INFINITE SERIES 1. Introduction The two basic concepts of calculus ...
Calculus Two Overview
25
Calculus Two Overview Andrew Shinko, under the advisement of Dr. Price (Image copyright spiriferous) December 4, 2005 Contents 1 Chapter Seven - Inverse Functions 2 1.1 One-to-One functions ......................... 2 1.2 Inverse Formulas ........................... 3 1.3 Basic Exponential and Logarithmic Problems ........... 4 1.4 Logarithmic and Exponential Derivatives ............. 5 1.5 Derivatives Using Trigonometric Substitutions ........... 6 1.6 LHospitalsRule ........................... 7 2 Chapter Eight - Techniques of Integration 7 2.1 Integrals Using Trigonometric Substitutions ............ 7 1
Transcript of Calculus Two Overview
Calculus Two Overview
Andrew Shinko, under the advisement of Dr. Price (Image copyright spiriferous)
December 4, 2005
Contents
1 Chapter Seven - Inverse Functions 21.1 One-to-One functions . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Inverse Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3 Basic Exponential and Logarithmic Problems . . . . . . . . . . . 41.4 Logarithmic and Exponential Derivatives . . . . . . . . . . . . . 51.5 Derivatives Using Trigonometric Substitutions . . . . . . . . . . . 61.6 L�Hospital�s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2 Chapter Eight - Techniques of Integration 72.1 Integrals Using Trigonometric Substitutions . . . . . . . . . . . . 7
1
2.2 Partial Fraction Decomposition . . . . . . . . . . . . . . . . . . . 102.3 Basic Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . 112.4 Comparison Theory . . . . . . . . . . . . . . . . . . . . . . . . . 12
3 Chapter Nine - Further Applications of Integration 123.1 Lengths of Curves . . . . . . . . . . . . . . . . . . . . . . . . . . 123.2 Areas of Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
4 Chapter Eleven - Parametric Equations and Polar Coordinates 144.1 Cartesian Equations - an Introduction . . . . . . . . . . . . . . . 144.2 Tangent Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154.3 Second Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . 164.4 Cartesian - Polar Coordinate Conversions . . . . . . . . . . . . . 164.5 Interpreting Polar Graphs . . . . . . . . . . . . . . . . . . . . . . 174.6 Drawing Polar Graphs . . . . . . . . . . . . . . . . . . . . . . . . 18
5 Chapter Twelve - In�nite Sequences and Series 195.1 Basic Forumulas and Limits . . . . . . . . . . . . . . . . . . . . . 195.2 Basic Convergence or Divergence . . . . . . . . . . . . . . . . . . 205.3 Integral Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215.4 Comparison Test . . . . . . . . . . . . . . . . . . . . . . . . . . . 215.5 Limit Comparision Test . . . . . . . . . . . . . . . . . . . . . . . 225.6 Alternating Series Test . . . . . . . . . . . . . . . . . . . . . . . . 225.7 Ratio Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235.8 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
1 Chapter Seven - Inverse Functions
1.1 One-to-One functions
Excercise 1.1.1 Determine if the given function, f(x) = 2x5 + x4 + 5x2 � 1,is one-to-one.
Solution 1.1.1 Setting up a table of values, we �nd
x -2 -1 0 1 2f(x) -29 3 -1 7 99
plotting the points, we �nd
2
21012
5
2.5
0
2.5
5
x
y
x
y
f(x)
Therefore, since f(1) and f(-1) > f(0), the function is not one-to-one
1.2 Inverse Formulas
Excercise 1.2.1 �nd the inverse formula of f(x) = ln((p2x+ 3)=e3) + 2a,
where a is not a variable.
Solution 1.2.1 The solution is displayed below
y = ln((p2x+ 3)=e3) + 2a
y � 2a = ln((p2x+ 3)=e3)
e(y�2a) = (p2x+ 3)=e3
e(y+3�2a) =p2x+ 3
e(2)(y+3�2a) = 2x+ 3
e(2)(y+3�2a) � 3 = 2x
(e(2)(y+3�2a) � 3)=2 = x
Using cancelation equations, we �nd:
x = ln((q2(e(2)(x+3�2a) � 3)=2 + 3)=e3) + 2a
x = (e(2)(ln((p2x+3)=e3)+2a+3�2a) � 3)=2
Once solved, both equations will equal x. Therefore:
f�1(x) = (e(2)(x+3�2a) � 3)=2
3
1.3 Basic Exponential and Logarithmic Problems
Excercise 1.3.1 Find the limit of the given function:
limx!5+
e25
(5�x)
Solution 1.3.1 To �nd the limit of the function, let t = 255�x . As x ! 5+,
t! �1therefore,
limt!�1
et = 0
Excercise 1.3.2 Di¤erentiate the following equation:
d
dx(e2x sin 4x + e2x cos 4x)
Solution 1.3.2 The solution is as follows. We use the product rule and chainrule in step two and simplify the equation.
