Download Sequences and Series

Chapter 1
Sequences and Series
1.1
Sequences and Their Limits
Example 1.1.1 (Bartle 3.1.2 (a) Page 55). If b ∈ R, then the sequence B := (b, b, b, . . .) is a constant
sequence of b. Thus we have the constant sequences 1 := (1, 1, 1, . . .) and 0 := (0, 0, 0, . . .).
Example 1.1.2 (Bartle 3.1.2 (b) Page 56). If b ∈ R, then B := (bn ) is the sequence B = (b, b2 , b3 , . . .).
Thus, for b = 1/2 we have the sequence
1
1
1 1 1
:
n
∈
N
=
,
·
·
·
,
,
,
·
·
·
,
2n
2 4 8
2n
Example 1.1.3 (Bartle 3.1.2 (c) Page 56). The sequence of even natural numbers (2n : n ∈ N) may be
defined inductively by
x1 := 2, xn+1 := xn + 2
or alternately by the definition
y1 := 2,
yn+1 := y1 + yn
Example 1.1.4 (Bartle 3.1.2 (d) Page 56). The Fibonacci sequence F := (fn ) is given by the inductive
definition
f1 := 1, f2 := 1, fn+1 := fn−1 + fn for (n ≥ 2)
Problem 1 (Bartle Exercise 1 for Section 3.1). Write the first five terms for the sequences whose nth terms
are defined by the formulae:
1. xn := 1 + (−1)n
2. xn := (−1)n /n
3. xn := 1/(n(n + 1))
4. xn := /(n2 + 2)
Solution:
1
1. (xn ) = 0, 2, 0, 2, 0, . . .
2. (xn ) = −1, 1/2, −1/3, 1/4, −1/5, . . .
3. (xn ) = 1/2, 1/6, 1/12, 1/20, 1/30, . . .
4. (xn ) = 1/3, 1/6, 1/11, 1/18, 1/27, . . .
Problem 2 (Bartle Exercise 2 for Section 3.1). Give the formula for the nth term of the following sequences:
1. 5, 7, 9, 11, . . .
2. 1/2, −1/4, 1/8, −1/16, . . .
3. 1/2, 2/3, 3/4, 4/5, . . .
4. 1, 4, 9, 16, . . .
Solution:
1. xn := 3 + 2n
2. xn := −(−1/2)n
3. xn := n/n + 1
4. xn := n2
Problem 3 (Bartle Exercise 3 for Section 3.1). Todo...
Example 1.1.5 (Bartle 3.1.6 (a) Page 57). Show that
1
lim
=0
n
We need to show that for every ε > 0 there exists a K = K(ε) ∈ N such that for all n ≥ K, the terms
1/n satisfy |1/n − 0| < ε.
By the Archimedean Property, for every ε > 0 there exists some natural, say K ∈ N, such that 0 < 1/K <
ε. Thus, for all n ≥ K we must have that 0 < 1/n ≤ 1/K < ε. Consequently
1
− 0 = 1 < ε
n
n
for all n ≥ K, and we can conclude that the sequence (1/n) converges with limit 0; that is, lim(1/n) = 0.
Example 1.1.6 (Bartle Example 3.1.6 (b) Page 58). Show that
1
=0
lim
n2 + 1
We may make use of the results of the previous example (Example 3.1.6 (a) Bartle) if we first notice that
if n ∈ N then
1
1
1
< 2 ≤
2
n +1
n
n
For every ε > 0 we may choose a K = K(ε) such that 0 < 1/K < ε, which holds under the Archimedean
Property. From the previous example we see that this implies that 1/n < ε for all n ≥ K, so that
1
1
1
n2 + 1 − 0 = n2 + 1 < n < ε
Hence, the sequence converges to a limit lim(1/(n2 + 1)) = 0.
2
Example 1.1.7 (Bartle Example 3.1.6 (c) Page 58). Show that
3n + 2
=3
lim
n+1
We first note that if n ∈ N then
3n + 2 − 3n − 3 −1 3n + 2
1
1
= n + 1 = n + 1 < n
n + 1 − 3 = n+1
Thus, we may choose K = K(ε) such that 0 < 1/K < ε. Then for any n ≤ K we have 1/n < ε, and
consequently
3n + 2
n + 1 − 3 < ε
Hence, the limit of the sequence is 3.
Example 1.1.8 (Bartle Example 3.1.6 (d) Page 58). Show that
√
√
lim( n + 1 = n) = 0
We first note that if n ∈ N then
√
√
√
√
√
√
( n + 1 + n)
1
1
1
( n + 1 − n) · 1 = ( n + 1 − n) · √
√ = √
√ ≤ √ <
n
n
( n + 1 + n)
n+1+ n
As in the previous examples, we may choose a K = K(ε) such that 0 < 1/K < ε. Then for any n ≥ K
we have 1/n < ε so that
√
√
|( n + 1 − n) − 0| < ε
√
√
√
√
Hence, the
√ sequence ( n + 12− n) converges to the 2limit lim( n + 1 − n) = 0. Notice that for a given
ε > 0, 1/ n < ε ⇔ 1/n < ε or rearranged n > 1/ε . Then the sequence converges to the limit 0 if we
take K > 1/ε2 ; for example, if ε = 1/10 we require K > 100.
Example 1.1.9 (Bartle Example 3.1.6 (e) Page 58). Since 0 < b < 1, we can write b = 1/(1 + a). So
bn = 1/(1 + a)n Since a = (1/b) − 1 > 0, with use of the Binomial Theorem, we can make the estimation,
(1 + a)n ≥ 1 + na. This implies that
|bn − 0| = bn ≤
1
1
1
≤
<
n
(1 + a)
1 + na
na
If we choose C = 1/a > 0, and knowing that the sequence (1/n) converges with limit lim(1/n) = 0, by
Theorem 3.1.10 Bartle, lim(bn ) = 0.
Example 1.1.10 (Bartle Example 3.1.11 (a) Page 60). We claim that if a > 0, then lim(1/(1 + na)) = 0.
Since a > 0, we have 0 < na < 1 + na. Taking the reciprocals we obtain 0 < 1/(1 + na) < 1/na. Thus
1
1 1
1 + na − 0 ≤ a n
So, with the earlier result lim(1/n) = 0, C = 1/a, and m = 1, we infer that
1
lim
=0
1 + na
3
Example 1.1.11 (Bartle Example 3.1.11 (b) Page 60). We claim that if 0 < b < 1, then lim(bn ) = 0.
