MA 4614 Convergence of Sequences Bounded Sequence Questions

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Attached is the full page of my homework, but I only need help for question 7(b), 9, 10, and 11. Handwriting would be good enough! The second one includes some concepts, definitions, and examples for reference if needed.

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Applied Analysis (MA-UY 4614) – Fall 2020
Homework 2 — Due Date: 10/09/2020
(1) Use the definition of convergence of sequences to verify the following limits:
r
1
(−1)n n
1
= 0.
(b) lim n
(a) lim 2
1+ −1 = .
n→∞
n→∞ n + 1
n
2
(2) Let {an } be a sequence in R with lim an = a. If a > 0, prove that there exists N ∈ N
n→∞
such that an > 0 for all n > N .
(3) Let {an } be a sequence in R satisfying |an − an+1 | ≥ c for some constant c > 0 for all
n ∈ N. Prove that {an } diverges.
(4) Let {an } and {bn } be sequences in R. If lim an = 0 and {bn } is bounded, show that
n→∞
lim an bn = 0.
n→∞
an − 1
(5) Suppose lim
= 0. Show that lim an = 1.
n→∞ an + 1
n→∞
√
(6) Let {an } be the sequence defined by an+1 = 2an + 3, a1 = 2. Show that {an } is convergent
and find the limit.
(7) Suppose a1 > a2 > 0. For n ≥ 2, set an+1 = 21 (an + an−1 ). Prove that
(a) {a2k+1 } is increasing and {a2k } is decreasing.
(b) {an } is convergent.
(8) Let A be a nonempty subset of R that is bounded above and let α = sup A. If α ∈
/ A,
prove that α is a limit point of A.
(9) Let {an } be a bounded sequence in R such that every convergent subsequence of {an } has
the same limit a. Prove that {an } converges to a.
(10) Find the upper and lower limits of each sequence:
n 2 − (−1)n n o
n 2n − 1
nπ o
(a)
.
(b)
sin
.
3n + 2
n
6
(11) Which of the following statements imply that {an } converges to α?
1
(a) For every integer m > 0, there is an integer N > 0 such that |an − α| < m when n > N.
(b) For each 0 < ε < 1, there is an integer N > 0 such that |an − α| < 3ε when n > N .
1
(c) For each 0 < ε < 1, there is an integer N > 0 such that |an − α| < ε when n > N .
1
(d) For each N > 0, there is ε > 0 such that |an − α| < N when n > N + ε.
(e) For each ε > 0, there is an integer N > 0 such that |an − α| < ε2 when n > N 2 .
(f) For each ε > 0, there is an integer N > 0 such that |an − α| < ε when n = N + k for all positive integer k. (g) For each ε > 0, there is an integer N > 0 such that |an − α| < ε when n = N + 2k for all positive integer k. Applied Analysis (MA-UY 4614), 2020 Fall CHAPTER 2 — SEQUENCES AND SERIES OF REAL NUMBERS 1. Convergent Sequences 1.1. Convergence. Introduce the concept of convergence. 1.1.1. Sequence of Numbers. A sequence of real numbers is a function f : N → R, pn = f (n) is the n-th term, denoted by {pn }. 1.1.2. Concept of Convergence. If a sequence {pn } approaches to number p as n tends to infinity, then {pn } is said to be convergent and p is the limit, denoted by lim pn = p. n→∞ • • Need a rigorous (quantitative) definition. sequence {pn } approaches to number p as n tends to infinity distance |pn − p| can be very small as n becomes very large Need two measurements. One measures the distance and the other measures the size of n. for any ε > 0, |pn − p| < ε as n > N , there exists positive integer N
1.1.3. Definition. A sequence {pn } is said to converge to number p if for any ε > 0 there exists
positive integer N so that
|pn − p| < ε as n > N.
•
If a sequence is convergent, then its limit is unique.
1.1.4. Examples.
1
1 lim
= 0.
n→∞ n2
For any ε > 0, choose integer N >
2
lim
n→∞
√
√
n + 1000 − n = 0.
√
√1 ,
ε
then as n > N ,
1
1
1
− 0 = 2 < 2 < ε. 2 n n N n + 1000 − √ n= √ 1000 1000 √ < √ . n n + 1000 + n SEQUENCES AND SERIES For any ε > 0, choose integer N >
√
n + 1000 −
√
10002
,
ε2
2
then as n > N ,
1000
1000
n < √ < √ < ε. n N 1.1.5. Questions. Are the following statements true or false? 1 2 3 4 If for any 0 < ε < 1 there is a positive integer N so that |an − a| ≤ 5ε when n > N + 2,
then {an } converges to a.
If for any ε > 0 there is a positive integer N so that |an − a| ≤ 5ε when n = N + 3, then
{an } converges to a.
If for each positive integer k with ε = k1 and N = k there is |an − a| < ε when n = N + 1, then the sequence {an } converges to a. If for each positive integer k with ε = k1 and N = k there is |an − a| < k1 when n = 2N , then the sequence {an } converges to a. 1.1.6. Bounded Sequence. A sequence {pn } is called bounded if there exists a positive constant M so that |pn | ≤ M for all n. • A convergent sequence is bounded. Proof. Suppose pn → p as n → ∞. By the definition of convergence, for ε = 1, there is positive integer N so that as n > N , |pn − p| < 1. Thus |pn | = |pn − p + p| ≤ |pn − p| + |p| < 1 + |p|, as n > N.
Let M = max{|p1 |, . . . , |pN |, 1 + |p|}. Then |pn | ≤ M for all n.