d
dx(e2x sin 4x + e2x cos 4x)
) (2 sin 4x)e2x sin 4x + (2x)(4)(cos 4x)e2x sin 4x + (2 cos 4x)e2x cos 4x + (2x)(4)(� sin 4x)=) 2 sin(4x)e2x sin 4x + 8x cos(4x)e2x sin 4x + 2 cos(4x)e2x cos 4x � 8x sin(4x)e2x cos 4x
=) 2 cos(4x)(4xe2x sin 4x + e2x cos 4x) + 2 sin(4x)(e2x sin 4x � 4xe2x cos 4x)
Excercise 1.3.3 Expand the following expression
ln((2x+ 3)3(5x3 + 4)8p
3x+ 3(2x))
Solution 1.3.3 The solution is as follows:
ln((2x+ 3)3(5x3 + 4)8p
3x+ 3(2x))
3 ln(2x+ 3) + 8 ln(5x3 + 4)� (0:5 ln(3x+ 3) + ln 2x)
Excercise 1.3.4 Find the solution(s) to the given equation:
2 ln(x) = ln 2 + ln(x� 4)
4
Solution 1.3.4 The solution is as follows:
2 ln(x) = ln 2 + ln(4� x)) lnx2 = ln[2(4� x)]) eln x
2
= eln[2(4�x)]
) x2 = 2x� 4) x2 + 2x� 8 = 0) (x+ 4)(x� 2) = 0) x = 2;�4
Excercise 1.3.5 Find the limit to the given problem:
limv!1
ln(2� v)� ln(5 + v) + 2
Solution 1.3.5 The solution is as follows
limv!1
ln(2� v)� ln(5 + v) + 2
) limv!1
[ln(2� v)� ln(5 + v)] + limv!1
2
) [ limv!1
ln(2� v5 + v
)] + 2
) [ limv!1
ln(2� v5 + v
)] + 2
) [ limv!1
ln(1)] + 2
) 2
1.4 Logarithmic and Exponential Derivatives
Excercise 1.4.1 Compute the derivitive of the following function:
f(x) = ln(4x2 + 2)
(7x� 24) + e�2x5
Solution 1.4.1 the solution is as follows:
f(x) = ln(4x2 + 2)
(7x� 24) + e�2x5
f 0(x) = ln(4x2 + 2)� ln(7x� 24) + e�2x5
f 0(x) =8x
4x2 + 2� 7
7x� 24 + e�2x5(�10x4)
f 0(x) =4x
2x2 + 1� 7
7x� 24 � 10x4(e�2x
5
)
5
Excercise 1.4.2 Compute the derivitive of the following function:
f(x) = 72x2
Solution 1.4.2 The solution is as follows:
f(x) = 72x2
f 0(x) = 72x2
ln 7(4x)
f 0(x) = 4x(ln 7)72x2
1.5 Derivatives Using Trigonometric Substitutions
Excercise 1.5.1 Find the derivitive of the given function. Please simplify whenpossible.
f(x) = cos�1(5x+ 1)
Solution 1.5.1 The solution, using trigonometric substitutions, is shown below
f(x) = cos�1(5x+ 1)
f 0(x) = � 1p1� (5x+ 1)2
(5)
f 0(x) = � 5p1� (25x2 + 10x+ 1)
f 0(x) = � 5p�25x2 � 10x
f 0(x) = � 5p(�5x)(5x+ 2)
Excercise 1.5.2 Find the derivitive of the given function. Please simplify whenpossible.
