Mathematics 242 – Principles of Analysis Solutions for Problem Set 5 Due: March 15, 2013

‘A’ Section

1. For each of the following sequences, determine three different subsequences, each converging to a different limit. For each one, express your three subsequences as x_{n_{k}} for a suitably chosen (strictly increasing) index sequence n_{k}, and give an explicit formula for n_{k} as a function of k:

(a) x_{n} = cos (\frac{π\sqrt{n}}{2})

Solution: If n_{k} = (4k)^{2}, then x_{n_{k}} = 1 for all k, so that subsequence converges to 1. If n_{k} = (2(2k + 1))^{2}, then x_{n_{k}} = −1 for all k, so that subsequence coverges to −1. Finally, if n_{k} = (2k − 1)^{2}, then x_{n_{k}} = 0 for all k so that subsequence converges to 0. There are infinitely many other correct examples too.

(b) x_{n} = \frac{n}{3} − [\frac{n}{3}] (as usual, [ ] denotes the greatest integer function)

Solution: Let n_{k} = 3k, then x_{3k} = k − k = 0 for all k, so that subsequence converges to 0. Let n_{k} = 3k + 1, then x_{3k+1} = k + 1/3 − k = 1/3 for all k, so that subsequence converges to 1/3. Finally, if n_{k} = 3k + 2, then the subsequence x_{3k+2} converges to 2/3.

2. Let x_{n} = n^{1/4}. For each of the following sequences, either express that sequence as a subsequence of the sequence x_{n_{k}} for some explicit (strictly increasing) index sequence n_{k}, or say why that is impossible:

(a) {2, 3, 4, 5, …}

Solution: This is the subsequence x_{n_{k}} for n_{k} = (k + 1)^{4}, k ≥ 1.

(b) \{\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, …\}

Solution: This is the subsequence x_{n_{k}} for n_{k} = (3k)^{2}, k ≥ 1.

(c) {1, 2, 4, 8, 16, 32, …}

Solution: This is the subsequence x_{n_{k}} for n_{k} = 2^{4k−4}, k ≥ 1.

3. Let x_{n} = sin (\frac{nπ}{4} ) and y_{n} = cos (\frac{nπ}{4} ).

(a) Find a (strictly increasing) index sequence n_{k} such that both x_{n_{k}} and y_{n_{k}} converge.

Solution: The sequence n_{k} = 8k + 1 is such an index sequence since

sin (\frac{(8k + 1)π}{4} ) = sin (2kπ + \frac{π}{4} ) = \frac{\sqrt{2}}{2}

and

cos (\frac{(8k + 1)π}{4} ) = cos (2kπ + \frac{π}{4} ) = \frac{\sqrt{2}}{2}

for all k ≥ 1.

(b) Find a second (strictly increasing) index sequence n_{k} such that both x_{n_{k}} and y_{n_{k}} diverge.

Solution: The sequence n_{k} = k is such a sequence.

(c) Find a third (strictly increasing) index sequence n_{k} such that one of x_{n_{k}} and y_{n_{k}} converges and the other diverges.

Solution: The sequence n_{k} = 4k is such an example since sin(kπ) = 0 for all k, but cos(kπ) = (−1)^{k} does not converge.

‘B’ Section

1. (True or False) – If the statement is true give a proof; if it is false give a counterexample.

(a) If x_{n} is a sequence of strictly negative numbers converging to 0, then x_{n} has a strictly increasing subsequence x_{n_{k}}.

Solution: This is TRUE. Here is one way to see it, constructing a strictly increasing subsequence inductively. We start with n_{1} = 1, so x_{n_{1}} = x_{1}. This is the base case. Now assume we have found x_{n_{1}} < x_{n_{2}} < … < x_{n_{k}} for 1 = n_{1} < n_{2} < ⋯ < n_{k}. Since x_{n} → 0, given ε = |x_{n_{k}}|, there exists N_{0} in N such that |x_{n} −0| < |x_{n_{k}}| for all n ≥ N_{0}. Take n_{k+1} = max(N_{0}, n_{k} + 1). Then we have |x_{n_{k+1}}| < |x_{n_{k}}|, so x_{n_{k+1}} > x_{n_{k}} (since they are both negative) and n_{k+1} > n_{k}. This shows we can continue to construct a strictly increasing subsequence.

(b) If x_{n} → 0, then x_{n} contains a strictly increasing subsequence or a strictly decreasing subsequence (or both).

