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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.

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.

Advanced Engineering Mathematics
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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#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
Transcribed Image Text: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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