Since 0 < b < 1, we can write b = 1/(1 + a). Here a = (1/b) − 1 so that a > 0. As such, we may make use
of Bernoulli’s inequality (or from the Binomial Theorem) (1 + a)n ≥ 1 + na. So 1/bn ≥ (1 + a)n ≥ 1 + na.
Taking the reciprocal, we obtain 0 < bn ≤ 1/(1 + a)n ≤ 1/(1 + na) < 1/na. Thus
1
1 1
− 0 ≤
|bn − 0| ≤ 1 + na
a n
So, with the earlier result lim(1/n) = 0, C = 1/a, and m = 1, we infer that
lim (bn ) = 0
Problem 4 (Bartle Exercise 4 for Section 3.1). For any b ∈ R, prove that lim(b/n) = 0.
We need to show that for every ε > 0 there exists some K = K(ε) such that for every n ≥ K, the terms
xn satisfy |b/n − 0| < ε.
By the Archimedean Property, for every ε > 0 we can find a K ∈ N such that 1/ε < K. Thus, for any b ∈ R
we have |b|/ε < K. Then for every n ≥ K we have |b|/ε < n. From this we obtain the desired result
b
− 0 < ε
n
for every n ≥ K. Hence, the sequence is convergent with limit lim(b/n) = 0.
Problem 5 (Bartle Exercise 5 (a) for Section 3.1). Use the definition of the limit of a sequence to establish
the following limit
n
lim
=0
n2 + 1
We need to show that for every ε > 0 we can find a K = K(ε) such that for every n ≥ K
n
n2 + 1 − 0 < ε
Notice that for every n ∈ N
n
1
n
n
n2 + 1 − 0 = n2 + 1 < n2 = n
As before, the Archimedean Property allows us to choose K = K(ε) such that 0 < 1/K < ε for every ε > 0.
Then for any n ≥ K we have 0 < 1/n ≤ 1/K < ε. Thus, for all n ≥ K
n 1
n2 + 1 < n < ε
Hence, the sequence n/(n2 + 1) converges with limit 0.
Problem 6 (Bartle Exercise 5 (b) for Section 3.1). Todo...
Problem 7 (Bartle Exercise 5 (c) for Section 3.1). Todo...
Problem 8 (Bartle Exercise 5 (d) for Section 3.1). Todo...
Problem 9 (Bartle Exercise 6 (a) for Section 3.1). Show that
1
=0
lim √
n+7
4
We need to show that for every ε > 0 we can find a K = K(ε) such that for every n ≥ K
1
√
n + 7 − 0 < ε
Notice that for every n ∈ N
1
1
1
√
= √ 1
−
0
< √ <
n+7
n
n
n+7
As before, the Archimedean Property allows us to choose K = K(ε) such that 0 < 1/K < ε for every ε > 0.
Then for any n ≥ K we have 0 < 1/n ≤ 1/K < ε. Thus, for all n ≥ K
1
1
√
< <ε
−
0
n+7
n
√
Hence, the sequence 1/( n) converges with limit 0.
Problem 10 (Bartle Exercise 6 (b) for Section 3.1). Show that
2n
=2
lim
n+2
Firstly, notice that
2n − 2n − 4 −4 2n
1
4
=
=
−
2
n + 2 = n + 2 < 4 · n
n + 2
n+2
Since (an ) = (1/n) is a sequence that converges with lim(1/n) = 0, and the constant C = 4 > 0, it follows
that
2n
lim
=2
n+2
Problem 11 (Bartle Exercise 6 (c) for Section 3.1). Todo...
Problem 12 (Bartle Exercise 6 (d) for Section 3.1). Todo...
Problem 13 (Bartle Exercise 7 for Section 3.1). Todo...
Problem 14 (Bartle Exercise 8 for Section 3.1). Todo...
Problem 15 (Bartle Exercise 9 for Section 3.1). Todo...
Problem 16 (Bartle Exercise 10 for Section 3.1). Todo...
Problem 17 (Bartle Exercise 11 for Section 3.1). Show that
1
1
=0
−
lim
n n+1
We must show that for every ε > 0 there exists a K = K(ε) ∈ N such that for all n ≥ K, the terms
xn := 1/n − 1/(n + 1) satisfy
1
1
<ε
−
−
0
n n+1
Firstly, note that for all n ∈ N
1
1
1
1
1 1 n − n + 1 − 0 = n − n + 1 ≤ n = n
5
By the Archimedean Property, for every ε > 0 there exists a K ∈ N such that 0 < 1/ε < K; that is, 1/K < ε.
Then for all n ≥ K we have 0 < 1/n < 1/K < ε. Thus
1
1
1
n − n + 1 − 0 < n < ε
By the ε-K definition of convergence, the sequence 1/n − 1/(n + 1) converges with limit
1
1
=0
lim
−
n n+1
Problem 18 (Bartle Exercise 12 for Section 3.1). Todo...
Problem 19 (Bartle Exercise 13 for Section 3.1). Todo...
Problem 20 (Bartle Exercise 14 for Section 3.1). Todo...
Problem 21 (Bartle Exercise 15 for Section 3.1). Todo...
Problem 22 (Bartle Exercise 16 for Section 3.1). Todo...
Problem 23 (Bartle Exercise 17 for Section 3.1). Todo...
Problem 24 (Bartle Exercise 18 for Section 3.1). Todo...
Problem 25 (Bartle Exercise 19 for Section 3.1). Todo...
1.2
Limit Theorems
Example 1.2.1 (Bartle 3.2.8 (a) Page 66). The sequence (n) is divergent. We can show this by contradiction:
Suppose, to the contrary, that the sequence (n) is convergent. But a convergent sequence is bounded in
R (Theorem 3.2.2 Bartle), so that there must exist some real number M > 0 such that n < M for all
n ∈ N. But this violates the Archimedean Property, so that the assumption that (n) is convergent must
be false. Hence (n) is divergent.
Example 1.2.2 (Bartle 3.2.8 (b) Page 67). Todo...
Example 1.2.3 (Bartle 3.2.8 (c) Page 67). We show that
2n + 1
=2
lim
n
Quite simply
lim
2n + 1
n
1
1
= lim 2 +
= lim(2) + lim
= 2+0= 2
n
n
Example 1.2.4 (Bartle 3.2.8 (d) Page 67). We show that
2n + 1
lim
=2
n+5
6
Here we factor out an n from the numerator and denominator and divide the limits of the numerator and
denominator:
lim(2 + 1/n)
2 + 1/n
2+0
2n + 1
= lim
=
=
=2
lim
n+5
1 + 5/n
lim(1 + 5/n)
1+0
Example 1.2.5 (Bartle 3.2.8 (e) Page 67). We show that
2n
lim
=
n2 + 1
Here we factor out an n2 from the numerator and denominator:
lim(2/n)
2n
2/n
0
lim
=
lim
=
=
=0
n2 + 1
1 + /n2
lim(1 + 1/n2 )
1+0
Example 1.2.6 (Bartle 3.2.8 (f) Page 67). We show that
sin n
=0
lim
n
The sequence (n) in the denominator is not convergent, so we are unable to apply the same techniques
of finding the quotient of the limits of the numerator and denominator that we have done previously.