Solution: This is FALSE. Counterexamples are the constant sequence x_{n} = 0 for all n ≥ 1, or any “eventually constant” sequence with x_{n} = 0 for all n ≥ n_{0} for some n_{0} ∈ N. (Comment: The statement would be true if we assumed x_{n} ≠ 0 for all n (or even all n ≥ n_{0} for some n_{0} ∈ N). Any such sequence contains either infinitely many positive terms or infinitely many negative terms, or both. If there are infinitely many negative terms, we get a strictly increasing subsequence of the negative terms by part (a) of this question. If there are infinitely many strictly positive terms, then there is a strictly decreasing subsequence, as can be seen by taking negatives, using part (a), then flipping signs again.)

(c) If x_{n} is a decreasing sequence with a bounded subsequence x_{n_{k}}, then x_{n} converges.

Solution: This is TRUE. Since the whole sequence is decreasing, so is the subsequence x_{n_{k}}. But that subsequence is bounded (below), say by a ∈ R. We claim that the whole sequence x_{n} is also bounded below by a. To see that, let n be any natural number. Since {n_{k}} is a strictly increasing sequence of natural numbers, it follows that it is not bounded above. So there is some k such that n_{k} ≥ n. But then since x_{n} is decreasing and a is a lower bound for the subsequence, a ≤ x_{n_{k}} ≤ x_{n}. It follows that x_{n} ≥ a for all n, and hence the whole sequence is bounded below. Then {x_{n}} converges as well by the Monotone Convergence Theorem. (You can also show that the limit of the whole sequence must be the same as the limit of the subsequence, but that was not required.)

2. Consider the sequence x_{n} = cos(n) (where we think of n as an angle expressed in radians).

(a) Prove that x_{n} has a convergent subsequence.

Solution: Since | cos(n)| ≤ 1 for all n ≥ 1, this is a bounded sequence. The statement to be proved is a direct consequence of the Bolzano-Weierstrass Theorem.

(b) In this part of the question we will show that x_{n} is not convergent, though. Suppose lim_{n→∞} cos(n) = a for some real number a. Using a trig identity for cos(n + 1) and considering lim_{n→∞}(cos(n + 1) − cos(n)), show that

\frac{a(cos(1) − 1)}{sin(1)} = lim_{n→∞} sin(n).

But then use the sequence lim_{n→∞}(sin(n + 1) − sin(n)) to deduce that a = 0, so lim_{n→∞} cos(n) = lim_{n→∞} sin(n) = 0. But this is a contradiction. Explain why to conclude the proof.

Solution: The addition formula for cos implies that cos(n + 1) = cos(n) cos(1) − sin(n) sin(1). If we assume that lim_{n→∞} cos(n) = a, then using parts of the “Big Theorem” and rearranging algebraically, we see

0 = lim_{n→∞}(cos(n + 1) − cos(n))
= lim_{n→∞} cos(n)(cos(1) − 1) − lim_{n→∞} sin(n) sin(1)
= a(cos(1) − 1) − sin(1) lim_{n→∞} sin(n).

Thus,

(1)  lim_{n→∞} sin(n) = \frac{a(cos(1) − 1)}{sin(1)}.

as claimed. Now the addition formula for sin shows

sin(n + 1) = sin(n) cos(1) + cos(n) sin(1)

Taking the limit as n → ∞ on both sides and substituting from (1), we get:

\frac{a(cos(1) − 1)}{sin(1)} = \frac{a(cos(1) − 1)}{sin(1)} cos(1) + a sin(1),

so

a(cos(1) − 1) = a(cos(1) − 1) cos(1) + a sin^{2}(1),

and hence (because cos^{2}(1) + sin^{2}(1) = 1),

a(cos(1) − 1) = a(1 − cos(1)).

The only way this can be true is if a = 0. But then this implies that

lim_{n→∞} sin(n) = 0 = lim_{n→∞} cos(n).

But that is impossible because it would say lim_{n→∞} cos^{2}(n) + sin^{2}(n) = 0 by parts (a) and (b) of the “Big Theorem.” However, we know by the basic identity for the trigonometric functions that

cos^{2}(n) + sin^{2}(n) = 1

for all n ≥ 1. Hence lim_{n→∞} cos^{2}(n) + sin^{2}(n) = 1 as well if the two limits exist. This contradiction shows that lim_{n→∞} cos(n) cannot exist.

3. A cluster point of a sequence x_{n} is a limit of a convergent subsequence x_{n_{k}}. (See question 1 on the A section for examples of sequences with several different cluster points.)

