23 Algebraic and Order Limit Theorems Practice

#todo/school : Do the weekend problems 📅 2024-10-13 ✅ 2024-10-13
#todo/school : Finish these up! 📅 2024-10-27 ✅ 2024-10-25

Read §2.3; Do problems 2.3.1-4, 6, 7, 9, 10, 12

2.3.1

Question

Let $x_{n} \geq 0$ for all $n \in N$ .
a. If $(x_{n}) \to 0$ , show that $(\sqrt{x_{n}}) \to 0$ .
b. If $(x_{n}) \to x$ , show that $(\sqrt{x_{n}}) \to \sqrt{x}$ .

Let's do some scratch work:

We want to show that, for $ε > 0$ :

| \sqrt{x_{n}} - 0 | < ε

Notice that since $\sqrt{x} \geq 0$ for any $x \in R$ then this is the same as:

\sqrt{x_{n}} < ε

Now we know that since $(x_{n}) \to 0$ then any $ε^{'} > 0$ has some $N \in N$ where:

| x_{n} | = x_{n} < ε^{'}

This implies that if we use $ε = \sqrt{ε^{'}} \leftrightarrow ε^{2} = ε^{'}$ then that would work.

We want $| \sqrt{x_{n}} - \sqrt{x} | < ϵ$ . Multiplying by $(\sqrt{x_{n}} + \sqrt{x})$ gives:

| x_{n} - x | < (\sqrt{x_{n}} + \sqrt{x}) ϵ

Because $x_{n} \to x$ then we want the RHS to be less than some function of $ϵ$ . It is bounded so then $| x_{n} | < M$ so then $\sqrt{x_{n}} > \sqrt{0}$ . So then:

(\sqrt{x_{n}} + \sqrt{x}) ϵ < \sqrt{x} ϵ

so you want to choose $ϵ (\sqrt{x})$ .

Proof
Since $x_{n} \geq 0$ for all $n \in N$ , then the Limits and Order (Order Limit Theorem) suggests that $lim x_{n} \geq 0$ , if it exists.

a. Let $ε > 0$ . Notice that $ε^{2} > 0$ , so since $(x_{n}) \to 0$ then $\exists N^{'}$ where for any $n \geq N^{'}$ then $| x_{n} | < ε^{2}$ . Then:

| x_{n} | = x_{n} < ε^{2} \Rightarrow \sqrt{x_{n}} = | x_{n} | < ε

as desired, if we choose $N = N^{'}$ .

b. Let $ϵ > 0$ . Notice that $(\sqrt{x}) ϵ > 0$ so since $(x_{n}) \to x$ then:

\begin{aligned} | x_{n} - x | & < (\sqrt{x}) ϵ \\ | \sqrt{x_{n}} - \sqrt{x} | | \sqrt{x_{n}} + \sqrt{x} | & < (\sqrt{x}) ϵ & < (\sqrt{x_{n}} + \sqrt{x}) ϵ \\ | \sqrt{x_{n}} - \sqrt{x} | & < ϵ \end{aligned}

Where in the second line we assume $\sqrt{x_{n}} > 0$ .

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2.3.2 (FIX b)

Question

Using only the Convergence of a Sequence definition, prove that if $(x_{n}) \to 2$ , then:
a. $(\frac{2 x_{n} - 1}{3}) \to 1$
b. $(\frac{1}{x_{n}}) \to \frac{1}{2}$

Scratch:

Let $ε > 0$ , we want to use that:

\begin{aligned} | \frac{2 x_{n} - 1}{3} - 1 | & < ε \\ | \frac{2 x_{n} - 4}{3} | & < ε \\ \frac{2 | x_{n} - 2 |}{3} & < ε \end{aligned}

It's obvious that if we start with $| x_{n} - 2 | < ε^{'}$ then $| x_{n} - 2 | < \frac{3}{2} ε^{'}$ so use $ε = \frac{3}{2} ε^{'}$ .

Let $ε > 0$ . We want to show that:

\begin{aligned} | \frac{1}{x_{n}} - \frac{1}{2} | & < ε \\ | \frac{2 - x_{n}}{2 x_{n}} | & < ε \\ \frac{| x_{n} - 2 |}{2 | x_{n} |} & < ε \end{aligned}

We know that since $| x_{n} - 2 | < ε^{'}$ then we need to use the Bounded For Sequences property to show that $| x_{n} | < M$ . This will allow us to get:

\begin{aligned} \frac{| x_{n} - 2 |}{2 | x_{n} |} & > \frac{| x_{n} - 2 |}{2 M} & | x_{n} | < M \Leftrightarrow \frac{1}{| x_{n} |} > \frac{1}{M} \\ > ε \end{aligned}

Thus we should choose $ε = 2 M ε^{'} > 0$ .