On first inspection it does not seem that we may make use of any algebraic manipulation to reduce the
sequence either. Instead, however, note that sin n is bounded as −1 ≤ sin n ≤ 1, so that
−
1
sin n
1
≤
≤
n
n
n
for all n ∈ N. Then, since (−1/n) and (1/n) are both convergent sequences with limit 0, the Squeeze
Theorem allows us to conclude that (sin n/n) is also convergent with limit 0.
Example 1.2.7 (Bartle 3.2.8 (g) Page 68). Todo...
Example 1.2.8 (Bartle 3.2.8 (h) Page 68). Todo...
1.3
Monotone Sequences
Example 1.3.1 (Bartle Page 71). The following sequences are increasing:
1. (1, 2, 3, . . . , n, . . .)
2. (1, 2, 2, 3, 3, 3, . . .)
3. (a, a2 , a3 , . . . , an , . . .) for a > 1
The following sequences are decreasing:
1. (1, 1/2, 1/3, . . . , 1/n, . . .)
7
2. (1, 1/2, 1/22, 1/23 , . . . , 1/2n−1, . . .)
3. (b, b2 , b3 , . . . , bn , . . .) for 0 < b < 1
Notice that in the above examples (an ) may be increasing for a > 1, but will be decreasing for 0 < a < 1.
The following sequences are not monotone:
1. (1, −1, 1, . . . , (−1)n+1 , . . .)
2. (−1, 2, −3, . . . , (−1)n n, . . .)
The following sequences are not monotone, but are ultimately monotone since they have m-tails which
are monotone:
1. (7, 6, 2, 1, 2, 3, 4)
2. (−2, 0, 1, 1/2, 1/3, 1/4, . . .)
Example 1.3.2 (Bartle 3.3.3 (a) Page 72). We shall make use of the Monotone Convergence Theorem
to show that1
√
lim(1/ n) = 0
To use the Monotone Convergence Theorem, we are required
to√show that the sequence is both monotone
√
and bounded. Firstly, for any
n
∈
N,
we
have
that
1/
n
>
1/
n + 1, and so the sequence is decreasing.
√
Also, we know that 0 < 1/ n < 1 for all
n
∈
N,
so
the
sequence
is bounded. Hence, by the Monotone
√
Convergence Theorem, the sequence (1/ n) is convergent.
To find the limit of the sequence we can try either one of the following two approaches:
√
1. We need to know the value of the infimum,
since the sequence is decreasing. We know that 1/ n > 0
√
for all n ∈ N, so that the set {1/ n : n ∈ N} has
√ a lower bound 0. The Completeness Property of
R thus ensures that an infimum of the set {1/ n : n ∈ N} exists. Then, for ε > 0 given, we can
choose K > 1/ε2 so that n ≥ K implies n > 1/ε2 , and consequently
1
1
√ = √ − 0 < ε
n
n
√
Hence, the limit of the sequence (1/ n) is given by
√
√
lim(1/ n) = inf{1/ n : n ∈ N} = 0
√
√
√
√
2. We know that (1/ n) converges,
so let lim(1/ n) = x. However, (1/ n) · (1/ n) = ((1√
√
sqrtn2 )) so that lim((1/ n)2 ) = lim(1/n), which implies that x2 = 0. Hence x = lim(1/ n) = 0.
Example 1.3.3 (Bartle 3.3.3 (b) Page 72). Is this example worth going through?
Example 1.3.4 (Bartle 3.3.4 (a) Page 73). Find the limit of the sequence Y = (yn ), which has been
defined inductively n ≥ 1 by
1
y1 := 1, yn+1 := (2yn + 3)
4
To use the Monotone Convergence Theorem, we need to show that the sequence is both monotone and
bounded:
8
Bounded : Firstly, the sequence is bounded below by 1. By simple calculation we see that y1 < y2 < 2, so
we make the claim that the sequence is bounded above by 2. Using Mathematical Induction:
1. The statement yn < 2 for all n ∈ N is true for n = 1.
2. Assume that the statement is true for some k ∈ N; that is, that yk < 2 for some k ∈ N. Then
(1/4)(2yk + 3) < (7/4) < 2 so that yk+1 < 2.
But since yk < 2 implies yk+1 < 2, by Mathematical Induction yn < 2 for all n ∈ N. Hence, the sequence
is bounded between 1 and 2.
Monotone: We claim that the sequence is monotonically increasing, so we need to show that yn < yn+1
for all n ∈ N. Again, using Mathematical Induction:
1. The statement yn < yn+1 for all n ∈ N is true for n = 1.
2. Assume that the statement is true for some k ∈ N; that is, that yk < yk+1 for some k ∈ N. Then
(1/4)(2yk + 3) < (1/4)(2yk+1 + 3) so that yk+1 < yk+2 . But since yk < yk+1 implies yk+1 < yk+2 ,
and k is arbitrary, the statement is true for all k.
Hence, by Mathematical Induction yn < yn+1 for all n ∈ N, and so the sequence is monotonically
increasing.
Convergence: Since the sequence is monotone and bounded, by the Monotone Convergence Theorem,
the sequence is convergent. We could find the limit by finding the supremum, but instead consider that
yn+1 = (1/4)(2yn + 3) is the nth term of the 1-tail Y1 of Y . But lim Y1 = lim Y . Thus if we let y := lim Y ,
then lim((1/4)(yn + 3)) = (1/4)(2 lim(yn ) + 3) = (1/4)(2y + 3). So (1/4)(2y + 3) = 3 from which we find
that y = 3/2.
Example 1.3.5 (Bartle 3.3.4 (b) Page 73). Show that the sequence Z = (zn ) converges to a limit of 2:
√
z1 := 1, zn+1 := 2zn for n ∈ N
By direct calculation 1 < z1 < z2 < 2. We then claim that Z is both monotonically increasing, and
bounded by 1 and 2. Thus it is required to show that 1 ≤ zn < zn+1 < 2 for all n ∈ N. Using
Mathematical Induction:
1. The statement 1 ≤ zn < zn+1 < 2 is true for n = 1.
2. Suppose
is true for some n = k > 1. Then 1 ≤ zk < zk+1 < 2 implies that
√ that
√ the statement
√
1 < 2 ≤ 2zk < 2zk+1 < 2; that is, 1 < zk+1 < zk+1 < 2. Since k > 1 is arbitrary, the
statement is true for all k > 1.
