# Assignment 3 Solutions

Question 1.  The Axiom of Completeness states that if $S \subseteq \mathbb{R}$ is a non-empty subset of real numbers that is bounded above, then the supremum of $S$ exists.

The Nested Interval Property states that if $I_n = [a_n, b_n]$ is a collection of nested, closed, non-empty intervals of real numbers, then the intersection $\displaystyle \bigcap_{n=1}^{\infty} I_n \neq \varnothing.$

The Bolzano-Weierstrass Theorem states that if $S \subseteq \mathbb{R}$ is a bounded, infinite subset of real numbers, then $S$ has at least one accumulation point.

Question 2.  Without these conditions, one could not even write down the expression $A/B$ or the expressions $a_n/b_n$.  Note: one can actually weaken the hypotheses of this theorem without altering the conclusion by requiring $b_n \neq 0$ for all but finitely many terms in $\{b_n\}$.

Question 3.  As discussed in class, this sequence can be thought of the product of two sequences, one of which we have proved converges to zero the other being bounded.  Hence, Theorem 1.13 applies and $a_n \to 0$.

Question 4.  Here is an example of a subsequence.  If $\{a_n\} = \{1/n\}$ then we can use $\{1, 1/3, 1/5, 1/7, \cdots \}$ as a subsequence.  Using sub-indices, this subsequence could be notated by setting

$a_{n_k} = a_{2k-1}$ for all $k \in \mathbb{N}$.

Question 5.  A sequence of real numbers is an increasing sequence if for every $n \in \mathbb{N}$ it follows that $a_{n+1} \geq a_n$.  Here are two examples of increasing sequences:

$a_n = 1^n$ (a rather boring but technically sound one)

$\displaystyle b_n = n$.

Question 6.  A sequence of real numbers is a decreasing sequence if for every $n \in \mathbb{N}$, it follows that $a_{n+1} \leq a_n$.  Here are threeexamples of decreasing sequences:

$a_n = 1^n$ (a rather boring but technically sound one)

$\displaystyle b_n = -n$ (this one diverges)

$\displaystyle c_n = \frac{1}{n}$ (this one converges)

Question 7.  A sequence of real numbers is monotone if it is either increasing or decreasing.

Question 8.  Personally, I think knowing that the supremum of the sequence exists is the most important part of the proof of MCT (in the case where the sequence is increasing).  Without this step, we would not know what proposed real number to use in checking that our bounded, monotone sequence converges.

Question 9.  (skipped)

## Proof Problems

Problem 1.  Let $a_n = (-1)^n$ and let $b_n = (-1)^{n+1}$.  Neither sequence converges (this can be proved by using results from section 1.4, for instance; noting that each sequence contains subsequences that converge to different numbers), but the sum of these sequences is given by

$c_n = a_n + b_n = (-1)^n + (-1)^{n+1} = 0 \, \text{ for all } n \in \mathbb{N}$.

Since $c_n = 0$ is constant, this sequence clearly converges to $0$.

Problem 2.  Suppose $\{a_n\}$ is a sequence of real numbers that converges to $A \in \mathbb{R}$.  We want to prove that the sequence of averages, $\{\alpha_n\}$, converges to $A$, too.  Here we have that each $\alpha_n$ is given by

$\displaystyle \alpha_n = \frac{a_1 + a_2 + \cdots + a_n}{n}$.

(Proof Idea).  Before sharing a proof, let’s first discuss an idea for the proof.  That is, let’s take some time to think about this problem in more intuitive terms rather than by using technical definitions.

We want to show that if we choose $n$ very large, then we can make $d(\alpha_n, A)$ very small.  If $n$ is very large, though, then the final terms in the numerator

$\displaystyle \alpha_n = \frac{a_1 + a_2 + \cdots + a_n}{n}$

all start to look very much like the number $A$ (since, after all, we’re assuming $a_n \to A$).  This means that, for very large values of $n$,

$\displaystyle \alpha_n \approx \frac{a_1 + a_2 + \cdots + a_{\text{??}}}{n} + \frac{A + A + \cdots + A}{n}$

I used a “?” symbol above because I don’t yet know how how many of the $n$ numerator terms “look like A” — I just know or sense that the “final ones” do.  In a formal proof, we can give an actual name to this index — we’ll call it $m$ or $k$ or something like that.

In any event, if $n$ is very large, then the first fraction must be very small.  After all, the numerator terms are fixed and so their sum is some fixed constant.  Hence, this number will be effectively zero.  In addition, the second term has $n - \text{??}$ terms on top and so can be expressed as

$\displaystyle \frac{(n-\text{??})A}{n} = A - \frac{\text{??}A}{n}$.

Again, because $n$ is very large and $A$ and the un-named index $\text{??}$ are fixed numbers, the last term above can be made arbitrarily small, and so is effectively zero.

