Mathematics 242 – Principles of Analysis Problem Set 4 Due: February 28, 2014

‘A’ Section

1. Using the ‘Big Theorem’ on limits of sequences (Theorem 2.2.5 in the text), find the limit of each of the following:

(a) x_{n} = \frac{(4^{n}+9^{n/2})^{2}}{2^{5n}}

(b) x_{n} = \sqrt{n}(\sqrt{25n + 3} − 5\sqrt{n}) (Hint: \sqrt{a} − \sqrt{b} = \frac{a−b}{\sqrt{a}+\sqrt{b}} – do you see why? Also, you may use the result of question 1 in the B section if needed.)

(c) x_{n} = \frac{1}{n^{4}} ∑_{k=0}^{4} \binom{n}{k} \frac{1}{3^{k}}. (The \binom{n}{k} are the binomial coefficients as in Problem Set 2.)

2. For each of the following statements, first say whether the statement is true. Then, if the statement is true, give a reason; if it is false give a counterexample.

(a) Let {x_{n}} and {y_{n}} be sequences. If it is known that {x_{n}} and {x_{n} − y_{n}} converge, then {y_{n}} converges too.

(b) If {x_{n}} and {y_{n}} both diverge, then so does {x_{n} · y_{n}}.

(c) If x_{n} → 1/1000, then there exists n_{0} ∈ N such that x_{n} > 0 for all n ≥ n_{0}.

(d) If x_{n} > 0 and \frac{x_{n+1}}{x_{n}} ≤ 1 for all n ≥ 1, then x_{n} → 0.

3. For each of the following sequences, first explain why the sequence is monotonic and bounded, then determine the limit:

(a) x_{n+1} = \sqrt{4 + x_{n}}, x_{1} = 1.

(b) x_{n+1} = \frac{3x_{n}+7}{8}, x_{1} = 2.

‘B’ Section

1. Let a > 0. Show, using the ε, n_{0} definition, that if x_{n} → a and x_{n} ≥ 0 for all n, then \sqrt{x_{n}} → \sqrt{a}. (Hint: See part (b) of question 1 in the A section.)

2. In this problem, you will show that the sequence x_{n} = ∑_{k=1}^{n} \frac{1}{k!} converges.

(a) First show that \frac{1}{k!} ≤ \frac{1}{2^{k−1}} for all integers k ≥ 1.

(b) Show that for all n ≥ 1,

1 + \frac{1}{2} + ⋯ + \frac{1}{2^{n−1}} = \frac{1 − \frac{1}{2^{n}}}{1 − \frac{1}{2}}

and deduce that x_{n} is bounded above by 2.

(c) Use part (b) to show that x_{n} converges.

3. In class, we proved the Monotone Convergence Theorem for sequences, the result stated as Theorem 2.3.3 in the text. In the proof we used the Least Upper Bound Axiom in a crucial way. It is an interesting fact that the implication goes the other way too. Namely if we assume that monotone increasing sequences of reals always converge to real numbers, it follows that that every set of reals that is bounded above has a least upper bound. You will show this in this problem using the following construction. Let A be any nonempty set of real numbers that is bounded above. The proof is based on constructing two sequences x_{n} and y_{n} by this inductive procedure:

(i) Let x_{1} ∈ A be any element of A, and let y_{1} be any upper bound for A. (ii) Assuming x_{k} and y_{k} have been constructed such that x_{k} ∈ A and y_{k} is an upper

bound for A, find x_{k+1} and y_{k+1} like this: Let m_{k} = \frac{x_{k}+y_{k}}{2} . If m_{k} is also an upper bound for A, then set x_{k+1} = x_{k} and y_{k+1} = m_{k}. Otherwise, there is some x ∈ A with m_{k} < x ≤ y_{k}, and in this case let x_{k+1} = x and y_{k+1} = y_{k}.

Note that an easy induction argument shows that this construction will produce sequences such that x_{n} ∈ A for all n ≥ 1 and y_{n} is an upper bound for A for all n ≥ 1.

Now prove the following statements to show that A has a least upper bound in R assuming the Monotone Convergence Theorem:

(a) Prove that both sequences {x_{n}} and {y_{n}} converge to elements of R, using the Monotone Convergence Theorem and its Corollary.

(b) Show that lim_{n→∞} x_{n} = lim_{n→∞} y_{n} by showing that |x_{n} − y_{n}| = \frac{|x_{1}−y_{1}|}{2^{n−1}} for all n ≥ 1.

(c) Let α be the common limit of the x_{n} and y_{n} sequences. Show that α = lub(A) by showing it satisfies the definition of a least upper bound.