Thus, by Induction, 1 ≤ zn < zn+1 < 2 for all n ∈ N; that is, the sequence is both bounded, and increasing.
Then, by the Monotone Convergence Theorem, the sequence converges to a limit z := sup{zn : n ∈ N}.
TODO: Calculate the limit of the sequence (do not necessarily need to calculate the supremum)...
Example 1.3.6 (Bartle 3.3.5 (a) Page 74). Todo...
Problem 26 (Bartle Exercise 1 for Section 3.4). Show that the sequence X = (xn ) defined inductively by
x1 := 8,
xn+1 :=
is bounded and monotone. Find the limit.
9
1
xn + 2
2
Bounded : By direct calculation (TODO: draw a graph) we claim that 4 ≤ xn ≤ 8. We prove this using
Mathematical Induction:
1. The statement 4 ≤ xn ≤ 8 is true for n = 1, since x1 = 8.
2. Suppose that 4 ≤ xk ≤ 8 is true for some k ∈ N, k > 1. Then 4 ≤ xk ≤ 8 ⇒ 4 ≤ (1/2)xk + 2 ≤ 6 ⇒
4 ≤ xk+1 ≤ 8. Since k is arbitrary, the statement is true for all k.
Hence, by Mathematical Induction we have that 4 ≤ xn ≤ 8 for all n ∈ N; that is, the sequence is bounded.
Monotone: We claim that the sequence is decreasing; that is, that xn > xn+1 for all n ∈ N. Again, by
Mathematical Induction:
1. xn > xn+1 is true for n = 1, since x1 = 8 > 6 = x2 .
2. Suppose that xk > xk+1 is true for some k ∈ N, k > 1. Then xk > xk+1 ⇒ (1/2)xk + 2 > (1/2)xk+1 +
2 ⇒ xk+1 > xk+2 . Since k is arbitrary, the statement is true for all k.
Hence, by Mathematical Induction, xn > xn+1 for all n ∈ N.
Convergence: Since the sequence is monotone and bounded, by the Monotone Convergence Theorem, the
sequence is convergent. Limit : Note that xn+1 = (1/2)xn + 2 is a relation between the nth term of the 1-tail
X1 = (x1+n ) of X = (xn ) and the nth term of X itself. But if X converges, then X1 must also converge and
have the same limit as X. If we define x := lim X then
1/2
1
1
lim(xn+1 ) = lim
= lim(xn ) + 2 = x + 2
xn + 2
2
2
Thus (1/2)x + 2 = x from which x = 4, so the limit of the sequence is given by lim X = 4.
Problem 27 (Bartle Exercise 2 for Section 3.3). Let x1 > 1 and let xn+1 := 2 − 1/xn for n ∈ N. Show that
(xn ) is bounded and monotone. Find the limit.
Bounded : It is easy to see that 1 < xn < 2 since 0 < 1/xn < 1 for xn . To prove this we use Mathematical
Induction:
1. 1 < xn < 2 is true for n = 1.
2. Suppose 1 < xk < 2 is true for all k > 1. Then 1 < xk < 2 ⇒ −1 > −1/xk > −1/2 ⇒ 1 < 2 − 1/xk <
3/2 < 2. Thus 1 < xk < 2 ⇒ 1 < xk+1 < 2 and since k is arbitrary, this is true for all k > 1.
By Induction 1 < xn < 2 for all n ∈ N; that is, (xn ) is bounded. Monotone: We claim that (xn ) is increasing.
We need to show that xn < xn+1 for all n ∈ N and prove this by Mathematical Induction.
1. xn < xn+1 is true for n = 1.
2. Suppose xk < xk+1 for all k > 1. Then xk < xk+1 ⇒ 2 − 1/xk < 2 − 1/xk+1 ⇒ xk+1 < xk+2 . Since
k > 1 is arbitrary, this is true for all k > 1.
By Induction xn < xn+1 for all n ∈ N; that is, (xn ) is monotonically increasing. Convergence: By the
Monontone Convergence Theorem, the sequence (xn ) is convergent. Limit : TODO...
√
Problem 28 (Bartle Exercise 3 for Section 3.3). Let x1 ≥ 2 and xn+1 := 1 + xn − 1 for n ∈ N. Show that
(xn ) is decreasing and bounded below by 2. Find the limit.
Bounded : To show that the sequence is bounded below by 2, we need to show that xn ≥ 2 for all n ∈ N.
Using Mathematical Induction
1. The statement xn ≥ 2 for all n ∈ N is true for n = 1 by the hypothesis.
√
√
2. Assume xk ≥ 2 for some k ∈ N, k > 1. Then xk ≥ 2 ⇒ 1 + xk − 1 ≥ 1 + 2 − 1 = 2. Thus
xk ≥ 2 ⇒ xk+1 ≥ 2, and since k is arbitrary, the statement is true for all k > 1.
10
Hence, by Mathematical Induction, xn ≥ 2 for all n ∈ N; that is, the sequence is bounded below by 22 .
Monotone: We claim that the sequence is decreasing. We need to show that xn ≥ xn+1 (or similarly that
xn − xn+1 ≥ 0) for all n ∈ N. Firstly
√
√
xn − xn+1 = xn − (1 + xn − 1) = (xn − 1) − xn − 1
√
√
But xn − 1 ≥ xn − 1 (more simply t ≥ t for any t ∈ R), so that xn ≥ xn+1 . Hence, the sequence is
monotonically decreasing.
Convergence: Since the sequence is monotone and √bounded, by the Monotone Convergence Theorem, the
sequence converges. Limit : Note that xn+1 := 1 + xn − 1 is a relation between the nth term of the 1-tail
X1 = (x1+n ) of the sequence X = (xn ) and the nth term of the sequence X itself. Then, since X converges,
X1 must also converge and converge to the same limit as X. Firstly, define x := lim(xn ) so that
p
√
√
lim(X1 ) = lim(x1+n ) = lim(1 + xn − 1) = 1 + lim(xn ) − 1 = 1 + x − 1
√
Since X1 and X converge to the same limit we have x = 1 + x − 1 from which x = 1 or x = 2, and since
x = 1 is not allowed (xn ≥ 2), the sequence has the limit lim X = 2.