1.2. Limit Theorems.
Study properties of limits.
1.2.1. Limit and Operations. If lim an = a and lim bn = b, then
n→∞
(a)
(b)
(c)
Proof.
n→∞
lim (an + bn ) = (a + b).
n→∞
lim (an · bn ) = (ab).
b
bn
= , where a 6= 0.
lim
n→∞ an
a
n→∞
(b)
For any ε > 0, there exists N > 0 such that
|an − a| < ε, |bn − b| < ε, when n > N.
Then when n > N ,
|an bn − ab| = |an bn − abn + abn − ab| ≤ |an − a||bn | + |a||bn − b| < ε(ε + |b|) + |a|ε, which can be arbitrarily small. SEQUENCES AND SERIES 3 1.2.2. Squeezing Principle. If an ≤ bn ≤ cn , then lim an = lim cn = a n→∞ n→∞ =⇒ lim bn = a. n→∞ Proof. For any ε > 0, there exists N > 0 so that
−ε < an − a < ε, −ε < cn − a < ε. Thus −ε < an − a ≤ bn − a ≤ cn − a < ε. 1.2.3. Examples. 1 1 Show that lim n n = 1. n→∞ lnn = 0. n→∞ n lim 1 Proof. Let an = n n − 1 ≥ 0. Then 1 1 n = (1 + an )n = 1 + nan + n(n − 1)a2n + · · · + an ≥ n(n − 1)a2n . 2 2 Thus 2 12 0 ≤ an ≤ . n−1 Then use the squeezing principle. 2 n3 Show that lim n = 0. n→∞ 2 Proof. By n n(n − 1) · · · (n − 3) n4 2 = (1 + 1) >
=
> 4
4
4!
2 4!
n
because n − 3 > 2 when n > 6. Thus
n
n
0< n3 1 < 24 4! . n 2 n 3 3n = 0. n→∞ n! Show that lim Proof. Use 3n 43 3 n 3n 3n 0< = < = . n! 3!4 · 5 · · · n 3!4n−3 3! 4 p 4 (lnn) = 0, when p > 0, α > 0.
n→∞
nα
Show that lim
Proof. Use Calculus,

p

p
p
p p
lnt α
(lnn)p
lnn
p
lnt
p
1
lim
= lim α
= lim
=
lim
=
lim
=0
n→∞
n→∞ n p
t→∞ t
nα
α t→∞ t
α t→∞ t

SEQUENCES AND SERIES
5
Show that lim

n→∞
4
1 n
1 + 2 = 1.
n
Proof. Use Calculus,

1
ln(1 + t2 )
2t
lim n ln 1 + 2 = lim+
= lim+
= 0.
n→∞
t→0
t→0 1 + t2
n
t

2. Monotone Sequences and Subsequences
2.1. Monotone-Bounded Principle.
It gives the convergence of a monotone and bounded sequence.
2.1.1. Monotone sequence.
•
•
Increasing sequence: a1 ≤ a2 ≤ . . . ≤ an ≤ . . .
Decreasing sequence: a1 ≥ a2 ≥ . . . ≥ an ≥ . . .
2.1.2. Monotone-Bounded Principle. A monotone and bounded sequence is convergent.
Proof. Assume that {an } is increasing and bounded above. Let a = sup{an : n ∈ N}. We show
that an → a as n → ∞.
For any ε > 0, since a−ε is not an upper bound of {an }, there exists N > 0 so that aN > a−ε.
Since {an } is increasing, we have when n > N ,
an ≥ aN > a − ε.
Thus
a − ε < an < a + ε, when n > N,
that is,
|an − a| < ε, when n > N.

