MATH 138
Calculus 2
Fall 2010
Solutions #7
1. First, we use induction to show 0
< a
n
≤
3 for all
n
≥
1. This is true for
n
= 1 since
a
1
= 3. If we assume that it is true for
n
=
k
, then we have
0
< a
k
≤
3
,
⇒
1
≤
4

a
k
<
4
⇒
1
4
<
1
4

a
k
≤
1
⇒
0
< a
k
+1
≤
3
since
a
k
+1
= 1
/
(4

a
k
). We have deduced that 0
< a
n
≤
3 is true for
n
=
k
+ 1.
Therefore, the statement that 0
< a
n
≤
3 is true for all
n
≥
1 by induction.
We now show that
a
n
is decreasing, i.e.,
a
n
+1
≤
a
n
for all
n
≥
1. This is true for
n
= 1
since
a
2
=
1
4

a
1
=
1
4

3
= 1
≤
3 =
a
1
. Assume that it is true for
n
=
k
, i.e.,
a
k
+1
≤
a
k
.
Then,
a
k
+2

a
k
+1
=
1
4

a
k
+1

1
4

a
k
=
a
k
+1

a
k
(4

a
k
+1
)(4

a
k
)
≤
0
since
a
k
+1
≤
a
k
, 0
< a
k
≤
3 and 0
< a
k
+1
≤
3. Hence,
a
n
+1
≤
a
n
is true for
n
=
k
+ 1.
Therefore,
a
n
+1
≤
a
n
for all
n
≥
1 by induction.
Since the sequence
a
n
is bounded below and decreasing (or bounded and monotonic),
it is convergent by the monotonic sequence theorem.
Let
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 Spring '07
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 Calculus, Harmonic Series, Mathematical Series, nth term test

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