Friday, March 6
Clicker Questions
Clicker Question 1
Partial sums that simplify
By finding a formula for sn , the nth partial sum of the series
∞
X
23/i − 23/(i+1) ,
i=1
calculate the value that the series converges to.
A. 7
A telescoping sum
B. 0
sn = 23/1 − 23/2 + 23/2 − 23/3 + · · ·
+ 23/(n−1) − 23/n + 23/n − 23/(n+1)
C. 8
D. −1
E. none of
the above
= 23/1 − 23/(n+1) . So
23/i − 23/(i+1) = lim 23/1 − 23/(n+1)
∞
X
i=1
n→∞
= 8 − 20 = 7.
Clicker Question 2
Dividing a complicated series into simpler ones
Using previous examples from this lecture, calculate
∞ X
3 3/i
4
3/(i+1)
− 2
2 −2
.
7
i + 8i + 15
i=1
A. −2
Series laws in action
B. 6.1
∞ X
3
C. 2.1
i=1
D. −6
E. none of
the above
7
3/i
2
−2
3/(i+1)
4
− 2
i + 8i + 15
∞
3 X 3/i
2 − 23/(i+1)
=
7
i=1
∞
−
4 X
10
2
10
i + 8i + 15
i=1
3
4 9
= ·7−
· = 3 − 0.9.
7
10 4
Clicker Question 3
Almost a geometric series
Calculate .790 = .790909090 . . .
9
9
9
9
7
+
+
+
+
+ ···.
=
10 100 10,000 1,000,000 100,000,000
A.
B.
C.
D.
E.
4
5
1
11
87
110
70
99
none of the above
From our formula
∞
X
9
1 r−1
7
+
10
100 100
i=1
7
9/100
+
10 1 − 1/100
7
9
=
+
10 100 − 1
7
1
=
+ .
10 11
=
Clicker Question 4
Staring a series at a different index
Let {an } be a sequence. Suppose you know that the series
∞
X
an = S converges to some value S. What can you say about
n=1
the series
∞
X
an ?
n=4
A. it converges only if
a1 > a2 > a3 > a4
B. it always converges,
to S − (a1 + a2 + a3 )
C. it always diverges
D. it converges only if
a1 , a2 , a3 are positive
E. there’s not enough
information to tell
Two related sums
The partial sums of
∞
X
an = a4 + a5 + · · ·
n=4
are all (a1 + a2 + a3 ) less than the
partial sums of
∞
X
an = a1 + a2 + a3 + a4 + a5 + · · · .
n=1