Proof
a. Let $ε > 0$ . Since $\frac{3}{2} ε > 0$ then there is some $N$ where for all $n \geq N$ then:

\begin{aligned} | x_{n} - 2 | & < \frac{3}{2} ε \\ \frac{2 | x_{n} - 2 |}{3} & < ε \\ | \frac{2 x_{n} - 4}{3} | & < ε \\ | \frac{2 x_{n} - 1}{3} - 1 | & < ε \end{aligned}

b. Let $ε > 0$ . Since $(x_{n})$ converges to 2 then $(x_{n})$ has a bound $M > 0$ where $| x_{n} | \leq M$ . Notice then that since $M > 0$ then $2 M ε > 0$ , so then by the definition of convergence:

\begin{aligned} | x_{n} - 2 | & < 2 M ε \\ \frac{| x_{n} - 2 |}{2 M} & < ε \\ \frac{| x_{n} - 2 |}{2 | x_{n} |} < \frac{| x_{n} - 2 |}{2 M} & < ε \\ | \frac{x_{n} - 2}{2 x_{n}} | & < ε \\ | \frac{1}{2} - \frac{1}{x_{n}} | & < ε \\ | \frac{1}{x_{n}} - \frac{1}{2} | & < ε \end{aligned}

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2.3.3

(Squeeze Theorem)

Show that if $x_{n} \leq y_{n} \leq z_{n}$ for all $n \in N$ and if $lim x_{n} = lim z_{n} = l$ then $lim y_{n} = l$ .

Proof

First, let's prove that $(y_{n})$ converges at all. Assume that $(y_{n})$ diverges. Then for any limit $L$ then $\exists ε > 0$ such that for any $N \in N$ then $\exists n \geq N$ where $| y_{n} - L | \geq ε$ . Then:

| y_{n} - L | \geq ε

We have two cases, taking $N = max (N_{x}, N_{z})$ to ensure we can use the definitions of the respective limits:

$y_{n} \geq L$ . Then $y_{n} - L \geq ε$ . Thus then $y_{n} \geq L + ε$ . Then since $y_{n} \leq z_{n}$ then $z_{n} \geq L + ε$ . Then $| z_{n} - L | \geq ε$ , suggesting that $z_{n}$ doesn't converge. But since $L$ was arbitrary, then that suggests that $z_{n}$ diverges, which is a contradiction.
$y_{n} \leq L$ . A similar argument to (1) with instead that $x_{n} \leq y_{n}$ shows that $x_{n}$ diverges, which is a contradiction.

Now let's suppose that $(y_{n})$ converges to some unknown limit $L$ . In this case we can use Limits and Order (Order Limit Theorem), mainly (2). Notice that $x_{n} \leq y_{n}$ for all $n \in N$ so then $lim x_{n} \leq lim y_{n}$ , so then $l \leq L$ . A similar argument using $y_{n} \leq x_{n}$ for all $n \in N$ shows that $lim y_{n} \leq lim z_{n}$ , so then $L \leq l$ . So then $l = L$ as expected.
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2.3.4

Question

Let $(a_{n}) \to 0$ and use the Algebraic Limit Theorem to compute each of the following limits (assuming the fractions are always defined):
a. $lim (\frac{1 + 2 a_{n}}{1 + 3 a_{n} - 4 a_{n}^{2}})$
b. $lim (\frac{(a_{n} + 2)^{2} - 4}{a_{n}})$
c. $lim (\frac{\frac{2}{a_{n}} + 3}{\frac{1}{a_{n}} + 5})$

Proof
a.

