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Math 140A Elementary Analysis Homework Questions 3
2
Sequences
2.7
Limits of Sequences
2, 3 For each of the following sequences, determine whether it converges. If it does, give the limit.
No proofs are required.
nπ 3n + 1
n
1
2 *
(b) bn =
(c) cn = n
(d) sin
(a) sn =
3n + 1
4n − 1
3
4
3 (b)
(l)
bn =
sin
n2 + 3
n2 − 3
nπ 2
(d)
(n)
tn = 1 +
2
n
sin
2nπ
3
(f)
sn = 21/n
2n +1 + 5
(p)
2n − 7
(h)
dn = (−1)n n
(r)
1
1+
n
(j)
7n3 + 8n
2n3 − 31
(t)
6n + 4
9n2 + 7
2
4 * Give an example of:
(a) A sequence ( xn ) of irrational numbers having a limit lim xn that is a rational number.
(b) A sequence (rn ) of rational numbers having a limit lim rn that is an irrational number.
√
5 (b) Determine the limit lim( n2 + n − n). No proof is required, but show all relevant algebra.
2.8
A Discussion About Proofs
2 Determine the limits of the following sequences, and then prove your claims.
(a)
(d)*
n
+1
2n + 4
dn =
5n + 2
an =
n2
(b)*
(e)*
7n − 19
3n + 7
1
sn = sin n
n
bn =
(c)
cn =
4n + 3
7n − 5
4 * Let (tn ) be a bounded sequence, i.e., there exists M such that |tn | ≤ M for all n, and let (sn ) be
a sequence such that lim sn = 0. Prove that lim(sn tn ) = 0.
8 Prove the following
√
(a) lim[ n2 + 1 − n] = 0
√
(b) lim[ n2 + n − n] = 21
√
(c) lim[ 4n2 + n − 2n] =
1
4
10 * Let (sn ) be a convergent sequence, and suppose that lim sn > a. Prove that there exists a
number N such that n > N implies sn > a.
1
Math 140A Elementary Analysis Homework Answers 3
2.7
Limits of Sequences
2
(a) Yes: lim sn = 0
(b) Yes: lim bn =
3
4
(c) Yes: lim cn = 0
(d) No: Sequence repeats pattern of eight terms ( √1 , 1, √1 , 0, − √1 , −1, − √1 , 0, . . .)
2
2
2
2
3 (b) Yes: lim bn = 1
(d) Yes: lim tn = 1
(f) Yes: lim sn = 1
(h) No: Sequence oscillates ± and absolute value grows unboundedly
(j) Yes: lim =
7
2
(l) No: similar to 2(d)
(n) No: similar to 2(d)
(p) Yes: lim = 2
(r) Yes: lim = 1
(t) Yes: lim = 0
4 Give an example of:
√
(a) xn =
2
n
is a sequence of irrational numbers, with limit 0
(b) There are several ways to answer this question: none are particularly straightforward, but
pick your favorite!
Explicitly Use, for example, Newton’s method for approximating solutions to equations:
r n +1 = r n −
f (r n )
f 0 (r n )
for the function f ( x ) = x2 − 2. This gives the recurrence relation
r n +1 = r n −
rn2 − 2
r2 + 2
= n
2rn
2rn
Starting with x1 = 2, this sequence uses only addition, multiplication and division to
create the future terms: thus all terms of the sequence are rational. It can be shown
that this is a positive, decreasing sequence and so,
√ by an argument from chapter 10,
converges. It is easy to see that the limit must be 2.
Series Many examples come from Taylor series, e.g.
∞
ln 2 =
n
(−1)n+1
(−1) j+1
= lim ∑
n→∞
n
j
n =1
j =1
∑
n
Another example: rn =
∑
j =1
n
i.e. rn =
(−1) j+1
j
j =1
∑
1
π2
has
limit
(often seen in Fourier Analysis).
j2
6
2
√
√
Implicitly Consider the sequence 2 + n1 → 2. For each n ≥ 1 we use the denseness
√ √
of the rational numbers to see that there exists an rational number rn ∈ ( 2, 2 + n1 ).