2.1.3. Examples.
1
2
1 If sequence {an } is defined by an+1 = 3 an + 2 , a1 = 1, find the limit of {an }.
Solution. First, we show that the limit exists. It is easily seen that a2 =
a1 < a2 < a3 < 2. Use induction to show that an < an+1 < 2, 7 , 6 a3 = for all n. It is true when n = 1. Suppose it is true when n = k, that is, ak < ak+1 < 2. Then 1 2 1 2 1 11 2 < 2. ak+1 = ak + < ak+1 + = ak+2 < · 2 + = 3 2 3 2 3 2 6 Thus, it true for n = k + 1. The limit exists. 23 , 18 and SEQUENCES AND SERIES 5 Letting a be the limit and n → ∞, we have 2 1 a= a+ . 3 2 Solving the equation gives a = 32 . 2 If sequence {an } is defined by an+1 = √ 2an + 12 , a1 = 1, find the limit of {an }. Solution. First, we show that the limit exists. It is easily seen that a2 = 1.91 . . ., a3 = 2.45 . . ., and a1 < a2 < a3 < 3. Use induction to show that an < an+1 < 3, for all n. It is true when n = 1. Suppose it is true when n = k, that is, ak < ak+1 < 3. Then √ √ 1 p 1 1 ak+1 = 2ak + < 2ak+1 + = ak+2 < 2 · 3 + < 3. 2 2 2 Thus, it true for n = k + 1. The limit exists. Letting a be the limit and n → ∞, we have √ 1 a = 2a + . 2 √ Solving the equation gives a = 2 + 32 . 3 The sequence {an } is defined by an+1 = 3an − 1, a1 = 1. Find the limit of {an }. Solution. Suppose an → a as n → ∞. Letting n → ∞ gives a = 3a − 1. Then a = 12 . But the sequence 2, 5, 14, 41, . . . is divergent. Thus it is important to show the convergence. 2.2. Bolzano-Weierstrass Theorem. It shows the existence of a convergent subsequence. 2.2.1. Subsequence. A sequence {ank }, n1 < n2 < . . ., is called a subsequence of {an }. For example, {a2k } and {a2k−1 }. • A sequence is convergent if and only if every subsequence is convergent. • Sequence {an } is convergent if and only if subsequences {a2k } and {a2k−1 } are convergent to the same limit. Proof. The necessary condition is obvious. Suppose a2k → a and a2k−1 → a. For any ε > 0,
there is N > 0 such that
|a2k − a| < ε, |a2k−1 − a| < ε, when k > N.
Thus,
|an − a| < ε, when n > 2N.
SEQUENCES AND SERIES
6

2.2.2. Nested Intervals Theorem. If {In } is a sequence of closed and bounded intervals with
∞

In+1 ⊂ In for all n ∈ N, then
In 6= ∅.
n=1
Proof. Let In = [an , bn ]. Then an ≤ an+1 ≤ bk and bn ≥ bn+1 ≥ ak for any n, k. Thus, {an } is
increasing and every bk is an upper bound; {bn } is decreasing and every ak is a lower bounded.
Therefore, {an } and {bn } are convergent, say to a and b, respectively. Then, a ≤ b. Thus,
[a, b] ⊂ In , n ∈ N, which implies
∞

[a, b] ⊂
In .
n=1

2.2.3. Bolzano-Weierstrass Theorem. Every bounded sequence has a convergent subsequence.
Proof. Let {an } be a bounded sequence. Then |an | ≤ M for some constant M > 0. We construct
a convergent subsequence {ank }:
•
•
•
Bisect interval [−M, M ]. At least one of the two sub-intervals contains infinitely many
terms of {an }. Denote it by I1 and pick an1 ∈ I1 .
Bisect I1 . Again, at least one of the two sub-intervals contains infinitely many terms of
{an }. Denote it by I2 and pick an2 ∈ I2 so that n2 > n1 .
Continuing the process, we get a subsequence {ank } satisfying:
M
ank ∈ Ik , I1 ⊃ I2 ⊃ · · · ⊃ Ik ⊃ · · · , |Ik | = k .
2
T∞
By the nested intervals theorem, k=1 Ik = {a} because |Ik | → 0. Since ank , a ∈ Ik ,
|ank − a| ≤
M
.
2k
Thus, ank → a as k → ∞.

