Back and Forth
I'm All Shook Up...
The 3 R'sCompared to What?!
ID, Please!
$1000$1000$1000$1000$1000
$800$800$800$800$800
$600$600$600$600$600
$400$400$400$400$400
$200$200$200$200 $200
ID, Please! - $800
DONE
The limit of this series equals 1, therefore by the nth term test, the series DIVERGES.
ID, Please! - $1000
DONE
32
1 1
1 1
n nn n n
¥ ¥
= =
=å å
This series can be written as
which is just a convergent p – series, since p > 1
Compared to What?! - $200
DONE
21
1
n n
¥
=å
Using the Direct Comparison Test to the
convergent series
Therefore, this series converges also.
2 2
2 2
5 7
1 1
5 7
n n
n n
+ >
<+
Compared to What?! - $400
DONE
33 2
3 22
18 6 7lim lim 01 88 6 7n n
nnn n
n nn
®¥ ®¥
+ - = = >+ -
Use the Limit Comparison Test with 21
1
n n
¥
=å
Since converges, so does the original series.21
1
n n
¥
=å
Compared to What?! - $600
DONE
32
1
1
n n
¥
=å
Using the Direct Comparison Test to the
convergent p – series
Therefore, this series converges also.
3 3
3 3
2
1 1
2
n n n
n n n
+ >
<+
Compared to What?! - $800
DONE
1 1
2 2
3 3
n n
nn n
¥ ¥
= =
æ öæö ÷ç÷ç ÷ç=÷ç ÷ç÷÷ ÷ç ÷è ø ççè øå å
Using the Direct Comparison Test to the
convergent geometric series
Therefore, this series converges also.
3 1 3
1 1
3 1 3
2 2
3 1 3
n n
n n
n n
n n
+ >
<+
<+
Compared to What?! - $1000
DONE
Using the Direct Comparison
Test, we first try
BUT this gives us a series that is less than a divergent series … Not helpful
Next, try using the Direct Comparison Test with
BUT this gives us a series that is more than a convergent series … Not helpful
So try using the Direct Comparison Test with something in between … say
2 2
ln
ln 1
n n
n n
nn n
<
< =
1
1
nn
¥
=
å
1
2
1
n n
¥
=
å
2 2
ln as long as 3
ln 1
1 n n
n
n n
>
>
>
1
32
1
n n
¥
=
å1
2
12
322 2
ln
ln 1
n n
n n
n n n
<
< =
Therefore, since the series is less than a convergent series, the original series is CONVERGENT ALSO!
The 3 R's - $200
DONE
By the Root Test,
1 1 1lim lim 1
2 1 2 1 2
n
nn n
n n
n n®¥ ®¥
æ ö æ ö+ +÷ ÷ç ç= = <÷ ÷ç ç÷ ÷÷ ÷ç çè ø è ø+ +
this series converges.
The 3 R's - $400
DONE
By the Ratio Test,
( )
( )
1
3 3
3
3
21 2
lim lim 2 12 1
n
nn n
n n
nn
+
®¥ ®¥
+= = >
+
this series diverges.
The 3 R's - $600
DONE
this series converges because the integral converges.
By the Integral Test,
( )1
11
10x xe dx e e
e
¥¥- - -=- = - - =ò
NOTE: you could have used the geometric series test since
1 1
1n
n
n n
ee
¥ ¥-
= =
æö÷ç= ÷ç ÷÷çè øå å
The 3 R's - $800
DONE
this series diverges because the integral diverges.
By the Integral Test,
( )22
22 2 2
ln 1 1ln
2 2
xx
x x
xdx u du u x
x
®¥¥ ®¥ ¥
= =
= = =ò ò
The 3 R's - $1000
DONE
By the Ratio Test,
( )( )( )
( ) ( )( )
1 !1 3 5 2 1 1 1 1
lim lim 1! 22 1 11 3 5 2 1
n n
nn n
n nn
®¥ ®¥
+×× + - +
= = <+ -
×× -
this series converges.
I'm All Shook Up... - $600
DONE
Using the Limit Comparison Test to the
divergent p – series
Therefore, since the p – series diverges, so does the original series.
12
3
32
3
12
lim lim 1 0
2n n
n nnn
nn n
®¥ ®¥
+= = >
+
12
1
1
n n
¥
=å
I'm All Shook Up... - $800
DONE
By the Ratio Test,
( )( )( ) ( )
( )
2 4 6 2 11 ! 2 1
lim lim 2 12 4 6 2 1
!n n
nn n
n nn
®¥ ®¥
×× ++ +
= = >×× +
this series diverges.
I'm All Shook Up... - $1000
DONE
1 1lim lim lim 1
11 11
n
n nn n nn
e
e ee
-®¥ ®¥ ®¥= = =
+ ++
By the nth term Test, since
this series diverges
Back and Forth - $200
DONE
Alternating Series Test
(condition 1):
(condition 2):
1lim 0n n®¥
=
1
1 1
1
n n
n n
+ >
<+
Since both conditions of the alternating series are met, the series converges.
Back and Forth - $400
DONE
Alternating Series Test
(condition 1):
Since the first condition of the alternating series test fails, then what we just did is the equivalent of the nth term test.
Therefore, this series diverges.
1lim
2 1 2n
n
n®¥=
-
Back and Forth - $600
DONE
Alternating Series Test
(condition 1):
(condition 2): To show it is decreasing, find the 1st derivative
The 1st derivate is 0 for x = 1, but negative everywhere else. Therefore, both conditions are met and the series converges.
2lim 0
1n
n
n®¥=
+
( )
( )( )( ) ( )( )
( ) ( )
2
2 2
2 22 2
1
1 1 2 1'
1 1
xf x
x
x x x xf x
x x
=+
+ - - += =
+ +
Back and Forth - $800
DONE
Alternating Series Test
(condition 1): 1
6
3lim limn n
nn
n®¥ ®¥= =¥
Since the first condition of the alternating series test fails, then what we just did is the equivalent of the nth term test.
Therefore, this series diverges.
Back and Forth - $1000
DONE
First, recognize this as an Alternating Series … since
(condition 1):
(condition 2):
Since both conditions of the alternating series are met, the series converges.
1lim 0n n®¥
=
1
1 1
1
n n
n n
+ >
<+
( ) ( ) ( ) ( )1
cos 1 1 1 1 1 1n
np¥
=
= - + + - + + - + +å
Directions for Changing the Game
• To change the questions and answers, just type over the problems…Use the “replace” feature to change the categories easily
• The daily doubles were originally set to category #1 for $800 and category #5 for $400
• To change the daily doubles you must– 1. Change the hyperlink for the links on the main board to go to the
appropriate question, therefore bypassing the daily double slide … (right click on category #1 for $800 and chose hyperlink, then edit hyperlink)
– 2. There is a rectangle drawn around each slide that is used to link the slide question with the answer (or answer back to the main board). To change this link, position your cursor at the edge of the slide to select the rectangle and edit the hyperlink by right clicking the rectangle. You need to edit the hyperlink on each Daily Double slide.
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