Exam 1 Solutions
1. Claim(1).
a
n
≤
a
n
+1
for all
n
≥
1.
Proof. By induction on
n
. Since
a
1
=
√
2
<
p
2 +
√
2 =
a
2
the base step is valid. Now assume
that
a
m
≤
a
m
+1
for some
m
≥
1. Then
a
m
+1
=
√
2 +
a
m
≤
p
2 +
a
m
+1
=
a
m
+2
so the induction step is also valid.
Claim(2).
a
n
<
2 for all
n
≥
1.
Proof. By induction on
n
. Since
a
1
=
√
2
<
2 the base step is valid. Assume that
a
m
<
2 for
some
m
≥
1. Then
a
m
+1
=
√
2 +
a
m
<
√
2 + 2 = 2
so the induction step is also valid.
By Claims (1) and (2), we know that (
a
n
) is monotone increasing and bounded above by 2.
Therefore
a
n
→
L
for some
L
∈
R
by the Monotone Convergence Theorem. Hence, by properties
of limits, we have
L
= lim
a
n
= lim
a
n
+1
= lim
√
2 +
a
n
=
p
2 + lim
a
n
=
√
2 +
L
and so
L
2

L

2 = 0. It follows that
L
= 2.
2. Suppose
a
n
→
L
and
a
n
→
L
0
. Given
>
0 there are
N
1
, N
2
∈
Z
such that

a
n

L

<
2
for all
n
≥
N
1
and

a
n

L
0

<
2
for all
n
≥
N
2
. So for any
n
≥
max
{
N
1
, N
2
}
we have

L

L
0

=

L

a
n
+
a
n

L
0
 ≤ 
L

a
n

+

a
n

L
0

<
2
+
2
=
.
Since
was arbitrary, it follows that

L

L
0

= 0. Hence
L
=
L
0
.
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 Fall '09
 WYNNE
 Logic, lim, Dominated convergence theorem, Monotone convergence theorem

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