2.2.4. Limit Point. Let E ⊂ R. A point p ∈ R is called a limit point of E if for every ε > 0
the interval (p − ε, p + ε) contains a point q ∈ E with q 6= p.
2.2.5. Proposition. A point p is a limit point of E if and only if there is a sequence {pn } of
different numbers in E that converges to p.
Proof. If p is a limit point of E, construct {pn } as follows:
• Let ε1 = 1 and pick p1 ∈ E so that 0 < |p1 − p| < ε1 . • Let ε2 = 1 |p1 − p| ≤ 1 and pick p2 ∈ E so that 0 < |p2 − p| < ε2 . 2 2 • Let ε3 = 1 |p2 − p| ≤ 12 and pick p3 ∈ E so that 0 < |p3 − p| < ε3 . 2 2 • Continuing the process, we get a sequence pn ∈ E so that 1 1 |pn+1 − p| < |pn − p| ≤ n → 0. 2 2 The converse is obviously true. SEQUENCES AND SERIES 7 2.2.6. Examples. Let E 0 be the set of all limit points of E. 1 2 3 4 5 6 E E E E E E = (0, 1), E 0 = [0, 1]. = {(−1)n : n ∈ N}, E 0 = ∅. = (−1)n n−1 : n ∈ N}, E 0 = {−1, 1}. n n−1 : n ∈ N}, E 0 = − √12 , √12 , −1, 1 . = n sin nπ 4 sin nπ : n ∈ N}, E 0 = ± sin kπ : k = 1, . . . , 8 . = { n−1 n 17 17 = Q ∩ (0, 1) = 12 , 31 , 23 , 14 , 34 , 51 , 52 , 35 , 45 , 16 , 65 , 71 , 27 , 37 , 74 , 57 , 67 , 18 , 83 , . . . . E 0 = [0, 1]. 2.2.7. Isolated Point. A point p ∈ E is called an isolated point of E if it is not a limit point of E. 2.2.8. Example. E = (−1)n n−1 : n ∈ N}. Every point in E is isolated. n 2.3. Upper and Lower Limits. Study the limits of convergent subsequences. 2.3.1. Bounds for limits of convergent subsequences. Let {pnk } be a convergent subsequence of {pn } with limit p0 . Obviously, inf{pn : n ≥ 1} ≤ p0 ≤ sup{pn : n ≥ 1}. Since p0 does not depend on first finitely many terms of {pn }, we have ak = inf{pn : n ≥ k} ≤ p0 ≤ sup{pn : n ≥ k} = bk , where bk and ak are finite numbers if and only if {pn } is bounded above and below, respectively, and in this case, bk is decreasing and ak is increasing. 2.3.2. Definition. The upper limit of {pn } is defined as lim pn = lim bk = inf sup{pn : n ≥ k}. n→∞ k→∞ k∈N The lower limit of {pn } is defined as lim pn = lim ak = sup inf{pn : n ≥ k}. n→∞ k→∞ k∈N 2.3.3. Theorem. Suppose lim pn < ∞. Then β = lim pn if and only if n→∞ (a) (b) n→∞ for every ε > 0 there is N > 0 so that pn < β + ε when n > N .
there is a subsequence pni → β.
Proof. (a) Since bk → β, for every ε > 0 there is N so that β − ε < bk < β + ε when k > N .
Thus, pn ≤ bk < β + ε, when n ≥ k > N .
(b) For εi > 0, by (a), we have pn < β + εi , n > Ni . By the definition of supremum, there is
pni , ni > Ni , so that pni > bNi +1 − εi ≥ β − εi . Thus, β − εi < pni < β + εi . Letting εi → 0 gives pni → β. This shows the necessary conditions. To show the sufficiency, need to show β = lim pn . n→∞ SEQUENCES AND SERIES By (a), 8 for ε > 0, there is N so that
bk = sup{pn : n ≥ k} < β + ε, Thus, β ≥ lim pn . n→∞ (b) when k > N.
implies that β ≤ lim pn .