√
Problem 29 (Bartle Exercise 4 for Section 3.3). Let x1 := 1 and xn+1 := 2 + xn for n ∈ N. Show that
(xn ) converges and find the limit.
By direct calculation 1 ≤ x1 < x2 < 2, so we claim that (xn ) is increasing and bounded as 1 ≤ xn < 2.
We show this by proving 1 ≤ xn < xn+1 < 2 for all n ∈ N. Using Mathematical Induction
1. The statement 1 ≥ xn < xn+1 < 2 is true for n = 1.
2. Suppose that
√ the
√ statement
√ is true for some n = k > 1. Then 1 ≥ xk < xk+1 < 2, which implies
that 1 < 3 ≥ 2 + xk < 2 + xk+1 < 2; that is, 1 < xk+1 < xk+2 < 2 and since k is arbitrary, the
statement is true for all k > 1.
Thus, by Induction, 1 < xn < xn+1 < 2 is true for all n ∈ N, and so the sequence is both monotonically
increasing and bounded. Then, by the Monotone Convergence Theorem, the sequence converges to a limit
x := lim(xn ).
TODO: Find the limit of the sequence...
√
√
Problem 30 (Bartle Exercise 5 for Section 3.3). Let y1 := p where p > 0 and yn+1 := p + yn for n ∈ N.
Show that yn converges and find the limit.
Monotone: We claim that (yn ) is increasing. We prove this by showing that yn < yn+1 .
1. yn < yn+1 for n = 1
2. Suppose yk < yk+1 for some k ∈ N, k > 1. Then yk < yk+1 ⇒
Since k is arbitrary this is true for all k.
√
√
p + yk < p + yk+1 ⇒ yk+1 < yk+2 .
Hence, by Mathematical Induction yn < yn+1 for all n ∈ N; that is, (yn ) is increasing (monotone).
√
√
Bounded : Since (yn ) is increasing it is bounded below by y1 = p. It is given that 1 + 2 p is an upper
p√
p
√
√
√
√
√
bound, thus yn < 1 + 2 p so that yn+1 = p + yn < p + 1 + 2 p = ( p + 1)2 = p + 1 < 1 + 2 p.
√
Indeed 1 + 2 p is an upper bound. Hence (yn ) is bounded.
Convergence: (yn ) converges
by the Monotone Convergence Theorem.
√
Limit : Again, yn+1 = p + yn is a relation between the nth term of the 1-tail Y1 = (y1+n ) of Y = (yn ) and
Y itself. Since Y converges Y√1 also converges and the have the same
p
√ limit, say y := lim Y . Since lim(y
√ n+1 ) =
p + lim(yn ) we have y = p + y which has roots y = 1/2 ± 1 + 4p/2 from which y = 1/2 + 1 + 4p/2
is the limit of Y .
2 Note that it would have been sufficient to show first that (x ) is increasing, and since x ≥ 2, x ≥ 2 for all n ∈ N; that
n
n
1
is, (x2 ) is bounded below by 2.
11
Problem 31 (Bartle Exercise 6 for Section 3.3). Let a > 0 and let z1 > 1. Define zn+1 :=
n ∈ N. Show that (zn ) converges and find the limit.
√
a + zn for
TODO...
Problem 32 (Bartle Exercise 7 for Section 3.3). Let x1 := a > 0 and xn+1 := xn + 1/xn for nN. Determine
whether (xn ) converges or diverges.
We show that (xn ) diverges by contradiction. Suppose, to the contrary, that (xn ) converges and lim(xn ) =
x. Note that xn+1 = xn + 1/xn is a relation between the nth term of the 1-tail (x1+n ) and (xn ) itself. Since
(xn ) converges (x1+n ) must also converge and converge to the same limit as (xn ); that is, lim(xn+1 ) =
lim(xn ) = x. Thus we have x = x + 1/x from which we obtain 0 = 1/x. Since there is no x ∈ R for which
this is satisfied, this contradicts our assertion that lim(xn ) = 0. Hence (xn ) does not converge.
Problem 33 (Bartle Exercise 8 for Section 3.3). Let A be an infinite subset of R that is bounded above and
let u := sup A. Show that there exists an increasing sequence (xn ) with xn ∈ A such that u = lim(xn ).
We have two cases to consider:
1. u ∈ A: Simply let xn = u for all xn ∈ A.
2. u ∈
/ A: Since u ∈
/ A we have that xn < u for all xn ∈ A. Let x1 ∈ A be arbitrary. Then (x1 + u)/2 < u
so that (x1 + u)/2 is not an upper bound of A, and hence there must exist some x2 ∈ ((x1 + u)/2, u)
such that ((x2 + u)/2) is not an upper bound of A. Continuing inductively there exists xk+1 ∈ A such
that (xk + u)/2 is not an upper bound of A. We conclude that (xk ) in A is increasing. From the
Monotone Convergence Theorem, the sequence (xk ) converges. Let lim(xk ) = x ≤ u. But
xk + u
x+u
≤ xk+1 ⇒
≤x
2
2
which implies that u ≤ x. Since we have that x ≤ u and u ≤ x, we conclude that x = u; that is,
lim(xk ) = sup A
Problem 34 (Bartle Exercise 9 for Section 3.3). Todo...
Problem 35 (Bartle Exercise 10 for Section 3.3). Todo...
Problem 36 (Bartle Exercise 11 for Section 3.3). Todo...
Problem 37 (Bartle Exercise 12 (a) for Section 3.3). Establish the convergence and find the limit of the
following sequence:
n+1 !
1
X = (xn ) =
1+
n
Firstly
n+1 n 1
1
1
1+
= 1+
1+
n
n
n
But, the sequence of each factor converges with limits
n 1
→ e,
1+
n
1
1+
→1
n
Since both of these two sequences converge, the sequence X converges, and since lim(X · Y ) = lim X · lim Y ,
the limit of X is
n+1 !
1
lim
1+
=e
n
12
Problem 38 (Bartle Exercise 12 (b) for Section 3.3). Establish the convergence and find the limit of the
following sequence:
2n !