f(x) = tan�1(e2x+1)
Solution 1.5.2 The solution, using trigonometric substitutions, is shown below
f(x) = tan�1(e2x+1)
f 0(x) =2(e2x+1)
(e2x+1)2 + 1
f 0(x) =2e2x+1
e4x+2 + 1
6
Excercise 1.5.3 Evaluate the given integral:
I =
Zx3p1� x8
dx
Solution 1.5.3 The solution, using trigonometric substitutions, is shown below
I =
Zx3p1� x8
dx let u = x4 and du = 4x3
I =1
4
Zdup1� u2
I =1
4sinu+ C
I =1
4sinx4 + C
1.6 L�Hospital�s Rule
Excercise 1.6.1 Find the limit of the given function:
limx!0
5x � 7xx
Solution 1.6.1 Using L�Hosiptal�s Rule, we �nd
limx!0
5x � 7xx
L0H
=limx!0
5x ln 5� 7x ln 71
= ln 5� ln 7
= ln5
7
2 Chapter Eight - Techniques of Integration
2.1 Integrals Using Trigonometric Substitutions
Excercise 2.1.1 Evaluate the given integral:Zx sec2 3xdx
7
Solution 2.1.1 The solution is as follows:
let u = x; du = dx, v = 3 tan 3x, dv = sec2 3x: Therefore,Zx sec2 3xdx = 3x tan 3x�
Z3 tan 3xdxZ
x sec2 3xdx = 3x tan 3x� 3 ln jsec 3xj+ C
Excercise 2.1.2 Evaluate the given integral:Zsecx csc2 xdx
Solution 2.1.2 The solution is as follows:
let u = secx, du = tanx secx, v = � cotx, dv = csc2 x: Therefore,Zsecx csc2 xdx = secx(� cotx) +
Zcotx tanx secxdx
= � cscx+Zsecxdx
= � cscx+ ln jsecx+ tanxj+ C
Excercise 2.1.3 Solve the given integral:Zsin5 x cos7 xdx
Solution 2.1.3 The solution is as follows:
Zsin5 x cos7 xdx
=
Zsinx(1� cos2 x)2 cos7 xdx, let u = cosx, du = sinx
=
Z(1� u2)2u7du
=
Z(u4 � 2u2 + 1)u7du
=
Z(u11 � 2u9 + u7)du
=1
12u12 � 2
10u10 +
1
8u8 + C
=1
12cos12 x� 1
5cos10 x+
1
8cos8 x+ C
8
Excercise 2.1.4 Solve the given integral:
I =
Zxp
1� 4x2dx
Solution 2.1.4 The solution is as shown:
I =
Zxp
1� 4x2dx, let x =
1
2sinu, dx =
1
2cosu
I =
Z 12 sinuq
1� 4( 12 sinu)2(1
2cosu)
I =1
4
Zsinu cosup1� sin2 u
I =1
4
Zsinu cosu
cosu
I =1
4
Zsinu
I = �14cosu+ C
I = �14
p1� 4x2 + C
Excercise 2.1.5 Integrate the given integral:Zdxpx2 + 25
9
Solution 2.1.5 The solution is as follows:
I =
Zdxpx2 + 25
, let x = 5 tanx, dx = 5 sec2 x
I =
Z5 sec2 xp
(5 tanx)2 + 25
I =
Z5 sec2 xp
25(tan2 x+ 1)
I =
Zsec2 xp
(tan2 x+ 1)
I =
Zsec2 x
secx
I =
Zsecx
I = ln jsecx+ tanxj+ C
I = ln
�����px2 + 25
5+x
5
�����+ CI = ln
����� (px2 + 25 + x)
5
�����+ CI = ln
���(px2 + 25 + x)���� ln j5j+ CI = ln
���(px2 + 25 + x)���+ C2.2 Partial Fraction Decomposition
Excercise 2.2.1 Evaluate the given integral:Z2x+ 5
(x+ 1)(x� 1)dx
Solution 2.2.1 Using partial fractions decomposition, we obtain the followingsolution:
I =
Z2x+ 5
(x+ 1)(x� 1)dx
I =
Z(
A
(x+ 1)+
B
(x� 1))dx
2x� 5 = A(x� 1) +B(x+ 1)
When x = �1, A = 7
2: When x = 1, B = �3
2
I =
Z(7=2
(x+ 1)� 3=2
(x� 1))dx
I =7
2ln(x� 1)� 3
2ln (x+ 1) + C
10
2.3 Basic Convergence
Excercise 2.3.1 Determine whether or not the given integral converges:Z 1
�1
x2
x6 + 16dx
Solution 2.3.1 The solution is shown below.