n→∞
2.3.4. Corollary 1. lim pn = β if and only if
n→∞
lim pn = lim pn = β.
n→∞
n→∞
2.3.5. Corollary 2. lim pn is the largest limit of convergent subsequences, and lim pn is the
n→∞
n→∞
smallest limit of convergent subsequences.
2.3.6. Examples.
1
2
3
4
5
pn =
n−1
.
n
lim pn = lim pn = 1.
n→∞
n
n→∞
lim pn = ∞, lim pn = 0.
n→∞
n→∞
√
√
3
3
pn = n−1
, lim pn = −
.
sin nπ
. lim pn =
n
3
n→∞
2
2
n→∞
8π
8π
pn = n−1
sin nπ
. lim pn = sin , lim pn = − sin .
n
17
n→∞
17 n→∞
17
Let {pn } be 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5, . . . The set of limits of convergent subsequences is N. The set of limit points is empty.
pn = (1 + (−1) )n.
3. Cauchy Sequences
Discuss completeness of real numbers.
3.0.1. Definition. A sequence {an } of numbers is called a Cauchy sequence if for every ε > 0,
there is N so that
|an − am | < ε, for all n, m ≥ N. 3.0.2. Theorem. A sequence of real numbers is a Cauchy sequence if and only if it is convergent. Proof. Sufficiency. Suppose that {an } is convergent to a. Then for ε > 0, there is N so that
ε
|an − a| < , when n ≥ N. 2 Therefore, ε ε |an − am | ≤ |an − a| + |a − am | < + = ε, when n, m ≥ N. 2 2 Necessity. Suppose that {an } is a Cauchy sequence. First, show that it is bounded. For ε = 1, there is a positive integer N1 so that |an − am | < 1, for all n, m ≥ N1 . SEQUENCES AND SERIES 9 Letting m = N1 gives |an | < 1 + |aN1 | when n ≥ N . Thus |an | ≤ M where M = max{1 + |aN1 |, |a1 |, . . . , |aN1 |}. for all n, Then by the Bolzano-Weierstrass theorem, {an } has convergent subsequence {ank }. Let ank → a. We show that an → a. Since {an } is Cauchy and ank → a, for ε > 0, there is N so that
|an − am | < ε when n, m > N,
and |ank − a| < ε when k > N.
Thus,
|an − a| ≤ |an − ank | + |ank − am | < 2ε, when n > N.

4. Series of Numbers
4.1. Convergence of Series.
∞
n
∞
X
X
X
an is said to be
ak be the partial sum sequence. Series
an be a series, and sn =
Let
n=1
n=1
k=1
convergent if its partial sum sequence {sn } is convergent.
4.1.1. Examples.
1
Geometric Series.
sn =
2
1−rn
1−r
∞
X
rn−1 = 1 + r + r2 + · · · ,
0 < r < 1. It is convergent because n=1 → 1 . 1−r ∞ X 1 Harmonic Series. . It is divergent. n n=1 s2k = s2k−1 + 1 2k−1 +1 + ··· + 1 1 1 > s2k−1 + 2k−1 · k = s2k−1 + .
k
2
2
2
k
,
2
2
Thus, s2k > and hence sn is not bounded.
∞
X
1
p-series.
. It is convergent when p > 1 and divergent when p ≤ 1.
np
n=1
4.1.2. Cauchy Criterion. Series
∞
X
an is convergent if and only if for every ε > 0 there is N
n=1
so that
m
X
ak < ε for all m ≥ n ≥ N. k=n Proof. It is equivalent to that the partial sum sequence {sn } is a Cauchy sequence. 4.1.3. Necessary Condition of Convergence. If ∞ X an is convergent, then lim an = 0. n=1 Proof. Letting m = n in the Cauchy criterion gives |an | < ε when n ≥ N . n→∞ SEQUENCES AND SERIES 4.1.4. Proposition 1. If an ≥ 0, then ∞ X 10 an is convergent if and only if {sn } is bounded. n=1 Proof. When an ≥ 0, {sn } is an increasing sequence. By the monotone-bounded principle, {sn } is convergent if and only if it is bounded. ∞ X 1 4.1.5. Example. The p-series is convergent when p > 1 and divergent when p ≤ 1.
np
n=1
When p > 1,
Z n
n Z k
n
X
X
1
1
p
1
1
1
1 − p−1 < . ≤1+ dx = 1 + dx = 1 + p p p k x p−1 n p−1 1 x k=2 k−1 k=1 When p ≤ 1 n n X X 1 1 → ∞. >
p
k
k
k=1
k=1
4.1.6. Proposition 2. If
∞
X
|an | is convergent, then
∞
X
an is convergent.
n=1
n=1
Proof. Use the Cauchy criterion and the inequality
m
X
ak ≤
k=n
m
X
|ak |.
k=n

