n 1
(4 parts)
18. Use the Extended Limit Comparison Test to determine the convergence or divergence of
the following series:


ln n
ln n
a)
b)
3
n
ne n
n 1
n 1



c)

n2

1
ln n

d)
n 1
ln n
n
19. If an  0 and lim  nan   L , where L is either a positive number or  , then what can you
n

say about
a ?
n
n 1

20. Suppose that

 a ,  b , and  c
n
n
n 1

converges and

n 1
c
n
n

are series with positive terms, and that
n 1
b
n
n 1
diverges.
n 1
a
b
a) If n1  n1 for all n  N , then show that
an
bn
{Hint:
aN 1 bN 1
,

aN
bN
so
a
aN 1   N
 bN

a
n
converges.
n 1
aN 1 bN 1

aN
bN

 bN 1 .

aN  2 bN  2

aN 1 bN 1
a
aN 1 aN 2 bN 1 bN 2
which means that aN  2   N



aN aN 1 bN bN 1
 bN
a
c
b) If n1  n1 for all n  N , then show that
an
cn
so
multiplying
gives

 bN  2 , so keep going.}


a
n
diverges.
n 1
(7 parts)
ln  nan 
 L (possibly  ).
n ln  ln n 
21. (Loglog Test) Suppose that an  0 for n  1 and lim
a) What happens if L  1?
ln  nan 
 1 which implies that ln  nan    ln  ln n  ,
ln  ln n 
ln n
and by exponentiating, you get that nan  ln n or that an 
. What do you
n

ln n
know about the series
?}
n
n 1
b) What happens if L  1?
ln  nan 
{Hint: This means that eventually
  p , where p  1 which implies that
ln  ln n 
1
or that
ln  nan    p ln  ln n  , and by exponentiating, you get that nan 
p
 ln n 
{Hint: This means that eventually

an 

1
n  ln n 
p
. What do you know about the series
 n  ln n 
1
p
for p  1 ?}
n2
22. Consider the sequence of partial sums of the harmonic series, sn  1  12  13  14 

Determine if the series

n 1
1
is convergent or divergent.
sn
{Hint: sn  1  12  13  14 
23.
Consider
the
64
sn  1  43  169  27

sequence
  34 
 1n .
n 1
of
 1n  1  1  1 
partial
 1  n , which implies that
sums

of the geometric series,
1
is convergent or divergent.
sn



. Determine if the series
n 1
{Hint: sn  1  
4
3
16
9

64
27

4 n1
3
1   43 
 34 n  1 .}


3


1  43
24. The Renaissance mathematician Pietro Mengoli proved the inequality
for x  1 .
a) Prove the inequality using algebra.
{Hint: The inequality is equivalent to
1 1
 .}
sn n
n
1
1
1
3
 

x 1 x x 1 x
1
1
2
1
1
2x
2x

 , and

 2
 2 .}
x 1 x 1 x
x 1 x 1 x 1 x
(5 parts)
b) Use the inequality in part a) to show that the harmonic series diverges.

1
{Hint: Suppose that
converges. Now group its terms after the first one into threes
n
n 1



to get that
1
1
n
n 1

that

n 1


n 1
1
1 
 1
 

 . But the inequality in part a) implies
 3n  1 3n 3n  1 
1
1 
 1




 3n  1 3n 3n  1 

25. Let’s examine the series

n 1
 ln n 


n 1


1
1
.}
3  3
3n
n
n 1
a
, a  0.
nb
a) Since ln x  x  1 for x  0 from Activity 2, problem #11, substitute x  t p with t , p  0
tp
to arrive at ln t  .
p
 
a
n
ln n 

nap
1
p
b) Use the previous inequality to get
 b  a b  a bap . What value(s) of p
b
n
n
pn
pn
will allow you to conclude convergence of the series? So what values of b will
guarantee convergence of the series?
tp
{Hint: Start by replacing t with n in ln t  .}
p
a
  ai  xbi 
n
26. Consider the sum
2
p
, where ai  and bi  are sequences of real numbers. It’s a
i 1
quadratic polynomial in x that is nonnegative. If we expand the sum, we get
n
n
 n 2 2  n

2
2 2
 ai  2ai bi x  bi x   
From the quadratic
bi  x   2
ai bi  x 
ai 2 .




i 1
i 1
 i1

 i1

2
formula, we get that if Ax  Bx  C  0 , then it’s discriminant, B 2  4 AC , must be less
than or equal to zero.
a) Set-up this discriminant inequality with our particular quadratic and find an inequality


 a b , a , b
n
involving

n
n
2
i
i i
i 1
i 1

i
2
.
i 1
(4 parts)

b) Suppose that
a
n
is a convergent series with an  0 . Use the previous inequality to
n 1

show that the series

n1
1 
1

 an    converges if   .
2
n 

1
c) Show that if   , the new series may be divergent by considering
2

 n  ln n 
1
.
2
n2
ln  an 
 L (possibly  ).
n ln n
27. (Log Test) Suppose that an  0 for n  1 and lim
a) What happens if L  1?
ln  an 
 1 which implies that ln  an    ln n , and by
ln n

1
1
exponentiating, you get that an  . What do you know about the series
?}
n
n
n1
b) What happens if L  1?
ln  an 
{Hint: This means that eventually
  p , where p  1 which implies that
ln n
1
ln  an    p ln n , and by exponentiating, you get that an  p . What do you know
n

1
about the series
for p  1 ?}
p
n
n 1
{Hint: This means that eventually


28. Use the Log Test of problem #27 to determine convergence or divergence of the following
series:

a)

n2
1
3ln n

b)

n2
1
2ln n

c)

n2
1
ln ln n 
3
(7 Parts)
29. Use the Loglog test of problem #21 to determine convergence or divergence of the
following series:

a)

n 1
n 2 n  n 2  1 !
n
2

b)
 n !
n2n
2
n 1
n
2
 2n  !
 n  n !

2
c)
2
n
n2
1
1 ln1n 

1
4
1
9
1
16
30. Let’s investigate the infinite product P  2  3  4 
. This can be written as
k
1
k2
with
k 2
n
the partial product pn 

1
k k 2 . Notice that pn  e
ln pn 
e
1 
 1 1 1
2
ln  2 4 39 416  n n 




1
 e4
ln 2 19 ln 3 
1
n2
ln n
.
k 2
So the infinite product’s convergence or divergence depends on the convergence or

ln k
divergence of the series
.
2
k
k 2
a) Show that the infinite product converges.


b) Use the estimate for the value of
k
ln k
2
to estimate the value of the infinite product.
k 2
(5 parts)