Mathematics 242 – Principles of Analysis Solutions for Exam 2 – March 28, 2014

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 statement is that any monotone bounded sequence of real numbers converges. If the sequence is monotone increasing, let a = lub{x_{n} | n ∈ N}. Then for all ε > 0, a − ε is not an upper bound for {x_{n} | n ∈ N}, so there exist n_{0} such that a − ε < x_{n_{0}} ≤ a. But then since {x_{n}} is monotone increasing, we have a − ε < x_{n_{0}} ≤ x_{n} ≤ a for all n ≥ n_{0}. This implies |x_{n} − a| < ε for all n ≥ n_{0}. Hence x_{n} → a.

B) (10) Let {x_{n}} be the sequence defined by x_{1} = 1 and

x_{n+1} = \sqrt{3x_{n} + 1}

for all n ≥ 1. Does this sequence converge? Why? If it does, what is the limit?

Solution: We have x_{1} = 1 and x_{2} = \sqrt{3 · 1 + 1} = 2. Assuming x_{k+1} > x_{k}, it follows that

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

Therefore, the sequence is monotone (strictly) increasing. Next, we claim that all the terms in the sequence are bounded above by 4. This is true for n = 1, 2 by the above. Furthermore, if x_{k} < 4, then x_{k+1} < \sqrt{3 · 4 + 1} = \sqrt{13} < 4 so the sequence is bounded above by 4. Therefore, it converges by the Monotone Convergence Theorem. The limit is found by letting n → ∞ in the recurrence. If a denotes the limit then a = \sqrt{3a + 1}, so a^{2} − 3a − 1 = 0. By the quadratic formula, a = \frac{3+\sqrt{13}}{2} (the negative sign gives a number < 0, so that cannot be the limit since x_{n} > 0 for all n).

II. A) (15) Show using the ε, n_{0} definition that

lim_{n→∞} \frac{4n + 1}{7n + 3} = \frac{4}{7}.

Solution: Given ε > 0, let n_{0} > \frac{5}{49ε} (such n_{0} exist since N is not bounded in R). This implies that \frac{5}{49n_{0}} < ε. Then for all n ≥ n_{0},

|\frac{4n + 1}{7n + 3} − \frac{4}{7}| = \frac{5}{49n + 21} < \frac{5}{49n} ≤ \frac{5}{49n_{0}} < ε.

This shows the limit is \frac{4}{7}.

B) (15) Show using the ε, δ definition that

lim_{x→1} \frac{4x + 1}{7x + 3} = \frac{1}{2}.

Solution: Given ε > 0, let δ = min(1, 6ε). Then for all x in the deleted neighborhood of 1 defined by 0 < |x − 1| < δ we have 0 < x < 2, so \frac{1}{|14x+6|} < \frac{1}{6} and |x − 1| < 6ε. Therefore

|\frac{4x + 1}{7x + 3} − \frac{1}{2}| = \frac{|x − 1|}{|14x + 6|} =< 6ε · \frac{1}{6} = ε.

This shows the limit is \frac{1}{2}.

III. (10) Suppose {x_{n}} is a sequence such that |x_{n} − 20| < 50 for all n ≥ 1. Show that there exists some number a ∈ [−30, 70] and a subsequence {x_{n_{k}}} such that x_{n_{k}} → a. State any “big theorems” you are using.

Solution: We have −30 < x_{n} < 70 for all n. This shows the sequence is bounded. Hence the Bolzano-Weierstrass theorem implies it has a convergent subsequence x_{n_{k}} → a. By the Order Limit theorem, it follows that −30 ≤ a ≤ 70.

IV. Give an example, or give a reason why there can be no such examples:

A) (10) A function f that is continuous at x = 0, but not continuous at any other x.

Solution: The function

f(x) = \{ \begin{matrix}x & if x ∈ Q \\ −x & if x ∉ Q \end{matrix}

is an example. lim_{x→0} f(x) = 0 = f(0), so f is continuous at x = 0. However, f is not continuous at any other c, since all neighborhoods of any c ≠ 0 contain x where f(x) is close to c and other x where f(x) is close to −c. Therefore lim_{x→c} f(x) does not exist if c ≠ 0.

B) (10) A sequence x_{n} such that x_{n} → 3, but for all n_{0} ∈ N there exist n ≥ n_{0} with x_{n} < 0.

Solution: There cannot be such a sequence. Let ε = 1, for instance. Then if x_{n} → 3, there must equal an n_{0} ∈ N such that |x_{n} − 3| < 1 for all n ≥ n_{0}. But |x_{n} − 3| < 1 implies 2 < x_{n} < 4 so all the terms for n ≥ n_{0} must be strictly positive.

C) (10) A function f and a c in the domain of f such that lim_{x→c^{+}} f(x) = lim_{x→c^{−}} f(x), but f is not continuous at c. Solution: Let

f(x) = \{ \begin{matrix}1 & if x ≠ 0 \\ 0 & if x = 0 \end{matrix}

Then for c = 0, we have lim_{x→0^{+}} f(x) = 1 = lim_{x→0^{−}} f(x). But f(0) = 0 ≠ 1. Therefore f is not continuous at 0. (This is a removable discontinuity.)

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 the original sequence: y_{n} = \frac{x_{1}+x_{2}+···+x_{n}}{n} . Show that \{y_{n}\} also converges to a.

Solution: Intuitively, the idea is that as n gets big, most of the terms in the numerator of y_{n} will be getting close to a, but a few terms at the start might not be that close. However, the n in the denominator means that those few terms at the start contribute less and less as n → ∞. But we need a more precise way to say this to get an actual proof. Here’s an 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. This shows that y_{n} → a as n → ∞.