1
X = (xn ) =
1+
n
Firstly
2n n 2
1
1
=
1+
1+
n
n
Since the sequence ((1 + 1/n)n ) converges with limit lim((1 + 1/n)n ) = e, we have that
2n ! n 2
1
1
lim
1+
= lim
1+
= e2
n
n
Problem 39 (Bartle Exercise 12 (c) for Section 3.3). Establish the convergence and find the limit of the
following sequence:
n 1
X = (xn ) =
1+
n+1
Firstly
1+
1
n+1
n
=
1+
1
n+1
But
1+
Thus
lim
1+
1
n+1
1
n+1
n
n+1 !
n =
·
1 + 1/(n + 1)
(1 + 1/(n + 1))n+1
=
1 + 1/(n + 1)
1 + 1/(n + 1)
→ e,
1+
1
n+1
→1
lim((1 + 1/(n + 1))n+1 )
e
= =e
lim(1 + 1/(n + 1))
1
Problem 40 (Bartle Exercise 12 (d) for Section 3.3). Establish the convergence and find the limit of the
following sequence:
n 1
X = (xn ) =
1−
n
Firstly
1−
Thus
lim
n−1
1
1
1
=
=
=
n
n
n/(n − 1)
1 + 1/(n − 1)
n lim(1)
1
1
=
=
1−
n
lim((1 + 1/(n − 1))n )
e
Problem 41 (Bartle Exercise 13 for Section 3.3). Todo...
Problem 42 (Bartle Exercise 14 for Section 3.3). Todo...
Problem 43 (Bartle Exercise 15 for Section 3.3). Todo...
Problem 44 (Bartle Exercise 16 for Section 3.3). Todo...
1.4
Subsequences and the Bolzano-Weirestrass Theorem
13
Example 1.4.1 (Bartle Page 78). Suppose we have the sequence
1
1 1
, ,..., ,...
X :=
1 2
n
Two examples of subsequences of X include the sequence of even indices
1
1 1
X ′ :=
, ,..., ,...
2 4
2k
and the sequence of odd indices
X ′′ :=
1
1 1
, ,...,
,...
1 3
2k − 1
Example 1.4.2 (Bartle Page 78). The m-tail of a sequence X is a subsequence of X, where the m-tail
corresponds to the sequence of indices
n1 = m + 1, n2 = m + 2, . . . , nk = m + k, . . .
Example 1.4.3 (Bartle 3.4.3 (a) Page 78). Show that the sequence (bn ) and subsequence (b2n ) are
convergent for 0 < b < 1, and that they share the limit
lim(bn ) = lim(b2n ) = 0,
0<b<1
We first show that the sequence converges and that the limit exists before finding the value of the limit.
We make use of the fact that a convergent sequence and its subsequences converge to the same limit.
Convergence: We first show that the sequence converges and that a limit exists. Let xn := bn . Then
0 < xn < 1 so that the sequence is bounded. Also, since 0 < b < 1, we know that xn+1 = bn+1 < bn = xn ,
so the sequence is decreasing (monotone). Hence, by the Monotone Convergence Theorem, the sequence
is convergent, and hence the limit must exist, say x := lim(xn ).
Limit : Consider the subsequence (x2n ) of even indexed terms of (xn ). Since it is a subsequence of (xn ),
it follows from Theorem 3.4.2 in Bartle, that (x2n ) converges and has the same limit as (xn ), namely
lim(x2n ) = lim(xn ) = x. Also, for each n ∈ N we have x2n = b2n = (bn )2 = x2n , from which it follows that
x = lim(x2n ) = lim(x2n ) = [lim(xn )]2 = x2
Solutions to x = x2 include x = 0 and x = 1. However, since the sequence (xn ) is decreasing, and is
bounded above by b < 1, it follows that x = lim(bn ) = 0.
Example 1.4.4 (Bartle 3.4.3 (b) Page 79). Show that3
lim(c1/n ) = 1,
c>1
First we show that the sequence converges and that a limit exists. Let zn := c1/n . Since 0 < 1/n < 1,
we have that zn > 1. Also, zn+1 = c1/n+1 < c1/n = zn . Thus the sequence is decreasing (monotone) and
bounded, so that by the Monotone Convergence Theorem the sequence is convergent and a limit exists,
say lim(zn ) = z.
Consider the subsequence (z2n ) which is also convergent with limit lim(z2n ) = z. However, z2n = c1/2n =
1/2
(c1/n )1/2 = zn for all n ∈ N, so that
z = lim(z2n ) = lim(zn1/2 ) = [lim(zn )]1/2 = z 1/2
14
Thus z 2 = z, and so we have the solutions z = 0 and z = 1. However, since zn > 1 for all n ∈ N, the limit
of the sequence (c1/n ) is lim(c1/n ) = 1.
Example 1.4.5 (Bartle 3.4.6 (a) Page 80). Consider the sequence X := ((−1)n ) = (−1, 1, −1, 1, . . .).
The subsequence of even-indexed terms X ′ := ((−1)2n ) = (1, 1, 1, . . .) converges with limit 1, while the
subsequence of odd-indexed terms X ′′ := ((−1)2n−1 ) = (−1, −1, −1, . . .) converges with limit −1. Since
the two subsequences converge with different limits, the sequence X diverges.
Example 1.4.6 (Bartle 3.4.6 (b) Page 80). The sequence Y := (1, 1/2, 3, 1/4, . . .) diverges since it is not
bounded.
Example 1.4.7 (Bartle 3.4.6 (c) Page 80). TODO...
Example 1.4.8 (Ross Section 11 Example 1 Page 49). Consider the sequence defined by sn = n2 (−1)n .
The sequence is given by (sn ) = (−1, 4, −9, 16, −25, 36, −49, 64, . . .). The positive, or even-indexed terms,
form a subsequence where nk = 2k so that snk = (2k)2 (−1)2k = 4k 2 . Thus, we have the subsequence
s2k = (4, 16, 36, 64, . . .).
Problem 45 (Bartle Exercise 1 for Section 3.4). Give an example of an unbounded sequence that has a
convergent subsequence.
The sequence
1
1
1, , 3, , . . .
2
4
is an unbounded sequence, but the subsequence of even indexed terms
1
1 1
′
, ,...,
,...
X = (x2n ) :=
2 4
2n
X :=
is convergent.
Problem 46 (Bartle Exercise 2 for Section 3.4). Show that
lim(c1/n ) = 1,
0<c<1
We show first that the sequence (c1/n ) with 0 < c < 1 is convergent. Firstly, since 0 < c < 1 we know
that 0 < c1/n < 1 for all n ∈ N so that (c1/n ) must be bounded. Secondly c1/n > c1/n+1 for 0 < c < 1 so that
the sequence is increasing (monotone). By the Monotone Convergence Theorem, the sequence is convergent
and a limit exists, say x := lim(c1/n .
