MATH 242 – Principles of Analysis Solutions for Exam 2 – April 1, 2011

I. A) (20) State and prove the Monotone Convergence Theorem for sequences. (You may

give the proof in the case that the sequence is monotone increasing.)

Solution: The Theorem says that every bounded monotone sequence in R converges to some real number a. If {x_{n}} is monotone increasing, let a = lub({x_{n} | n ≥ 1}). then for all ε > 0, a − ε is not an upper bound for {x_{n} | n ≥ 1}. Hence there exists some n_{0} such that a − ε < x_{n_{0}} ≤ a. Since the sequence is monotone increasing, for all n ≥ n_{0}, we have a − ε < x_{n_{0}} ≤ x_{n} ≤ a. But this implies that |x_{n} − a| < ε for all n ≥ n_{0}. Hence x_{n} → a.

B) (10) Suppose that x_{n} is a sequence of strictly negative numbers and x_{n+1}/x_{n} ≤ 1 for all n ≥ 1. Show that lim_{n→∞} x_{n} = a exists in R and satisfies a ≤ 0.

Solution: Since the x_{n} < 0, from the inequality x_{n+1}/x_{n} ≤ 1, we get x_{n+1} ≥ x_{n} for all n ≥ 1. (The inequality reverses when we multiply by the negative number x_{n}.) This implies the sequence is monotone increasing. Since it is also bounded above by 0, part (A) shows that x_{n} → a for some a ≤ 0.

II. (20) Suppose {x_{n}} is a sequence such that |x_{n} − 10| < 100 for all n ≥ 1. Show that there exists some number a ∈ [−90, 110] and a subsequence {x_{n_{k}}} such that x_{n_{k}} → a. State any “big theorems” you are using.

Solution: Since |x_{n} − 10| < 100 for all n, it follows that −100 < x_{n} − 10 < 100, so −90 < x_{n} < 110. Hence |x_{n}| ≤ 110 for all n, so this is a bounded sequence. By the Bolzano-Weierstrass theorem, there must be a convergent subsequence x_{n_{k}} → a for some a ∈ R. The limit must also lie in the interval [−90, 110] because of the “order limit theorem” (Corollary 2.2.8 in our text). This result says: If a ≤ x_{n} ≤ b for all n, and x_{n} → c, then a ≤ c ≤ b also.

III. (20) Show using the ε, δ definition that lim_{x→2} x^{2} − x = 2.

Solution: (“prep”: |x^{2} − x − 2| = |x − 2||x + 1|. If |x − 2| < 1, then 1 < x < 3, so |x + 1| satisfies 2 < |x + 1| < 4.)

proof: Given ε > 0, let δ = min (1, \frac{ε}{4}). Then for all x with 0 < |x − 2| < δ, 1 < x < 3, so |x + 1| < 4. Therefore, for all such x,

|x^{2} − x − 2| = |x − 2||x + 1| < \frac{ε}{4} · 4 = ε.

This shows lim_{x→2} x^{2} − x = 2.// IV. True/False. For each true statement, give a short proof. For each false statement, give a counterexample or reason.

A) (10) If {x_{n}} diverges, then for all a ∈ R, there exist ε_{0} > 0 and n_{0} ∈ N such that |x_{n} − a| ≥ ε_{0} for all n ≥ n_{0}.

Solution 1: This is FALSE. A counterexample is the sequence x_{n} = (−1)^{n}. If a = 1 (or a = −1), note there is no n_{0} and no ε_{0} > 0 such that |x_{n} − a| ≥ ε_{0} for all n ≥ n_{0} because |(−1)^{n} − 1| = 0 for all even n. (Similarly, |(−1)^{n} + 1| = 0 for all odd n.)

Solution 2: We can also see that this is FALSE if we think of the meaning of the statement: For all a ∈ R, x_{n} ↛ a (that is, x_{n} does not converge to a.). In symbolic form, by negating the definition of the statement x_{n} → a, this says

∀a, ∃ε_{0} > 0, ∀n_{0}, ∃n ≥ n_{0}, s.t. |x_{n} − a| ≥ ε_{0}.

(Note: the quantifier on the n is ∃, not ∀. The given statement would certainly imply that the sequence does not converge to a, but it is not necessarily true for divergent sequences, as the example in the first solution shows.)

B) (10) It is possible to find a function f : R → R that is continuous at c = \sqrt{2}, with f(\sqrt{2}) = 0, but satisfying f(x) = 1 for all rational x.

Solution 1: This is FALSE. If \{x_{n}\} is any sequence of rational numbers with x_{n} → \sqrt{2}, then by a standard property of continuous functions, f(x_{n}) → f(\sqrt{2}). But f(x_{n}) = 1 for all n, and f(\sqrt{2}) = 0. So this is not possible.

Solution 2: We can also see that this is FALSE if we apply the definition of continuity directly. Aiming for a contradiction, suppose there were such a function. Then for all ε > 0, there would exist δ > 0 such that |f(x)−f(\sqrt{2})| < ε for all x with |x−\sqrt{2}| < δ. If f(\sqrt{2}) = 0 and we take ε < 1, then the interval defined by |y − 0| < ε does not contain 1. But every interval defined by |x − \sqrt{2}| < δ for δ > 0 contains rational numbers x and for those x, f(x) = 1. This is a contradiction which shows no such f exist.

C) (10) The sequence defined by x_{1} = 1 and x_{n} = \sqrt{3x_{n−1} + 1} for n ≥ 2 is monotone increasing.

Solution: This is TRUE. We have x_{1} = 1 and x_{2} = \sqrt{4} = 2. so x_{2} > x_{1}. This is the base case for an induction. Assume x_{k+1} ≥ x_{k} for a given k. Then by the induction hypothesis,

x_{k+2} = \sqrt{3x_{k+1} + 1} ≥ \sqrt{3x_{k} + 1} = x_{k+1}.

Therefore {x_{n}} is increasing by induction.

Extra Credit. (10) Assume that {x_{n}} is a sequence that converges to a. Construct a new sequence \{y_{n}\} by making y_{n} the average of the first n terms in \{x_{n}\}: y_{n} = \frac{x_{1}+x_{2}+···+x_{n}}{n} . Show that y_{n} also converges to a.

Solution: Here’s the idea. By the triangle inequality, note that

|y_{n} − a| = |\frac{x_{1} + x_{2} + ⋯ + x_{n}}{n} − a|

= \frac{1}{n}|x_{1} + x_{2} + ⋯ + x_{n} − na|

≤ \frac{1}{n} (|x_{1} − a| + ⋯ + |x_{n} − a|) .

Since {x_{n}} converges to a, recall that we know the sequence is bounded. This implies that there exists M such that |x_{n} − a| ≤ M for all n as well. Now, given ε > 0, there exists N such that |x_{n} − a| < ε/2 for all n ≥ N. So if n ≥ N, continuing from the last line above,

\begin{matrix}≤ \frac{1}{n}(|x_{1} − a| + ⋯ + |x_{N−1} − a|) + \frac{n − N + 1}{n} \frac{ε}{2} \\ < \frac{NM}{n} + \frac{ε}{2}. \end{matrix}

Since we can think of the N that works here as fixed, but n is still allowed to grow, note that we can now find n large enough so that \frac{NM}{n} < \frac{ε}{2} as well by taking n > \frac{2NM}{ε}. In other words, given ε > 0, we take n_{0} = max(N, \frac{2NM}{ε}). Then n ≥ n_{0} implies |y_{n} − a| < ε.