#7 c Section 4.1 • Convergenc=1 EXERCISESExercises marked with * are used in later sections, and exercises markehave hints or solutions in the back of the book.1. Mark each statement Trục or False. Justify each answer.(a) If (s,) is a sequence and s; =S, then i =j.(b) If s, → s, then for every e> 0 there exists Ne N such that n2|s, - s|< E.(c) If s, → k and 1, → k, then s, = t, for allne N.(d) Every convergent sequence is bounded.2. Mark each statement True or False. Justify each answer.(a) If s, → 0, then for every ɛ > 0 there exists Ne N such that n2.Sn < E.(b) If for every ɛ > 0 there exists Ne N such that n>N implies smS→ 0.(c) Given sequences (s,) and (a,), if for some s e R, k> 0 and -have |s, - s| < kla, for all n> m, then lim s, = s.(d) If s, → s and s, →1, then s = t.3. Write out the first seven terms of each sequence.(-1)"(b) b, =(а) а, — п? *(c) C, = coS32n +1(d) d, =3n -14. Find k> () and m eN so that 6n' + 17n s kn' for all integers n2 m.5. Find k> 0 and m e N so that n - 7n 2 kn' for all integers n2 m.6. Using only Definition 4.1.2, prove the following.(a) For any real number k, lim, z (k/n) = 0.*(b) For any real number k > 0, lim,, →» (1/n) = 0.(d) lim sin n= 04n+1(c) lim= 4n+3(e) limn+3= 0(f) lim-= 0n +n-3n+2-137. Using any of the results in this section, prove the following.5n - 61(а) lim-(b) lim03-2n - 7n2+3n0 =6n +3n= 3=0 *n+1(с) lim(d) lim2n -5

Name: #7 c Section 4.1 • Convergenc=1 EXERCISESExercises marked with * are u
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Description: #7 c Section 4.1 • Convergenc=1 EXERCISESExercises marked with * are used in later sections, and exercises markehave hints or solutions in the back of the book.1. Mark each statement Trục or False. Justify each answer.(a) If (s,) is a sequence and s;

we know, lim(1/n) =0 if n is tend to infinity

Answered: 6n +3n = 3 2n² -5 (c) lim

Math

Advanced Math

6n +3n = 3 2n² -5 (c) lim

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Concept explainers

Percentage

A percentage is a number indicated as a fraction of 100. It is a dimensionless number often expressed using the symbol %.

Algebraic Expressions

In mathematics, an algebraic expression consists of constant(s), variable(s), and mathematical operators. It is made up of terms.

Numbers

Numbers are some measures used for counting. They can be compared one with another to know its position in the number line and determine which one is greater or lesser than the other.

Subtraction

Before we begin to understand the subtraction of algebraic expressions, we need to list out a few things that form the basis of algebra.

Addition

Before we begin to understand the addition of algebraic expressions, we need to list out a few things that form the basis of algebra.

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Question

#7 c

Section 4.1 • Convergenc=
1 EXERCISES
Exercises marked with * are used in later sections, and exercises marke
have hints or solutions in the back of the book.
1. Mark each statement Trục or False. Justify each answer.
(a) If (s,) is a sequence and s; =S, then i =j.
(b) If s, → s, then for every e> 0 there exists Ne N such that n2
|s, - s|< E.
(c) If s, → k and 1, → k, then s, = t, for allne N.
(d) Every convergent sequence is bounded.
2. Mark each statement True or False. Justify each answer.
(a) If s, → 0, then for every ɛ > 0 there exists Ne N such that n2.
Sn < E.
(b) If for every ɛ > 0 there exists Ne N such that n>N implies sm
S→ 0.
(c) Given sequences (s,) and (a,), if for some s e R, k> 0 and -
have |s, - s| < kla, for all n> m, then lim s, = s.
(d) If s, → s and s, →1, then s = t.
3. Write out the first seven terms of each sequence.
(-1)"
(b) b, =
(а) а, — п? *
(c) C, = coS
3
2n +1
(d) d, =
3n -1
4. Find k> () and m eN so that 6n' + 17n s kn' for all integers n2 m.
5. Find k> 0 and m e N so that n - 7n 2 kn' for all integers n2 m.
6. Using only Definition 4.1.2, prove the following.
(a) For any real number k, lim, z (k/n) = 0.
*(b) For any real number k > 0, lim,, →» (1/n) = 0.
(d) lim sin n
= 0
4n+1
(c) lim
= 4
n+3
(e) lim
n+3
= 0
(f) lim-
= 0
n +n-3
n+2
-13
7. Using any of the results in this section, prove the following.
5n - 6
1
(а) lim-
(b) lim
03-
2n - 7n
2+3n
0 =
6n +3n
= 3
=0 *
n+1
(с) lim
(d) lim
2n -5

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