Mathematics 242 – Principles of Analysis 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})

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

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} for some explicit (strictly increasing) index sequence n_{k}, or say why that is impossible:

(a) {2, 3, 4, 5, …}

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

(c) {1, 2, 4, 8, 16, 32, …}

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.

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

(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.

‘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}}.

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

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

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.

(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.

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!)

(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.