For all a > 0 and b> 1, the inequalities In (n) ≤n", n < b are true for n sufficiently large (this can be proved usingL'Hôpital's Rule).Let N be this sufficiently large n. Write an inequality comparing (In (n)) and n for n > N.(Express numbers in exact form. Use symbolic notation and fractions where needed.)inequality: 11IncorrectDraw a conclusion about the convergence ofΣBecause00It is not possible to use the Direct Comparison Test to determine the convergence or divergence ofn=1(n(n))81- =1nconverges,00Σ#=1(In (n))*(In (n))also converges by the Direct Comparison Test.Because is a harmonic series, it diverges. Therefore,1100(In (n))(In (n))diverges by the Direct Comparison Test.

The main objective is to write an inequality comparing (ln(n))8 and n. The expression ln n is…

Answered: For all a > 0 and b> 1, the…

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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8:49 3rd Term HomeWork... 2. (a) Given that a b means Va+46 , where the positive root is taken, determine (i) the value ofI*2 (2 marks) (i) whether the operation denoted by * is commutative. Justify your answer. (b) (i) Solve the inequality 3-2x>5. (2 marks) Represent your answer in (b) (i) on the number line shown below. (c) Statement one: Two adult tickets and three children tickets cost $43.00. Statement two: One adult ticket and one ticket for a child cost $18.50. (i) Let x represent the cost of an adult ticket and y the cost of a ticket for a child. Write TWO equations in x and y to represent the information above. (2 marks) (i) Solve the equations to determine the cost of an adult ticket.
D. Suppose real sequences (sn) and (tn) are bounded. (That is, that their ranges are bounded sets.) i. Show the sequence given by (sn + tn) is bounded. ii. For any real number α, show that the sequence (α⋅sn) is bounded.
According to the Bolzano-Weierstrass theorem, every bounded sequence of real numbers has a subsequence that converges to a real number. True or false: Every bounded sequence of integers has a subsequence that converges to an integer. True False
Real Analysis For each n in the positive integers, let Pn=<xn,yn> be a point in R2. Show that {Pn} from n=1 to infinity converges to P=<x,y> in R2 if and only if {xn} from n=1 to infinity and {yn} from n=1 to infinity converge in R1 to x and y respectively. Please be thorough in your answer, explaining each portion to me. Thank you.
Suppose real sequences (sn) and (tn) are bounded (That is, that their ranges are bounded sets.)1. Show the sequence given by (sn + tn) is bounded2. For any real number α, show that the sequence (α⋅sn) is bounded
41. The function A defined by Answer A(x) = 1 + is called an Airy function after the English mathematician and astronomer Sir George Airy (1801-1892). a. Find the domain of the Airy function. Answer x⁹ 2-3+2-3-5-6-2-3-5-6-8-9 b. Graph the first several partial sums on a common screen. c. T Use a computer algebra system that has built-in Airy functions to graph A on the same screen as the partial sums in part (b) and observe how the partial sums approximate A.
Suppose that the sequence x0, x₁, x2... is defined by x0 = 0, x₁ = 7, and xk+2 = xk for k≥0. Find a general formula for xk. Be sure to include parentheses where necessary, e.g. to distinguish 1/(2k) from 1/2k. xk = 0
D. Suppose real sequences (sn) and (tn) are bounded. (That is, that their ranges are bounded sets.)i. Show the sequence given by (sn + tn) is bounded.ii. For any real number α, show that the sequence (α⋅sn) is bounded.
17. Evaluation of proofs See the instructions for Exercise (19) on page 100 from Section 3.1. (a) For each natural number n with n > 2, 2" > 1 +n. Proof. We let k be a natural number and assume that 2k > 1 + k. Multiplying both sides of this inequality by 2, we see that 2k+1 > 2+ 2k. However, 2 + 2k > 2 + k and, hence, 2k+1 > 1+ (k + 1). By mathematical induction, we conclude that 2" >1+ n.
2) Let f be the rational counting function : Solve for integer f(n) = f(5) + f (3)
Genralize if the numbers 1 to n are placed around a circle in some order what is the best possible lower bound on k such that some set of three consecutive number sum to at least k
Determine if the sequence is monotonic and if it is bounded. 2"5" an = n! nal Select the correct answer below and, if necessary, fill in the answer box(es) to complete your choice. OA. (a) is monotonic because the sequence is nonincreasing. The sequence has a least upper bound when n = but is unbounded because it has no lower bound. (Simplify your answer.) OB. (a) is not monotonic. The sequence is unbounded with no upper or lower bound. OC. (a) is not monotonic. The sequence is bounded by a greatest lower bound of when n. (Simplify your answers.) and a least upper bound OD. (a) is monotonic because the sequence is nondecreasing. The sequence has a greatest lower bound of but is unbounded because it has no upper bound. (Simplify your answer.)

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