I =
Z 1
�1
x2
x6 + 16dx
= 2
Z 1
0
x2
x6 + 16dx u = x2, du = 3x2dx
=2
3
Z 1
0
du
u2 + 16v = 4u, dv = 4du
=8
3
Z 1
0
dv
16v2 + 16
=8
48
Z 1
0
dv
v2 + 1
=1
6tan�1 v
=1
6tan�1
u
4
=1
6tan�1
x2
4
Therefore, to see the convergence, we take the following limit:
=
�1
6limt!1
(tan�1(x2
4))
�t0
=1
12�
Excercise 2.3.2 Determine if the given integral converges. Show your work.
Z 2
�1
1
x3dx
Solution 2.3.2 The solution is shown:To solve this integral, we split up the integral where there is in�nite disconituity,
11
which is at x=0 Z 2
�1
1
x3dx =
Z 0
�1
1
x3dx+
Z 2
0
1
x3dx
=
Z 0
�1
1
x3dx+
Z 2
0
1
x3dx
��limt!0=
(1
2x2)
�t�1
= limt!0=
�� 1
2t2+1
2
�= �1, diverges
Therefore, sinceR 0�1
1x3 dx diverges,
R 2�1
1x3 dx does.
2.4 Comparison Theory
Excercise 2.4.1 Use the comparision theory to determine whether or not thegiven integral converges Z 1
1
2x5
7x8 + 3x3 + 5dx
Solution 2.4.1 The solution is as follows:
for x � 1,2x5
7x8 + 3x3 + 5dx <
1
x3Z 1
1
1
x3, p = 3
Therefore,R11
2x5
7x8+3x3+5dx is convergent by the comparison theory
3 Chapter Nine - Further Applications of Inte-gration
3.1 Lengths of Curves
Excercise 3.1.1 Set up, but do not evaluate, an integral for the length of thegiven curve
y = 2x4 � x3 + 7x 0 < x < 1
Solution 3.1.1 The solution is shown below:dy
dx= 8x3 � 3x2 + 7
L =
Z 1
0
p1 + (8x3 � 3x2 + 7)2dx
L =
Z 1
0
p1 + 64x6 � 48x5 + 9x4 + 112x3 � 42x2 + 49dx
L =
Z 1
0
p64x6 � 48x5 + 9x4 + 112x3 � 42x2 + 50dx
12
Excercise 3.1.2 Set up, but do not evaluate, an integral for the length of thegiven curve
y = sin(x) 0 < x < 2�
Solution 3.1.2 The solution is shown below:
dy
dx= cosx
L =
Z 2�
0
p1 + (cosx)2dx
L =
Z 2�
0
p1 + cos2 xdx
3.2 Areas of Curves
Excercise 3.2.1 Find the area of the given curve rotated about the x-axis
y =2
3x3 0 < x < 3
Solution 3.2.1 The solution is shown below:
dy
dx= 2x2
S = 2�
Z 3
0
2
3x3q1 + (2x2)
2dx
S =4
3�
Z 3
0
x3p1 + 4x4dx; u = 1 + 4x4; du = 16x3
S =1
12�
Z 324
1
pudu
S =1
12�
�2
3u3=2
�3241
S =5831
18� (note: due to the complexity of the solution, the previous step is an acceptable answer).
Excercise 3.2.2 Find the area of the given curve rotated about the x-axis
x = y2 + 4 0 < y < 2
13
Solution 3.2.2 The solution is shown below:dx
dy= 2y
S = 2�
Z 2
0
yp1 + (2y)2dy
S = 2�
Z 2
0
yp1 + 4y2dy, u = 1 + 4y2, du = 8ydy
S =�
4
Z 17
1
pudu
S =�
4
�2
3u3=2
�171
S =1
4�
�34
3
p17� 2
3
�
Excercise 3.2.3 Set up, but do not evaluate, the area of the given curve rotatedabout the y-axis
y =
px
7+ 3 0 < y < 5
Solution 3.2.3 The solution is shown below:
x2 = (7y � 21)x = (7y � 21)2
x = 49y2 � 294y + 441dx
dy= 98y � 294
S = 2�
Z 5
0
(49y2 � 294y + 441)p1 + (98y � 294)2dy
4 Chapter Eleven - Parametric Equations andPolar Coordinates
4.1 Cartesian Equations - an Introduction
Excercise 4.1.1 Find the Cartesian equation of the given curve.
