Download Wed, Nov 20

Clicker Question 1

The shortest interval which the Integral Test
guarantees us will contain the value of

n
is:

2
3/ 2
(
n

3
)
n 1
–
–
–
–
–
A. 0 to 1
B. 0 to ½
C. ½ to 1
D. ½ to 5/8
E. The series diverges.
Clicker Question 2

The shortest interval which the above result
guarantees us will contain the value of
1/(1 + n2) is:
–
–
–
–
–
A. [0, 2]
B. [0, /2]
C. [/4, 1]
D. [/4, /4 + ½]
E. [/4, /4 + 1]
Clicker Question 3

The series
–
–

n2

4
3/ 2
n 1 (n  3)
A. converges
B. diverges
Alternating Series (11/20/13)



If the terms of a series alternate sign and if
the terms themselves are approaching 0,
then the series converges.
Example: 1 – 1/2 + 1/3 - 1/4 + 1/5 - …
must converge!
(Guess the name of this series.)
To what, you ask? Not obvious, since this
series is not geometric. Experiment a bit?
The Alternating Series Test


Theorem. If b1  b2  b3 … is a
decreasing sequence of positive numbers
and if
limnbn = 0, then the series (-1)(n+1)bn
converges.
Proof outline:
–
–
The even partial sums s2n are monotone
increasing and bounded above by b1, so…?
Each odd partial sum s2n+1 = s2n + bn+1 , so…?
Clicker Question 4


The series
 (1)
n 1
–
–
–
–
( n 1)
3n
2n 2  1
A. converges to a number smaller than 0
B. converges to a number smaller than 3
C. converges to a number bigger than 3
D. diverges
Assignment for Friday

Read Section 11.5 to the bottom of
page 729 and do Exercises 1 – 13 odd,
32, 33, and 35.