Mathematics 311
Solutions to Preliminary Exam 2
Introduction to Analysis
Instructor:
Marius Ionescu
1. Let
a
n
=
√
n
.
(a)
(5pts)
Prove that for every
ε >
0 there is
N
≥
1 such that

a
n
+1

a
n

< ε
if
n
≥
N
.
Solution:
Let
ε >
0. Then

a
n
+1

a
n

< ε
is equivalent with

√
n
+ 1

√
n

< ε
. Simplifying,
this inequality is equivalent with
1
√
n
+ 1 +
√
n
< ε.
Note that the previous expression is positive so we do not need the absolute value bars. Since
1
√
n
+ 1 +
√
n
<
1
2
√
n
if we choose
N
= (1
/
(2
ε
))
2
, it follows that

a
n
+1

a
n

< ε
if
n
≥
N
.
(b)
(5pts)
Is
a
n
a Cauchy sequence? Why?
Solution:
a
n
is not a Cauchy sequence. To prove this, suppose by contradiction that
a
n
is
Cauchy. Then
a
n
must be convergent by Theorem 6.4. The sequence
a
n
is, however, divergent.
Thus it can not be Cauchy.
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Solutions to Preliminary Exam 2
Introduction to Analysis
2. (a)
(7pts)
Show that if
∑
a
n
converges absolutely, then so does
∑
a
2
n
. Is this true without the hypothesis of
absolute convergence (prove or give a counterexample)?
Solution:
If the series
∑
a
n
converges absolutely it converges. Then lim
n
→∞
a
n
= 0. Then
there is
N
≥
1 such that

a
n

<
1 for all
n
≥
N
. It follows that
a
2
n
≤ 
a
n

for all
n
≥
N
. The
comparison test for positive series implies that
∑
∞
n
=
N
a
2
n
converges. From the tail theorem we
conclude that
∑
a
2
n
converges.
The conclusion fails without the hypothesis of absolute convergence. Consider
∑
(

1)
n
√
n
. This
series is convergent by the Cauchy test. The series
∑
1
n
, however, diverges (it is a
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 '08
 RAMAKRISHNA
 Math, Mathematical analysis, Cauchy, Preliminary Exam

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