Math 25: Advanced Calculus
UC Davis, Spring 2011
Math 25 — Solutions to Homework Assignment #7
1. Prove that the sequence
a
n
=
1
·
3
·
5
· · ·
(2
n

1)
2
·
4
·
6
· · ·
(2
n
)
converges.
Proof.
We will apply the monotone convergence theorem. Note that since
2
n

1
2
n
<
1 we have that
a
n
+1
< a
n
.
So the sequence
{
a
n
}
is decreasing monotonically. Clearly, the sequence is bounded below by zero. So, by
the monotone convergence theorem, it must converge.
2. Define a sequence
{
x
n
}
by
x
1
=
√
3
,
x
2
=
q
3 +
√
3
,
x
n
+1
=
√
3 +
x
n
.
Prove that the sequence converges and find its limit. For a small bonus credit, answer the same question
when 3 is replaced an arbitrary integer
k
≥
2.
Proof.
We show that the sequence converges by applying the monotone convergence theorem.
We will
proceed by induction. First, we claim that
x
n
<
3 for all
n
. Note that
√
3
<
3. Now, suppose that
x
n
<
3.
Then,
x
n
+1
=
√
3 +
x
n
≤
√
3 + 3 =
√
6
<
3
,
which shows the the sequence
x
n
is bounded above.
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 Winter '11
 Wiley
 Differential Equations, Calculus, Equations

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