x =pt� 5; y = t� ln t+ 8
Solution 4.1.1 The solution is shown below:
t = (x� 5)2
t = x2 � 10x+ 25y = x2 � 10x+ 25� ln(x2 � 10x+ 25) + 8y = x2 � 10x+ 33� ln(x2 � 10x+ 25)
14
Excercise 4.1.2 Find the Cartesian equation of the given curve. Graph it un-der the given parameters and indicate the direction of travel.
x = 5 cos �; y = 4 sin � 0 < x < �
Solution 4.1.2 The solution is shown below:
(1
5x)2 + (
1
4y)2 = cos2 x+ sin2 y = 1
1
25x2 +
1
16y2 = 1
52.502.55
4
3
2
1
0
x
y
x
y
The graph moves in a counter-clockwise direction, starting and ending at (5,0)and ending at (-5,0).
4.2 Tangent Lines
Excercise 4.2.1 Find the tangent to the curve at the point corresponding tothe parameter value
x =1
2t4 + 4; y = 3t2 + 4t� 5 t = 2
Solution 4.2.1 The solution is as follows:
dy
dx=
6t+ 4
2t3 + 4
at t = 2;dy
dx=16
20=4
5(x; y) = (12; 15)
y =4
5(x� 12) + 15
y =4
5x+ 5
2
5
Excercise 4.2.2 Find the equation to the tangent(s) to the curve at the givenpoint
x = 2 sin t; y = � cos t (1;p3)
15
Solution 4.2.2 The solution is as shown:
dy
dx=
sin t
2 cos t
the point (1;p3) corresponds to a t =
�
3value. To �nd the slope, we plug in that value
dy
dx=
sin �32 cos �3
=
p3
2
y =
p3
2(x� 1) +
p3
y =
p3
2x+
p3
2
4.3 Second Derivatives
Excercise 4.3.1 Find d2ydx2
x = t2 � 5 y = 4t3 � 5t2 + 3
Solution 4.3.1 The solution is as shown
dy
dx=
12t2 � 10t2t
dy
dx= 6t� 5
d2y
dx2=
ddt (6t� 5)
2td2y
dx2=
6
2td2y
dx2=
3
t
4.4 Cartesian - Polar Coordinate Conversions
Excercise 4.4.1 Find the the Cartesian points of the given polar coordinates:
(5;�
3)
Solution 4.4.1 The solution is shown below:
x = 5 cos�
3
x =5
2
y = 5 sin�
3
y =5
2
p3
16
Excercise 4.4.2 Find the polar coordinates (where 0 < � < 2�) of the givenCartesian points.
(4; 4p3)
Solution 4.4.2 The solution is shown below:
r =
q42 + (4
p3)2
r =p16 + 48
r = 8
� = tan�1(4p3
4)
� =�
3
Therefore, the polar coordinates are (8; �3 ) and (�8;4�3 )
4.5 Interpreting Polar Graphs
Excercise 4.5.1 Which graph shows the polar equation r = 1� sin �?
1.510.500.511.5 0
0.5
1
1.5
2
2.5
3
xy
xy
A
10.500.51 0
0.5
1
1.5
2
xy
xy
B
10.500.51
0.5
0
0.5
1
1.5
x
y
x
y
C
10.500.51 0
0.5
1
1.5
x
y
x
y
D
Solution 4.5.1 The answer is choice B.
Excercise 4.5.2 Which graph shows the polar equation r = sin 4�?
0.500.51
1
0.5
0
0.5
1
x
y
x
y
A
0.500.51
0.5
0
0.5
x
y
x
y
B
0.750.50.2500.250.50.75
0.75
0.5
0.25
0
0.25
0.5
0.75
x
y
x
y
C
0.750.50.2500.250.50.75
0.75
0.5
0.25
0
0.25
0.5
0.75
x
y
x
y
D
Solution 4.5.2 The answer is choice C.
