#2.15) I have to explain each step of the solved problems in the picture. Bounded monotonic sequences2n – 72.15.Prove that the sequence with nth u,(a) is monotonic increasing, (b) is bounded above, (c) isЗп + 2bounded below, (d) is bounded, (e) has a limit.(a) {u„} is monotonic increasing if up+1 > up, n = 1, 2, 3, . . . Now2(п + 1) — 72n – 72n - 5if and only if2n – 73(n + 1) + 2 = 3n + 22n + 5Зп + 2or (2n – 5) (3n + 2) > (2n – 7) (3n + 5), 6n² – 11n – 10 > 6n² – 11n – 35, i.e., –10> -35, which is true. Thus,by reversal of steps in the inequalities, we see that {u,} is monotonic increasing. Actually, since –10>-35, thesequence is strictly increasing.(b) By writing some terms of the sequence, we may guess that an upper bound is 2 (for example). To provethis we must show that u, < 2. If (2n – 7)/(3n + 2) < 2, then 2n – 7 < 6n + 4 or –4n < 11, which is true.Reversal of steps proves that 2 is an upper bound.(c) Since this particular sequence is monotonic increasing, the first term –1 is a lower bound; i.e., u,n = 1, 2, 3, . .. Any number less than – 1 is also a lower bound.-1,(d) Since the sequence has an upper and a lower bound, it is bounded. Thus, for example, we can writeTu,l < 2 for all n.(e) Since every bounded monotonic (increasing or decreasing) sequence has a limit, the given sequence has22 -7/nlimn- 3 + 2/n2n – 7a limit. In fact, limn0 3n + 23

Given a sequence un = 2n-73n+2 a) It is monotonically increasing if un≤un+1 ∀n∈ℕ We need to show…

Answered: 2n – 7 Prove that the sequence with nth…

2n – 7 Prove that the sequence with nth u, = (a) is monotonic increasing, (b) is bounded above, (c) is Зп + 2 bounded below, (d) is bounded, (e) has a limit.

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter10: Sequences, Series, And Probability

Section10.1: Infinite Sequences And Summation Notation

Problem 74E

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Question

#2.15) I have to explain each step of the solved problems in the picture.

Bounded monotonic sequences
2n – 7
2.15.
Prove that the sequence with nth u,
(a) is monotonic increasing, (b) is bounded above, (c) is
Зп + 2
bounded below, (d) is bounded, (e) has a limit.
(a) {u„} is monotonic increasing if up+1 > up, n = 1, 2, 3, . . . Now
2(п + 1) — 7
2n – 7
2n - 5
if and only if
2n – 7
3(n + 1) + 2 = 3n + 2
2n + 5
Зп + 2
or (2n – 5) (3n + 2) > (2n – 7) (3n + 5), 6n² – 11n – 10 > 6n² – 11n – 35, i.e., –10> -35, which is true. Thus,
by reversal of steps in the inequalities, we see that {u,} is monotonic increasing. Actually, since –10>-35, the
sequence is strictly increasing.
(b) By writing some terms of the sequence, we may guess that an upper bound is 2 (for example). To prove
this we must show that u, < 2. If (2n – 7)/(3n + 2) < 2, then 2n – 7 < 6n + 4 or –4n < 11, which is true.
Reversal of steps proves that 2 is an upper bound.
(c) Since this particular sequence is monotonic increasing, the first term –1 is a lower bound; i.e., u,
n = 1, 2, 3, . .. Any number less than – 1 is also a lower bound.
-1,
(d) Since the sequence has an upper and a lower bound, it is bounded. Thus, for example, we can write
Tu,l < 2 for all n.
(e) Since every bounded monotonic (increasing or decreasing) sequence has a limit, the given sequence has
2
2 -7/n
lim
n- 3 + 2/n
2n – 7
a limit. In fact, lim
n0 3n + 2
3

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