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CM221A
ANALYSIS I
Solutions to Sheet 1
1. By the Archimedean Property (see course notes), we have
δ

1
< n
for some
n
∈
N
. Multiplying both parts of this inequality by
n

1
δ
, we obtain
n

1
< δ
.
2. Obviously, each
a
n
is positive and, given this, 0
< a
n
+1
< a
n
. The sequence,
being monotonically decreasing and bounded, converges to a limit which
satisﬁes
a
=
a/
(
a
+ 2) (by passing to the limit in the identity deﬁning
a
n
+1
).
Hence
a
= 0.
3.
lim
n
→∞
2
n
3
+
n
2
+
n

n
5
n
3
+ 2
n

1 + 2
n
5
= lim
n
→∞
2
n

2
+
n

3
+
n

4

1
n

2
+ 2
n

4

n

5
+ 2
.
Evaluating the limits in the numerator and denominator, we obtain

1
2
.
lim
n
→∞
2
n
+ 3
3
n
+ 2
= lim
n
→∞
(2
/
3)
n
1 + 3
×
2

n
1 + 2
×
3

n
= lim
n
→∞
(2
/
3)
n
1 + 3 lim
n
→∞
×
2

n
1 + 2 lim
n
→∞
3

n
= 0
.
4. Let
a
n
=
n
k
b

n
. Then
a
n
+1
/a
n
=
b

1
±
n
+1
n
²
k
. Since
n
+1
n
→
1 as
n
→ ∞
, we
see that there is a number
c <
1 such that
a
m
+1
/a
m
< c
for all suﬃciently
large
m
. It follows that there exists a
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This document was uploaded on 01/31/2011.
 Spring '09

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