(a) Show that there exists a sequence x_{n} whose set of cluster points is all of R_{+} = {x ∈ R | x > 0}. (Hint: Look at the solution for the Extra Credit problem on Exam 1, which shows how to get a sequence containing all the positive rational numbers. Think rational approximations to decimals to get a subsequence converging to any given positive a ∈ R. However, note that you will need to be careful to produce an actual subsequence x_{n_{k}} with a strictly increasing index sequence n_{k}. In fact, this example can be “jazzed up” to get a sequence whose set of cluster points consists is all of R!)

Solution: As shown in the solution to the Extra Credit problem there is a sequence {x_{n}} whose terms contain all the positive fractions p/q with p, q integers ≥ 1. We claim that every positive real number is cluster point of that sequence. Note that because of the “zig-zag” way that sequence is constructed, in the sequence {x_{n}}, all the fractions p/q with p + q = ℓ occur before those with p + q = ℓ + 1, etc. In other words, the sum of the numerator and the denominator of x_{n} is an increasing function of n. Now, if p/q is a positive fraction in lowest terms, then we can also consider the fractions (pk)/(qk) for all k ≥ 1. These form a subsequence of the x_{n} sequence since the sum pk + qk = (p + q)k is strictly increasing as a function of k. Hence the indices n_{k} that give (pk)/(qk) = x_{n_{k}} must also be strictly increasing. That subsequence converges to a = p/q since all the fractions (pk)/(qk) reduce to p/q, so it’s actually a constant sequence. Next, any positive irrational number can be expressed with an infinite (nonrepeating) decimal expansion

a = d_{k}d_{k−1} … d_{0}.f_{1}f_{2} ⋯

(where the d_{i} and the f_{j} are the decimal digits). As we saw in class, that expansion is equivalent to a monotone increasing sequence of rational approximations with denominators that are powers of 10:

d_{k} … d_{0}/1, d_{k} … d_{0}f_{1}/10, d_{k} … d_{0}f_{1}f_{2}/100, ….

The notation means something like this: If a = \sqrt{2}, for instance, where \sqrt{2} = 1.4142 ⋯, then our sequence of rational approximations is

1/1, 14/10, 141/100, 1414/1000, ….

When we add one more decimal digit to get the next term in a sequence approximating a given positive real a, the corresponding fraction will definitely occur later in the x_{n} sequence, since the sum of the numerator and the denominator always increases when we include an additional digit. This shows that we have a subsequence converging to a. Hence a is a cluster point of the x_{n} sequence.

(b) Show that if a_{m} is a convergent sequence of cluster points of a given sequence x_{n}, then a = lim_{m→∞} a_{m} is also a cluster point of the x_{n} sequence.

Solution: We let a_{m} be a convergent sequence of cluster points of x_{n} with a = lim_{m→∞} a_{m}. Since each a_{m} is a cluster point of the x_{n} sequence, there is a subsequence of x_{n} converging to a_{m}. From the subsequence converging to a_{1}, select any x_{n_{1}} with |x_{n_{1}} − a_{1}| < 1. Then from the subsequence converging to a_{2}, select any x_{n_{2}} with n_{2} > n_{1} and |x_{n_{2}} − a_{2}| < \frac{1}{2}, then from the subsequence converging to a_{3}, select x_{n_{3}} with |x_{n_{3}} − a_{3}| < \frac{1}{3} and n_{3} > n_{2}. By an induction argument, we can always continue this process since for any ℓ ≥ 1, there are infinitely many index values n_{ℓ} for which |x_{n_{ℓ}} − a_{ℓ}| < \frac{1}{ℓ} (all the indices giving terms in the subsequence converging to a_{ℓ} that are distance \frac{1}{ℓ} or less from a_{ℓ}). In this way we get a subsequence \{x_{n_{ℓ}}\} (indexed by ℓ ∈ N) with n_{ℓ} strictly increasing and |x_{n_{ℓ}} − a_{ℓ}| < \frac{1}{ℓ} for all ℓ ≥ 1. We claim that the subsequence x_{n_{ℓ}} converges to a. To see this note that given any ε > 0, there exists ℓ_{0} ∈ N such that \frac{1}{ℓ} < \frac{ε}{2} and |a_{ℓ} − a| < \frac{ε}{2} for all ℓ ≥ ℓ_{0}. Then by the triangle inequality, for all ℓ ≥ ℓ_{0},

|x_{n_{ℓ}} − a| = |x_{n_{ℓ}} − a_{ℓ} + a_{ℓ} − a| ≤ |x_{n_{ℓ}} − a_{ℓ}| + |a_{ℓ} − a| < \frac{ε}{2} + \frac{ε}{2} = ε.

This shows that the subsequence {x_{n_{ℓ}}} converges to a, so a is also a cluster point of the {x_{n}} sequence.