17
4.6 Drawing Polar Graphs
Excercise 4.6.1 Sketch the curve and �nd the area it encloses
r = cos 3�
Solution 4.6.1 The solution is shown below:
10.750.50.2500.250.5
0.75
0.5
0.25
0
0.25
0.5
0.75
x
y
x
y
r = cos3�
A =1
2
Z �
0
cos2 3�d�
A = 3
Z �6
0
cos2 3�d�
A =3
2
Z �6
0
(1 + cos 6�)d�
A =3
2
�� +
1
6sin 6�
��6
0
A =�
4
Excercise 4.6.2 Graph the curve and �nd the area one loop of the graph en-closes
r = 3 sin 5�
Solution 4.6.2 The solution is shown below
2.51.2501.252.5
2.5
1.25
0
1.25
x
y
x
y
r = 3 sin 5�
18
A =1
2
Z �5
0
9 sin2 5�d�
A =9
2
Z �5
0
sin2 5�d�
A =9
4
Z �5
0
(1� cos 10�)d�
A =9
4
�� � 1
10sin 10�
��5
0
A =9
20�
5 Chapter Twelve - In�nite Sequences and Se-ries
5.1 Basic Forumulas and Limits
Excercise 5.1.1 Determine whether or not if the sequence converges. If it does,�nd its limit
an =2n
5n+1
Solution 5.1.1 The solution is shown below:
an =2n
5n+1=1
5(2
5)n
1
5limn!1
(2
5)n =
1
5(0) = 0
The given sequence converges with r= 25
Excercise 5.1.2 Find a formula for the general term, an, assuming the patternof the given terms continues.
f1;�12;1
4;�18;1
16;� 1
32::::g
Solution 5.1.2 The solution is shown below:Since each term is being multiplied by � 1
2 except for the �rst term, an = (�12 )n�1
19
5.2 Basic Convergence or Divergence
Excercise 5.2.1 Determine if the given series converges or diverges. If it isconvergent, �nd its sum.
1Xn=1
n2 � nn2 + n
Solution 5.2.1 The solution is shown below:1Xn=1
n2 � nn2 + n
limn!1
n2 � nn2 + n
= limn!1
1� 1n
1 + 1n
= 1 6= 0
Therefore, the series1Xn=1
n2�nn2+n diverges by the test for divergence
Excercise 5.2.2 Determine if the given series converges or diverges. If it isconvergent, �nd its sum.
1Xn=1
2n�1
5n
Solution 5.2.2 The solution is shown below:
1
5
1Xn=1
2n�1
5n�1
The series is geometric with an = 1 and r = 25 . Since
�� 25
�� < 1, the given seriesconverges.
an1� r =
1
1� 25
=5
3(1
5) =
1
3
Thus, the given series converges to 13 .
Excercise 5.2.3 Determine if the given series converges or diverges. If it isconvergent, �nd its sum.
1Xn=1
�n
3n+1
Solution 5.2.3 The solution is as shown:
1
3
1Xn=1
�n
3n
The series is a geometric series with r = �3 . But, jrj > 1, so the given series
diverges.
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5.3 Integral Test
Excercise 5.3.1 Use the integral test to determine of the given series convergesor diverges.
1Xn=1
1
n2 + 1
Solution 5.3.1 The solution is shown below:First, we must determine if the series meets the conditions of the integral test.Let f(x) = 1
x2+1
1. x > 0 for all x 2 (1;1)
2. f 0(x) =�2x
2x2 + x4 + 1< 0
so, the given series satis�es the integral test.
f(x) = limt!1
Z t
1
1
x2 + 1
f(x) = limt!1
�tan�1 x
�t1
f(x) =�
4
Since f(x) converges, the given integral,1Xn=1
1n2+1 , also converges by the integral
test.
5.4 Comparison Test
Excercise 5.4.1 Determine if the series converges or diverges.1Xn=1
1
n2 + n
Solution 5.4.1 The solution is shown below:Using the comparison test, we �nd
0 <1
n2 + n<1
n2
let an =1Xn=1
1
n2 + n, and bn =
1Xn=1
1
n2
the series bn is a convergent p=2 series.
Since an < bn and bn =1Xn=1
1n2 converges, then the series an =
1Xn=1
1n2+n also
converges by the comparison test.
