MATH201 INTRODUCTION TO ANALYSIS
Worksheet for week 5: Sequences and series
Lecture sessions 1/2
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1. Let xn =
2n2 +3n−1
3n2 −3n+1 ,
n ∈ N be a given sequence.
(a) Establish the inequality |xn − 32 | <
Solution. |xn −
<
5(3n−1)
3(3n2 −3n)
(b) Let ε =
1
100 .
<
2
3|
=
2
2n +3n−1
| 3n
2 −3n+1
5(3n−1)
3(n−1)(3n−1)
=
−
5
3(n−1) for all n ∈ N.
2
6n2 +9n−3−6n2 +6n−2
|
3| = |
3(3n2 −3n+1)
5
3(n−1)
[2]
for all n ∈ N.
Determine an N ∈ N so that |xn − 23 | <
1
100 ,
for all n > N .
5
Solution. Since for all n > N , 3(n−1)
< 3(N5−1) , then we will find an N ∈ N such that
500
So N > 3 + 1. Thus N = 168 would be an appropriate choice.
[1]
5
3(N −1)
<ε=
1
100 .
2. If (bn ) is a bounded sequence and lim an = 0, prove directly from the definition of limit that lim an bn
= 0.
[3]
Proof. Since (bn ) is bounded, then there exists a real number M > 0 such that |bn | ≤ M for all n ∈ N.
ε
lim an = 0, we conclude that if ε > 0 is given, then there exists a K ∈ N such that |an | < M
for all
n ≥ K. Therefore |an bn | ≤ M |an | < ε for all n ≥ K. Since ε > 0 is arbitrary, it follows that lim an bn
= 0.
Why can’t we apply Theorem 3.2.3 to solve this problem?
[1]
Solution. Although (bn ) is bounded, it may not converge. For example, (bn ) = ((−1)n ). So we can’t
apply Theorem 3.2.3 to solve this problem.
3. Let (xn ) be a sequence of positive numbers such that lim( xxn+1
) = 32 . Prove that (xn ) is not a convergent
n
sequence.
[3]
Proof. First we show (xn ) is unbounded. For any real number M > 0, if we take an ε0 > 0 such that
xn+1
xn+1
3
3
3
2 − ε0 > 1, then there exists a K = K(ε0 ) ∈ N such that for all n ≥ K, | xn − 2 | < ε0 . So xn > 2 − ε0
for all n ≥ K. Therefore, if n ≥ K, we have xn+1 > ( 32 − ε0 )xn > · · · > ( 23 − ε0 )n−K+1 xK . i.e.,
xn ≥ ( 32 −ε0 )n−K xK for all n ≥ K. Note that ( 32 −ε0 )n−K xK > M ⇔ n > log( xMK ( 23 −ε0 )K )/ log( 32 −ε0 ).
Thus for n > max{ log( xMK ( 23 − ε0 )K )/ log( 32 − ε0 ), M }, we have xn > M . Hence (xn ) is unbounded. It
follows that (xn ) is not a convergent sequence.
4. (optional) Let bn = (1+ n1 )n+1 for each n ∈ N. Show that (bn ) is a monotonic decreasing sequence. (see §3.3)
1 n+1
(1+ n
)
1
(1+ n+1
)n+2
n2 +2n+1 n+2 n
( n(n+2) )
n+1 = (1
Proof.
bn
bn+1
=
=
(n+1)n+1 (n+1)n+2
nn+1 (n+2)n+2
2
(n+1) n+2 n
= ( n(n+2)
)
n+1
1
n
n+2
n
n
=
+ n(n+2)
)n+2 n+1
≥ (1 + n(n+2)
) n+1
= n+1
n · n+1 = 1.
So for all n ∈ N, bn ≥ bn+1 , it means (bn ) is a monotonic decreasing sequence.