All together, these ideas suggest that as $n$ becomes very large, the numbers $\alpha_n$ start to look very much like $A$.  $\square"$

(Rigorous Proof).  (coming soon!)

Problem 3.  The first sequence can be handled by an application of Theorem 1.13, and so converges to zero.  The second sequence can be handled by rewriting it as

$\displaystyle \frac{n}{n^2-3} = \frac{\frac{1}{n}}{1-\frac{3}{n^2}}$

and then appealing to our theorems governing the arithmetic of sequences of real numbers

Problem 4.  Here is a sketch of the proof.

Suppose $\{a_n\}$ is a bounded sequence of real numbers.  If this sequence has finite range, then we may build a constant (and hence convergent) subsequence.  In particular, if the real number $a$ appears in the original sequence infinitely often, we may set

$a_{n_k} = \text{ the } k\text{-th term in } \{a_n\} \text{ that equals } a$

Suppose, then, that $\{a_n\}$ has infinite range.  This sequence is then infinite and bounded, and so by the Bolzano-Weierstrass Theorem it contains at least one accumulation point, $a \in \mathbb{R}$.  We can now construct a subsequence $a_{n_k}$ that converges to $a$.

By definition of accumulation point, there exists an element of the sequence $\{a_n\} = \{a_1, a_2, a_3, \cdots \}$ that lies in the interval $(a-1, a+1)$.  Call this point $a_{n_1}$.

By definition of accumulation point, there also exists an element of the “remaining sequence” $\{a_{n_1+1}, a_{n_1+2}, \cdots \}$ that lies in the interval $(a-1/2, a+1/2)$.  Call this point $a_{n_2}$.

In general, there exists an element of the “remaining sequence” $\{a_{n_{k-1}+1}, a_{n_{k-1}+2}, \cdots \}$ that lies in the interal $(a-1/k, a+1/k)$, and we can call this point $a_{n_k}$.

By construction, $\displaystyle d(a_{n_k}, a) < 1/k$ for all $k \in \mathbb{N}$.  Hence, given any $\varepsilon > 0$ we can choose $k > 1/\varepsilon$ to find that $\displaystyle d\left(a_{n_k}, a\right) < \varepsilon$.

Note: we need to say something about “remaining sequences” (or something similar) to ensure that sequence we are constructing is, in fact, a subsequence.  In particular, our construction above forces $n_{k-1} < n_{k}$ since each successive term selected for our subsequence was chosen from a smaller or “remaining sequence.”  For example, $a_{n_4}$ would be chosen from the sequence

$\displaystyle a_{n_4} \in \{a_{{n_3}+1}, a_{{n_3}+2}, a_{{n_3}+3}, \cdots \}$.

Problem 5.  Suppose $\{x_n\}$ and $\{y_n\}$ are two convergent sequences of real numbers that both converge to $x_0 \in \mathbb{R}$.  We claim that the sequence defined by

$z_{2n} = x_n \text{ and } z_{2n-1} = y_n$

also converges to $x_0 \in \mathbb{R}$.

Proof.  Let us denote the indices of our “spliced sequence” using the variable $k \in \mathbb{N}$.  We then have that

$z_k = x_{k/2} \text{ if } k \text{ is even}$

$z_k = y_{(k+1)/2} \text{ if } k \text{ is odd }$.

Let $\varepsilon > 0$ be given.  Choose $k \geq K > \max\left\{2N_1, 2N_2+1\right\}$ where $N_1, N_2 \in \mathbb{N}$ are indices for the sequences $\{x_n\}$ and $\{y_n\}$, respectively.  These indices are chosen so that

$\displaystyle d\left(x_n, x_0\right) = \left|x_n-x_0\right| < \varepsilon \text{ whenever } n \geq N_1$

$\displaystyle d\left(y_n, x_0\right) = \left|y_n-x_0\right| < \varepsilon \text{ whenever } n \geq N_2$

Suppose $k \geq K > \max\left\{N_1, 2N_2-2\right\}$.  If $k = 2n$ is even, then it follows that $2n \geq 2N_1 \iff n \geq N_1$ and so the first inequality above holds.  In particular, this means that

$\displaystyle d\left(z_k, x_0\right) = d\left(x_{n}, x_0\right) < \varepsilon$.

Similarly, if $k = 2n+1$ is odd, then it follows that $2n+1 \geq 2N_2 +1\iff n \geq N_2$ and so the second inequality above holds.  In particular, this means that

$\displaystyle d\left(z_k, x_0\right) = d\left(y_{n}, x_0\right) < \varepsilon$.

Problem 6.  There was a typo on this question.  As most of you surmised, the first term should have been

$a_1 = \sqrt{6}$.

With this change, one can prove by induction that the sequence $\{a_1\}$ is bounded, and one can also prove by induction that it is increasing.  The Monotone Convergence Theorem then applies and lets us calculate the limit of the sequence as done in the rest of the problem.