First, we show that it is monotone, in particular, increasing.
This is done by
induction. The base case is to show that
x
1
≤
x
2
.
Is this true? Let’s see:
x
1
= 1
,
x
2
=
√
1 + 1 =
√
2
>
1
.
True. Next, we assume that
x
n
≤
x
n
+1
.
1
Hint: Try an alternating sequence, which alternates between positive and negative terms.
8
DR. JULIE ROWLETT
Then, we must show that
x
n
+1
≤
x
n
+2
.
Well, let’s try:
x
n
+2
=
1 +
x
n
+1
,
but
x
n
+1
≥
x
n
so
1 +
x
n
+1
≥
√
1 +
x
n
=
x
n
+1
.
Follow the inequalities and you’ll see that
x
n
+2
≥
x
n
+1
.
Next, we’ll show that the sequence is bounded. Look at
x
1
.
Well, clearly
x
1
<
2
.
Will this bound work for all the rest? Let’s try.
x
2
=
√
1 + 1
<
2
.
Since we’ve shown the base case (
n
= 1) and the next case (
n
= 2)
,
it makes sense to
try to prove the bound holds for all the remaining terms by induction. So, assume
that
x
n
<
2
.
Then,
x
n
+1
=
√
1 +
x
n
<
√
1 + 2 =
√
3
<
2
.
So, by induction, we have shown that 2 is an upper bound for the sequence.
Hence, by the monotone convergence theorem, the sequence converges. What is
the limit? This is fun.
If
x
n
→
x
where
x
is the limit, then, by homework problem 16.11,
x
n
+1
→
x
also. So,
lim
n
→∞
x
n
=
x
= lim
n
→∞
x
n
+1
= lim
n
→∞
√
1 +
x
n
.
Since on the far right side, the only term changing in the limit is
x
n
→
x,
by
Theorem 17.1,
x
=
√
1 +
x
⇒
x
2

x

1 = 0
.
The solutions to this equation are
1

√
5
2
and
1 +
√
5
2
.
Since the terms in the sequence are bounded below by 1
,
the limit is also bounded
below by 1
,
by Theorem 17.4. Hence, the correct limit is the positive root, and so
lim
n
→∞
x
n
=
1 +
√
5
2
.
3.
Cauchy Sequences
Theorem 3.1.
A sequence of real numbers converges if and only if it is a Cauchy
sequence.
Proof:
Assume the sequence of real numbers converges
{
s
n
}
n
∈
N
converges to
s.
Then, let
>
0
.
There exists
N
∈
N
such that

s
n

s

<
2
for all
n
≥
N.
Hence, for all
n, m
≥
N,

s
n

s
m
 ≤ 
s
n

s

+

s

s
m

<
2
+
2
=
.
That is the definition of a Cauchy sequence (MEMORIZE)¿
Next, assume that the sequence of real numbers is Cauchy. One reason it must
converge is by the construction of the real numbers as the set of all rationals and
MATH 117 LECTURE NOTES FEBRUARY 17, 2009
9
limits of Cauchy sequences of rational numbers. Another way is the following. We’ll
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 Fall '08
 Akhmedov,A
 Math, Mathematical analysis, Cauchy sequence, DR. JULIE ROWLETT