√
By the squeeze theorem we clearly have rn → 2.
5 (b) lim(
2.8
p
n2 + n − n2
n
1
1
n2 + n − n) = lim √
= lim √
= lim √
=
2
2
−
1
2
n +n+n
n +n+n
1+n +1
A Discussion About Proofs
2
(a) lim an = 0. Let e > 0 be given. Let N = 1e . Then n2 + 1 > n2 , whence
n
n
1
− 0 < 2 = ≤ e
n > N =⇒ 2
n +1
n
n
7
(b) lim bn = 73 . Let e > 0 be given. Let N = max{1, 106
9e − 3 }. Then 3N + 7 ≥
106
3e ,
whence
7n − 19 7 3(7n − 19) − 7(3n + 7) 106
=
n > N =⇒ − =
3n + 7
3
3(3n + 7)
3(3n + 7)
106
<
≤e
3(3N + 7)
(c) lim cn = 47 . Let e > 0 be given. Let N = f rac149e + 57 . Then 7N − 5 ≥
1
7e ,
whence
4n + 3 4 7(4n + 3) − 4(7n − 5) 1
=
n > N =⇒ − = 7n − 5 7
7(7n − 5)
7(7n − 5)
1
<
≤e
7(7N − 5)
16
(d) lim dn = 52 . Let e > 0 be given. Let N = max{1, 25e
− 25 }. Then 5N + 2 ≥
2n + 4 2 5(2n + 4) − 2(5n + 2) 16
=
n > N =⇒ − =
5n + 2 5
5(5n + 2)
5(5n + 2)
16
<
≤e
5(5N + 2)
(e) lim sn = 0. Let e > 0 be given. Let N = 1e . Then
1
1
1
n > N =⇒ sin n − 0 ≤ <
=e
n
n
N
4 Since sn → 0, there exists N such that n > N =⇒ |sn | <
n > N =⇒ |sn tn − 0| = |sn | |tn | ≤ M |sn | < e
Thus lim(sn tn ) = 0.
3
e
.
| M|
But then,
16
5e ,
whence
8
(a) Let e > 0 be given. Let N =
1
2e .
Then
p
n > N =⇒ n2 + 1 − n = √
1
n2
+1+n
<√
1
n2
+n
=
1
1
<
=e
2n
2N
√
Thus lim[ n2 + 1 − n] = 0.
(b) Let e > 0 be given. Let N =
1
8e .
Then
p
n > N =⇒ n2 + n − n −
√
n
1 2n − ( n2 + n + n) 1 √
= √
−
=
2 n2 + n + n 2 2( n2 + n + n ) √
2
n − n + n
n
√
<
=
4n
4n(n + n2 + n)
1
1
1
√
<
=
<
=e
8n
8N
4( n + n2 )
√
Thus lim[ n2 + n − n] = 21 .
(c) Let e > 0 be given. Let N =
1
64e .
Then
p
n > N =⇒ 4n2 + n − 2n −
√
1 1 4n − ( 4n2 + n + 2n) n
√
= √
−
=
4 4n2 + n + 2n 4 4( 4n2 + n + 2n) √
2n − 4n2 + n
n
√
<
=
16n
16n(2n + 4n2 + n)
1
1
<
=e
<
64n
64N
√
Thus lim[ 4n2 + n − 2n] = 14 .
10 Let lim sn = s > a and let e =
n > N =⇒ |sn − s| <
s− a
2 .
Then ∃ N such that
s−a
a−s
s−a
=⇒
< sn − s <
2
2
2
In particular,
sn − s >
a−s
s−a
a−s
=⇒ sn − a >
+s−a =
=e>0
2
2
2
Hence n > N =⇒ sn > a.
4