I have attached the solution my question is 1b) In the solution I don’t understand why it said if n belongs to z with n>=0And then it says if n belongs to z with n property¹ there exists n E N such that n >doesn't exist.1b) If n E Z with n ≥ 0, then 2n + 1 ≥ 1 and12n +1If ne Z with n < -1, then 2n +1 < -1 and0 <-1 ATherefore, for all ne Z we have -1 <1-E2€11c) For all n EN\ {0}, we see that n -n21:12n + 112n + 11have= 1, then 1 = max B = sup B. Since for n = -1 we have2n +1-1 = min B = inf B.<1.20; sinceTherefore 0 inf A. Since 0 A, min A< 0.<1; then B is bounded.1n< 1, we have1n--> 1-n- 1/120.nSince for n = 0 we12n + 1= -1, thenTherefore C is bounded from below. Since for n = 1, n¹Archimedean property. Let a, b e R, a, b>0. Then there exists neN such that na > b.2017 Politecnico di Torino= 0, we have 0 = min C = inf C. Exercise 1. Tell whether the following sets are bounded from below and/or from above, specifyinginfimum, supremum and minimum and maximum, if they exist.{2+1 neN}.2n1b) B=nez}.2n +1c) C = {n-1: ne N\ {0}}.(-1)0 = {1+ (=17: n=N}.n+1a) A =d) D:e) E=:E = {neN: 2+1€N}.E2nn-1ƒ) F = {[1 + (-1)³] ¹ = ¹; n € N\ {0}}.:n19) G={₁2: neN\ {0}}.g)1-2-nh) H = {-n: neN}.n²i) I =91-{nEN: 32 +3 >1}.3n+1ne N\{0}}.13) J = {2 cos (nm) + : ne Nj)n

Answered: Image /qna-images/answer/c3bd54e6-e580-4614-a46c-b5de747910d5.jpg

Answered: = {₁ b) B= 1 2n +1 : nez}.

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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Question

I have attached the solution my question is 1b) In the solution I don’t understand why it said if n belongs to z with n>=0 And then it says if n belongs to z with n

property¹ there exists n E N such that n >
doesn't exist.
1b) If n E Z with n ≥ 0, then 2n + 1 ≥ 1 and
1
2n +1
If ne Z with n < -1, then 2n +1 < -1 and
0 <
-1 A
Therefore, for all ne Z we have -1 <
1-E
2€
1
1c) For all n EN\ {0}, we see that n -
n
21:
1
2n + 1
1
2n + 1
1
have
= 1, then 1 = max B = sup B. Since for n = -1 we have
2n +1
-1 = min B = inf B.
<1.
20; since
Therefore 0 inf A. Since 0 A, min A
< 0.
<1; then B is bounded.
1
n
< 1, we have
1
n--> 1-
n
- 1/120.
n
Since for n = 0 we
1
2n + 1
= -1, then
Therefore C is bounded from below. Since for n = 1, n
¹Archimedean property. Let a, b e R, a, b>0. Then there exists neN such that na > b.
2017 Politecnico di Torino
= 0, we have 0 = min C = inf C.

Exercise 1. Tell whether the following sets are bounded from below and/or from above, specifying
infimum, supremum and minimum and maximum, if they exist.
{2+1 neN}.
2n
1
b) B=
nez}.
2n +1
c) C = {n-1: ne N\ {0}}.
(-1)
0 = {1+ (=17: n=N}.
n+1
a) A =
d) D
:
e) E=
:
E = {neN: 2+1€N}.
E
2n
n-1
ƒ) F = {[1 + (-1)³] ¹ = ¹; n € N\ {0}}.
:
n
1
9) G={₁2: neN\ {0}}.
g)
1-2-n
h) H = {-n: neN}.
n²
i) I =
91-{nEN: 32 +3 >1}.
3n+1
ne N\{0}}.
1
3) J = {2 cos (nm) + : ne N
j)
n

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no I understand what you meant but that was not my question when he divided it into two parts I and II how can i know that it is the correct way to divide these two parts why can it not be n >= 0 and n<= -3 for example how can i know those values for sure.

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