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
