MATH 74 HOMEWORK 10  DUE MONDAY, APRIL 30
1.
We have 30
3
= 27
,
000 so that 10
4
≤
30
3
<
10
5
and thus
a
3
= 4.
Similarly 30
4
= 810
,
000 so that 10
5
≤
30
4
<
10
6
and thus
a
4
= 5.
If
p
∈
N
0
, the set of numbers
x
satisfying 10
p
≤
x <
10
p
+1
is precisely the set of
all
p
+ 1digit numbers. We thus see that
a
n
is the number of digits in 30
n
minus
one.
2.
Fix
n
∈
N
. By definition of the sequence we have 10
a
n
≤
30
n
. Applying
L
and
using properties from last week’s homework we conclude
a
n
L
(10)
≤
nL
(30); using
the fact that
L
(10) = 1 and dividing by
n
we conclude
L
(30)
≥
a
n
n
. Similarly, since
30
n
<
10
a
n
+1
we conclude
nL
(30)
<
(
a
n
+ 1)
L
(10) and thus
L
(30)
<
a
n
+1
n
.
3(a).
Yes,
b
n
≤
L
(30) for all
n
.
3(b).
No. We computed
a
3
= 4 and
a
4
= 5, so that
b
3
=
4
3
and
b
4
=
5
4
. Since
b
3
> b
4
the sequence is not increasing.
3(c).
Yes,
c
n
≥
L
(30) for all
n
.
3(d).
No. We have
c
2
=
a
2
+1
2
=
2+1
3
=
3
2
, but
c
3
=
a
3
+1
3
=
5
3
. Since
c
3
> c
2
the
sequence is not decreasing.
4.
By Exercise 2 we know
b
n
≤
L
(30)
≤
c
n
for any
n
∈
N
, and thus
0
≤
L
(30)

b
n
≤
c
n

b
n
,
n
∈
N
.
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 Fall '07
 COURTNEY
 Math, Ann, 1digit, 0 L

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