4.1.7. Proposition 3. If
∞
X
an and
∞
X
bn are convergent, then
n=1
n=1
∞
∞
∞
X
X
X
bn .
an + c 2
(c1 an + c2 bn ) = c1
n=1
n=1
n=1
5. Convergence Tests of Series
5.1. Series of Positive Terms.
5.1.1. Comparison Test.
(a)
If 0 ≤ an ≤ bn , then
∞
X
bn < ∞ implies n=1 (b) ∞ X an < ∞, and n=1 ∞ X an = ∞ implies n=1 ∞ X n=1 bn = ∞. ∞ ∞ X X an If an , bn > 0, and lim
= L > 0, then
an < ∞ if and only if bn < ∞. When n→∞ bn n=1 n=1 ∞ ∞ X X L = 0, bn < ∞ implies an < ∞. n=1 The proof of (a) n=1 follows from the Cauchy criterion, and (b) follows from (a). SEQUENCES AND SERIES 5.1.2. Examples. ∞ X n + 1 21 1 2 3 4 5 n=1 ∞ X 2n4 +3 . n+1 2n4 +3 12 ∼ 1 3 n2 11 . Convergent. . sin √2n sin √2n √ √ . 3 n+2 n1 → 23 . Divergent. 3 n + 2 n=1 P Suppose n an , an > 0, converges. Determine the convergence of the series:
X
a2n . a2n < an because an → 0. Convergent. n X √ an . May or may not be convergent. n X√ √ √ √ an sin an . an sin an /an → 1. Convergent. n 6 7 8 X 1+ 7 p 4 an n −1 (1+x)7 −1 x = 7 and compare with 15 . n4 n + a n n 1 Suppose 0 < an < √n . Determine the convergence of the series: √ ∞ X √ an √ . n+a√nn < 15 . Convergent. n4 n+ n n=1 ∞ X an + 1 an +1 √ > 1 . Divergent.
√ . n+
2n
n
n
+
n
n=1
. Use limx→0
5.1.3. Integral Test. If f (x) ≥ 0 is decreasing in [1, ∞), then
Z ∞
∞
X
f (n) < ∞ if and only if f (x) dx < ∞. 1 n=1 Proof. Since f (x) is decreasing, the following inequalities hold, Z n Z n+1 n X f (k) ≤ f (1) + f (x)dx. f (x)dx ≤ 1 1 k=1 5.1.4. Example. Determine the convergence of the series ∞ X n=2 Solution. Z 2 n    1 1−p 1 . n(lnn)p 1−p 1−p (lnn) − (ln2) p 6= 1 1 dx =  x(lnx)p ln(lnn) − ln(ln2) p = 1. which is convergent when p > 1 and divergent when p ≤ 1.
5.1.5. Ratio and Root Tests. For series
∞
X
n=1
(a)
r < 1 =⇒ ∞ X n=1 an converges. 1 an+1 or r = lim ann . Then n→∞ an n→∞ an , an > 0, let r = lim
SEQUENCES AND SERIES
(b)
r > 1 =⇒
∞
X
12
an diverges.
n=1
1
Proof.
(a)
Suppose r = lim ann . Then for ε > 0 so that 0 < r + ε < 1, there is N > 0 such that
n→∞
1
ann < r + ε, when n > N.
Thus
an < (r + ε)n , when n > N.
By comparison with the geometric series, it is convergent.
1
(b) can be proved similarly by using 1 < r − ε < ann when n > N and comparison.

Ratio and root tests can be applied to test series that are exponentially decreasing, but can
not be used to test series that are polynomially decreasing.
5.1.6. Example. Find the values of x such that the series
∞
X
2n
n=1
n2
xn is convergent.
2n n n1
1
x
→ 2|x| = r < 1. |x| < . 2 n 2 P∞ 1 P∞ (−1)n 1 When r = 1, x = ± 2 , the series n=1 n2 and n=1 n2 are convergent. Thus, |x| ≤ 21 . Gaoyong Zhang, Department of Mathematics, Courant Institute of Mathematical Sciences, New York University. E-mail address: gaoyong.zhang@nyu.edu Purchase answer to see full attachment Tags: convergent Bounded sequence odd number convergent subsequence Arbitray element User generated content is uploaded by users for the purposes of learning and should be used following Studypool's honor code & terms of service.