Problem 10.1.8. The purpose of this problem is to show that12) - In (n + 1))nexists.(a) Let xnshow1-00-1lim 1+ +n→∞2(1 + 1⁄2 + 3 +121133 +· + ¹) − In (n + 1). Use the following diagram toX1 ≤ X2 ≤ X3 ≤…..3+(b) Let zn = ln (n + 1) − ( ¼½ + 3 +that 2₁ ≤ 22 ≤ 23 ≤..4nn+1+ „¹₁). Use a similar diagram to shown+1=(c) Let yn1 - Zn. Show that (n) and (yn) satisfy the hypotheses of thenested interval property and use the NIP to conclude that there is a realnumber y such that n ≤ y ≤ yn for all n.(d) Conclude that limn→∞ ((1 + 2 + 3 + ... + ¹) − In (n + 1)) = y.

In this question, we need to show that: limn→∞1+12+13+⋯+1n-lnn+1 exists. (a) Given that…

Problem 10.1.8. The purpose of this problem is to show that ( ( ₁ + 1/2 + 1 ² + ··· + — — ) - 1+ 3 n exists. (a) Let xn show 1- = 0 lim n→∞ (1 + + + 1 2 X1 ≤ x₂ ≤ x3 ≤... + ¹) − In (n + 1). Use the following diagram to 3 4 In (n + 1) 1)) n n+1 (b) Let zn = ln (n + 1) − ( ½ + 3 + · · · •+1). Use a similar diagram to show that 2₁ 22 23 ≤.... (c) Let yn = 1 - Zn. Show that (xn) and (yn) satisfy the hypotheses of the nested interval property and use the NIP to conclude that there is a real number y such that n ≤ y ≤ yn for all n. (d) Conclude that limn→∞ ((1 + 2 + 3 + · · + ½¹⁄) − In (n + 1)) = y.

Problem 10.1.8. The purpose of this problem is to show that ( ( ₁ + 1/2 + 1 ² + ··· + — — ) - 1+ 3 n exists. (a) Let xn show 1- = 0 lim n→∞ (1 + + + 1 2 X1 ≤ x₂ ≤ x3 ≤... + ¹) − In (n + 1). Use the following diagram to 3 4 In (n + 1) 1)) n n+1 (b) Let zn = ln (n + 1) − ( ½ + 3 + · · · •+1). Use a similar diagram to show that 2₁ 22 23 ≤.... (c) Let yn = 1 - Zn. Show that (xn) and (yn) satisfy the hypotheses of the nested interval property and use the NIP to conclude that there is a real number y such that n ≤ y ≤ yn for all n. (d) Conclude that limn→∞ ((1 + 2 + 3 + · · + ½¹⁄) − In (n + 1)) = y.

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter2: Equations And Inequalities

Section2.6: Inequalities

Problem 80E

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