Winter 2011
Assignment 7 – Solutions
1. Suppose
{
a
n
}
is the sequence deﬁned recursively by
a
1
= 0
a
n
+1
=
√
5 + 2
a
n
for
n
= 1
,
2
,
3
....
Prove that
{
a
n
}
converges and ﬁnd its limit.
The ﬁrst few terms of the sequence are:
a
1
= 0
,
a
2
=
√
5
≈
2
.
23
,
a
3
=
p
5 + 2
√
5
≈
3
.
07
,
a
4
=
q
5 + 2
p
5 + 2
√
5
≈
3
.
34
,
a
5
≈
3
.
41
. So we
would like to show that the sequence is increasing and bounded above, by
say 4 (it is clear that 0 is a lower bound). We will do this by induction.
For the base case, we have
a
1
= 0
≤
a
2
≈
2
.
23
≤
4
.
Then, we assume that the following is true:
a
n
≤
a
n
+1
≤
4
.
Now we would like to show that
a
n
+1
≤
a
n
+2
≤
4
. Start with the assump
tion:
a
n
≤
a
n
+1
≤
4
⇐⇒
5 + 2
a
n
≤
5 + 2
a
n
+1
≤
5 + 2(4) = 13
⇐⇒
√
5 + 2
a
n
≤
p
5 + 2
a
n
+1
≤
√
13
⇐⇒
a
n
+1
≤
a
n
+2
≤
√
13
<
4
Therefore, this is true for all
n
so the sequence is increasing and bounded,
and by the Monotonic Sequence Theorem, it converges. If
{
a
n
}
converges
to
L
, then we must have
lim
n
→∞
a
n
= lim
n
→∞
a
n
+1
=
L
. Taking the limit of the
deﬁnition of
{
a
n
}
gives:
lim
n
→∞
a
n
+1
= lim
n
→∞
√
5 + 2
a
n
lim
n
→∞
a
n
+1
=
q
5 + 2 lim
n
→∞
a
n
L
=
√
5 + 2
L
Squaring and rearranging, we get
L
2

2
L

5 = 0
which gives
L
= 1
±
√
6
.
Since
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 Winter '07
 Anoymous
 Math, Calculus

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