We must find the value of the limit. Consider the subsequence (x2n ) = (c1/2n ) of even-indexed terms of the
sequence (xn ) = (c1/n ). Since (xn ) converges, (x2n ) must also converge and converge to the same limit as
(xn ), namely x. But
lim(x2n ) = lim(c1/2n ) = lim((c1/n )1/2 ) = (lim(c1/n ))1/2
Thus we have that x = x1/2 , fomr which we have the solutions x = 0, x = 1. But (xn ) = (c1/n ) is increasing
so that x = lim(c1/n ) = 1.
Problem 47 (Bartle Exercise 3 for Section 3.4). Let (fn ) be the Fibonacci sequence defined by
f1 := 1,
f2 := 1,
fn+1 := fn−1 + fn for n ≥ 2
Let xn := fn+1 /fn . Given that lim(xn ) = L exists, determine the value of L.
15
From the definition of the Fibonacci sequence fn+1 = fn−1 + fn , from which fn+2 = fn + fn+1 . Then
xn =
fn+2
fn + fn+1
fn
1
fn+1
=
=
=
+1=
+1
fn
fn+1
fn+1
fn+1
xn
Since lim(xn ) = L, we have
1
1
lim(xn ) = lim
+1 ⇔L= +1
xn
L
We have two roots to L2 − L − 1 = 0, but since fn ≥ 1 for all n ∈ N, we choose the positive root
√
1+ 5
L=
2
Problem 48 (Bartle Exercise 4 (a) for Section 3.4). Show that the following sequence is divergent
1
n
(xn ) := 1 − (−1) +
n
We can show that the sequence (xn ) diverges if we can find two subsequences that converge to different
limits.
The subsequence of even-indexed terms is given by
1
1
(x2k ) = 1 − (−1)2k +
=
2k
2k
The subsequence of odd-indexed terms is given by
(x2k−1 ) = 1 − (−1)2k−1 +
1
2k − 1
=
2+
1
2k − 1
But (x2k ) → 0 and (x2k−1 ) → 2. Hence, since the subsequences (x2k ), (x2k−1 ) of the sequence (xn ) converge
to different limits, the sequence (xn ) diverges.
Problem 49 (Bartle Exercise 4 (b) for Section 3.4). Show that the following sequence is divergent
nπ (xn ) := sin
4
Note that
(√
)
√
o
n
nπ
2
2
:n∈N =
, 1, 0, −
, −1
sin
4
2
2
So the subsequence (x8k ) = (sin 2πk) converges to 0, while the subsequence (x8k+2 ) = (sin((8k + 2)/4))
converges to 1. Hence, since the subsequences (x8k ), (x8k+2 ) of the sequence (xn ) converge to different
limits, the sequence (xn ) diverges.
Problem 50 (Bartle Exercise 5 for Section 3.4). Todo...
Problem 51 (Bartle Exercise 6 (a) for Section 3.4). Let (xn ) = n1/n for n ∈ N. Show that xn+1 < xn if
and only if (1 + 1/n)n < n and infer that the inequality is valid for n ≤ 3. Conclude that (xn ) is ultimately
decreasing and that x := lim(xn ) exists.
Firstly xn+1 < xn ⇒ (n + 1)1/(n+1) < n1/n . Then
(n + 1)1/(n+1) < n1/n
(n + 1)n < nn+1
(n + 1)n
<n
n
n n
n+1
<n
n
n
1
1+
<n
n
16
Thus xn+1 < xn ⇔ (1 + 1/n)n < n); that is, the sequence is decreasing (1 + 1/n)n < n. By Example 3.3.6
Bartle we see that (1 + 1/n)n < 3 so that for all n ≥ 3 we have (1 + 1/n)n < 3 ≤ n and thus xn+1 < xn
for all n ≥ 3. Since the 3-tail of (xn ) is decreasing, (xn ) is ultimately decreasing and since (xn ) is bounded
below by x1 = 1, the sequence is convergent by the Monotone Convergence Theorem. Hence x := lim(xn )
exists.
Problem 52 (Bartle Exercise 6 (b) for Section 3.4). Use the fact the the subsequence (x2n ) of the sequence
(xn ) = (n1/n ) in the previous problem also converges to x to conclude that x = 1.
Firstly, x2n := (2n)1/2n = (21/n · n1/n )1/2 , so that lim(x2n ) = [lim(21/n ) · lim(n1/n )]1/2 . If we let
x := lim(n1/n ) then lim(x2n ) = x1/2 . Since (x2n ) is a subsequence of (xn ) it also converges and has the same
limit as (xn ). Thus x = x1/2 , and since x ≥ 1, we have x = 1.
Problem 53 (Bartle Exercise 7 (a) for Section 3.4). Establish the convergence and find the limit of the
2
sequence ((1 + 1/n2 )n ).
Notice that the above sequence is a subsequence of the sequence xn = (1 + 1/n)n , which is convergent
2
with limit e. Thus, the above subsequence, which we may write xnk = (1 + 1/k 2 )k with nk = k 2 , is also
convergent and converges to the same limit:
n 1
lim
1+
→e
n
Problem 54 (Bartle Exercise 7 (b) for Section 3.4). Establish the convergence and find the limit of the
sequence ((1 + 1/2n)n ).
Note that (x2n ) := ((1 + 1/2n)2n ) is the subsequence of even-indexed terms of the convergent sequence
(xn ) := ((1 + 1/n)n ) with limit e. Since (x2n ) is a subsequence
and has the same
p of (xn ) it also converges
√
limit: lim(x2n ) = lim(xn ) = e. Thus lim((1 + 1/2n)n ) = lim( (1 + 1/2n)2n ) = e.
Problem 55 (Bartle Exercise 7 (c) for Section 3.4). Establish the convergence and find the limit of the
sequence ((1 + 2/n)n ).
TODO...
Problem 56 (Bartle Exercise 7 (d) for Section 3.4). Todo...
Problem 57 (Bartle Exercise 8 (a) for Section 3.4). Todo...
Problem 58 (Bartle Exercise 8 (b) for Section 3.4). Todo...
Problem 59 (Bartle Exercise 9 for Section 3.4). Todo...
Problem 60 (Bartle Exercise 10 for Section 3.4). Todo...
Problem 61 (Bartle Exercise 11 for Section 3.4). Todo...
Problem 62 (Bartle Exercise 12 for Section 3.4). Todo...
Problem 63 (Bartle Exercise 13 for Section 3.4). Todo...
Problem 64 (Bartle Exercise 14 for Section 3.4). Todo...
Problem 65 (Bartle Exercise 15 for Section 3.4). Todo...
Problem 66 (Bartle Exercise 16 for Section 3.4). Todo...
1.5
The Cauchy Criterion
17
Example 1.5.1 (Bartle 3.5.2 (a) Page 85). Show that the sequence (1/n) is a Cauchy sequence.