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5.5 Limit Comparision Test
Excercise 5.5.1 Determine if the following series converges or diverges
1Xn=1
n� 1n2 + 2n
Solution 5.5.1 The solution is shown below:Using the limit comparison test, we determine if the given series converges ordiverges.
let an =n� 1n2 + 2n
and bn =1
n
limn!1
n�1n2+2n1n
= limn!1
n(n� 1)n2 + 2n
= limn!1
n2 � nn2 + 2n
= 1 > 0
Since the harmonic series1Xn=1
1n diverges, the given series
1Xn=1
n�1n2+2n diverges by
the limit comparison test.
5.6 Alternating Series Test
Excercise 5.6.1 Determine if the series converges or diverges.
1Xn=1
(�1)npn+ 1
Solution 5.6.1 The solution is shown below:Using the Alternating series test and having bn = 1p
n+1, we �nd
1. bn > 0 for all x 2 (1;1)2.fbng is decreasing3. lim
n!11pn+1
= 0
Therefore, by the Alternating Series Test, the series1Xn=1
(�1)npn+1
converges.
Excercise 5.6.2 Determine if the series absolutely converges, conditionally con-verges, or is divergent.
1Xn=1
(�1)n 2n� 3n+ 1
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Solution 5.6.2 The solution is shown below:Using the test for divergence, we �nd
limn!1
2n� 3n+ 1
= 2
Hence, the series1Xn=1
2n�3n+1 diverges by the test for divergence
Excercise 5.6.3 Determine if the series absolutely converges, conditionally con-verges, or is divergent.
1Xn=1
(�1)nn+ 1
Solution 5.6.3 The solution is shown below:Using the Alternating series test, we �nd,
1. 1(n+1) > 0 for all x 2 [1;1)
2. 1(n+2) <
1(n+1) or (n+ 1) < (n+ 2) or 0 < 1
3. limn!1
1(n+1) = 0
Since the series1Xn=1
(�1)nn+1 satis�es the AST, we can test for absolute convergence
and conditional convergenceUsing the limit comparison test, we �nd,
let an =1
(n+ 1)and bn =
1
n
limn!1
(
1(n+1)
1n
) = limn!1
n
n+ 1= 1 > 0
Since the harmonic series1Xn=1
1n diverges, the series
1Xn=1
1n+1 does by the limit
comparison test
Therefore, the given series1Xn=1
(�1)nn+1 is conditionally convergent.
5.7 Ratio Test
Excercise 5.7.1 Find the radius of convergence and the integral of convergencefor the given series.
1Xn=1
xnpn
n!
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Solution 5.7.1 The solution is shown below:1Xn=1
xnpn
n!
Using the ratio test, we �nd,
limn!1
������xn+1
pn+1
(n+1)!
xnpn
n!
������ = jxj limn!1
pn+ 1pn(n+ 1)
= 0(jxj) > 1 = 0 > 1
hence, by using the ratio test, the radius of convergence, R, of the given seriesis 1The interval of convergence is (�1;1)
Excercise 5.7.2 Find the radius of convergence and the integral of convergencefor the given series.
1Xn=1
(�1)nn2nxn
Solution 5.7.2 The solution is shown below:1Xn=1
(�1)nn2nxn
Using the ratio test, we �nd,
limn!1
���� (n+ 1)2n+1xn+1n2nxn
���� = jxj limn!1
2(n+ 1)
n= 2(jxj) > 1 = jxj > 1
2
hence, by using the ratio test, the radius of convergence, R, of the given seriesis 1
2To �nd the interval of convergence, we look at the endpoints.
when x =1
2;
1Xn=1
(�1)nn2n 12
n
=
1Xn=1
(�1)nn
when x = �12;1Xn=1
(�1)nn2n(�12)n =
1Xn=1
n
for both endpoints, limn!1
n = 1, so the enpoints � 12 , and
12 diverge by the test
for divergence.Hence, the integral of convergence for the given series is (� 1
2 ;12 ).
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5.8 Power Series
Excercise 5.8.1 Find a power series representation for the function.
f(x) =1
x� 4
Solution 5.8.1 The solution is shown below:
f(x) =1
x� 4
f(x) = (�14)1
1� x4
f(x) = �14
1Xn=0
�x4
�nor �
1Xn=0
xn
4n+1
25