If ε > 0 is given, then we choose H = H(ε) ∈ N such that H > 2/ε. Then, if n, m ∈ N are such that
n, m ≥ H we have n, m ≥ H > 2/ε so that 1/n < 1/H and 1/m < 1/H. Thus
1
− 1≤ 1 + 1 < ε+ ε =ε
n m n m
2 2
Since ε > 0 is arbitrary, (1/n) must be a Cauchy sequence.
Example 1.5.2 (Bartle 3.5.2 (b) Page 85). Show that the following sequence is not a Cauchy sequence
X = (xn ) := (1 + (−1)n )
Firstly, we need to be aware of the negation of the definition of a Cauchy sequence: there exists a ε > 0
such that for every H there exists at least one n > h and one m > h such that |xn − xm | ≥ ε0 .
If n is even, then xn = 1 + 1 = 2, and xn+1 = 1 − 1 = 0. If we choose n > H ∈ N and let m = n + 1, then
|xn − xm | = 2. With ε = 2 we have satisfied the negation of a Cauchy sequence. Hence, the sequence is
not a Cauchy sequence.
Example 1.5.3 (Bartle 3.5.6 (a) Page 87). TODO...
Example 1.5.4 (Bartle 3.5.6 (b) Page 87). TODO...
Example 1.5.5 (Bartle 3.5.6 (c) Page 87). TODO...
Problem 67 (Bartle Exercise 1 for Section 3.5). Give an example of a bounded sequence that is not a Cauchy
sequence.
The sequence ((−1)n ) is an example of a sequence that is bounded (by −1 and 1), but is not a Cauchy
sequence since it does not converge.
Problem 68 (Bartle Exercise 2 (a) for Section 3.5). Show directly from the definition that the following is
a Cauchy sequence.
n+1
X=
n
Let xn := (n + 1)/n = 1 + 1/n. Then if n > m we have |xn − xm | = |1/n − 1/m| ≤ 1/n + 1/m.
If ε > 0 is given, then we may choose H = H(ε) ∈ N such that H > 2/ε. Thus, if n, m ≥ H, then
1/n, 1/m ≤ 1/H < ε/2, which implies that |xn − xm | < ε/2 + ε/2 = ε. Hence, X is a Cauchy sequence.
Problem 69 (Bartle Exercise 2 (b) for Section 3.5). Todo...
Problem 70 (Bartle Exercise 3 for Section 3.5). Show directly from the definition that the sequence
X = (xn ) := ((−1)n )
is not a Cauchy sequence.
18
Briefly, if n is even we have xn = 1 and xn+1 = −1. If we choose n > H ∈ N and let m = n + 1, then
|xn − xm | = 1 + 1 = 2. Hence, the sequence ((−1)n ) is not a Cauchy sequence.
In more detail, we first define xn := (−1)n . Then if n > m we have |xn − xm | = |(−1)n − (−1)m | ≤
|(−1)n | + |(−1)m | = 2. Thus, if we choose ε0 = 2, then for every H(ε0 ) ∈ N there exists n > m ≥ H(ε0 such
that |xn − xm | ≤ ε0 . Hence, ((−1)n ) is not a Cauchy sequence.
Problem 71 (Bartle Exercise 4 for Section 3.5). Show directly from the definition that if (xn ) and (yn ) are
Cauchy sequences, then (xn + yn ) and (xn yn ) are Cauchy sequences.
If ε > 0 is given, then we choose H = H(ε) ∈ N such that H > 2/ε. Since (xn ) and (yn ) are Cauchy
sequences, for every n, m ≥ H we have that |xn − xm | < ε/2 and |yn − ym | < ε/2.
Let zn := xn + yn . Then for all n, m ≥ H we have
ε ε
|zn − zm | = |xn + yn − (xm + ym )| = |(xn − xm ) + (yn − ym )| ≤ |xn − xm | + |yn − ym | < + = ε
2 2
Hence, the sequence (xn + yn ) is a Cauchy sequence.
√
Problem 72 (Bartle Exercise 5 for Section 3.5). If xn := n, show that (xn ) satisfies lim |xn+1 − xn | = 0,
bu that it is not a Cauchy sequence.
√
Firstly, xn+1 − xn = n + 1 − sqrtn. We can find the limit by considering
√
√
√
√
√
√
( n + 1 + n)
1
1
1
( n + 1 − n) · 1 = ( n + 1 − n) · √
√ = √
√ ≤ √ <
n
n
( n + 1 + n)
n+1+ n
We choose a K = K(ε) such that 0 < 1/K < ε. Then for any n ≥ K we have 1/n < ε so that
√
√
|( n + 1 − n) − 0| < ε
√
√
√
√
√
1 − n)
To show
Hence, the sequence ( n + 1 − n) converges to the limit lim( n + √
√ = 0. √
√ that
√ ( n) is
not a Cauchy sequence we need only choose m = 4n, then xm − xn = 4n − n = 2 n − n = n for all
n ∈ N.
Problem 73 (Bartle Exercise 6 for Section 3.5). Todo...
Problem 74 (Bartle Exercise 7 for Section 3.5). Todo...
Problem 75 (Bartle Exercise 8 for Section 3.5). Show directly that a bounded, monotone increasing sequence
is a Cauchy sequence.
Let X = (xn ) be a bounded sequence in R. The Completeness Property of R ensures that the supremum
of the set {xn : n ∈ N} exists, say u := sup{xn : n ∈ N}. Since X is bounded increasing, by the Monotone
Convergence Theorem X converges and lim X = sup{xn : n ∈ R} = u. Then for ε > 0 given, there exists an
H ∈ N such that u − ε < xH ≤ u. If m ≥ n ≥ H then u − ε < xH ≤ xn ≤ u and u − ε < xH ≤ xm ≤ u
where xn ≤ xm . Thus −u < −xm ≤ −u + ε so that
(u − ε) + (−u) < xm − xn < (u) + (−u + ε) ⇒ −ε < xn − xm < ε
Hence |xm − xn | < ε, and since ε > 0 is arbitrary X is a Cauchy sequence.
Problem 76 (Bartle Exercise 9 for Section 3.5). Todo...
Problem 77 (Bartle Exercise 10 for Section 3.5). Todo...
Problem 78 (Bartle Exercise 11 for Section 3.5). Todo...
Problem 79 (Bartle Exercise 12 for Section 3.5). Todo...
Problem 80 (Bartle Exercise 13 for Section 3.5). Todo...
Problem 81 (Bartle Exercise 14 for Section 3.5). Todo...
19