\begin{aligned} lim (\frac{1 + 2 a_{n}}{1 + 3 a_{n} - 4 a_{n}^{2}}) & = \frac{lim (1 + 2 a_{n})}{lim (1 + 3 a_{n} - 4 a_{n}^{2})} \\ = \frac{lim 1 + lim 2 a_{n}}{lim 1 + lim 3 a_{n} - lim 4 a_{n}^{2}} \\ = \frac{1 + 2 lim a_{n}}{1 + 3 lim a_{n} - 4 lim (a_{n} \cdot a_{n})} \\ = \frac{1 + 2 (0)}{1 + 3 (0) - 4 (0)^{2}} \\ = \frac{1}{1} \\ = 1 \end{aligned}

\begin{aligned} lim (\frac{(a_{n} + 2)^{2} - 4}{a_{n}}) & = lim (\frac{a_{n}^{2} + 4 a_{n} + 4 - 4}{a_{n}}) \\ = lim (\frac{a_{n}^{2} + 4 a_{n}}{a_{n}}) \\ = lim (a_{n} + 4) \\ = lim a_{n} + lim 4 \\ = 0 + 4 \\ = 4 \end{aligned}

\begin{aligned} lim (\frac{\frac{2}{a_{n}} + 3}{\frac{1}{a_{n}} + 5}) & = lim (\frac{\frac{2}{a_{n}} + 3}{\frac{1}{a_{n}} + 5} \cdot \frac{a_{n}}{a_{n}}) \\ = lim (\frac{2 + 3 a_{n}}{1 + 5 a_{n}}) \\ = \frac{lim (2 + 3 a_{n})}{lim (1 + 5 a_{n})} \\ = \frac{lim 2 + lim 3 a_{n}}{lim 1 + lim 5 a_{n}} \\ = \frac{2 + 3 lim a_{n}}{1 + 5 lim a_{n}} \\ = \frac{2 + 3 (0)}{1 + 5 (0)} \\ = \frac{2}{1} \\ = 2 \end{aligned}

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2.3.6

Question

Consider the sequence given by $b_{n} = n - \sqrt{n^{2} + 2 n}$ . Taking $(\frac{1}{n}) \to 0$ as given, and using both the Algebraic Limit Theorem and the result from 2.3.1, show $lim b_{n}$ exists and find the value of the limit.

Proof
Notice that:

\begin{aligned} b_{n} & = n - \sqrt{n^{2} + 2 n} \\ = (n - \sqrt{n^{2} + 2 n}) \cdot \frac{n + \sqrt{n^{2} + 2 n}}{n + \sqrt{n^{2} + 2 n}} \\ = \frac{n^{2} - n^{2} - 2 n}{n + \sqrt{n^{2} + 2 n}} \\ = \frac{- 2 n}{n + \sqrt{n^{2} + 2 n}} \\ = \frac{- 2}{1 + \sqrt{\frac{n^{2}}{n^{2}} + \frac{2}{n}}} & Divide by n on num. + den. \\ = \frac{- 2}{1 + \sqrt{1 + \frac{2}{n}}} \end{aligned}

So using the Limit Laws (Algebraic Limit Theorem) in combination with out 2.3.1 to handle the $\sqrt{}$ :

\begin{aligned} lim b_{n} & = lim (\frac{- 2}{1 + \sqrt{1 + \frac{2}{n}}}) \\ = \frac{- 2}{1 + \sqrt{lim (1 + \frac{2}{n})}} \\ = \frac{- 2}{1 + \sqrt{1 + 2 lim \frac{1}{n}}} \\ = \frac{- 2}{1 + \sqrt{1 + 2 \cdot 0}} \\ = - 1 \end{aligned}

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2.3.7

Question

Give an example to each of the following, or state that such a request is impossible by referencing the proper theorems:
a. Sequences $(x_{n}), (y_{n})$ which both diverge, but whose sum $(x_{n} + y_{n})$ converges.
b. Sequences $(x_{n}), (y_{n})$ where $(x_{n})$ converges, $(y_{n})$ diverges, and $(x_{n} + y_{n})$ converges.
c. A convergence sequence $(b_{n})$ with $b_{n} \neq 0$ for all $n$ such that $(1 / b_{n})$ diverges.
d. An unbounded sequence $(a_{n})$ and a convergent sequence $(b_{n})$ with $(a_{n} - b_{n})$ bounded.
e. Two sequences $(a_{n}), (b_{n})$ where $(a_{n} b_{n})$ and $(a_{n})$ converge but $(b_{n})$ does not.

Proof
a. Have $(x_{n}) = n$ and $(y_{n}) = - n$ . Clearly both diverge but $(x_{n} + y_{n}) = (n - n) = 0$ clearly converges to 0.

b. This is impossible. If $(x_{n})$ then it must have a bound. Meanwhile $(y_{n})$ diverges, so no such bound exists for $| y_{n} |$ . That implies that eventually $| x_{n} | < | y_{n} |$ for all future $n$ , implying that $(x_{n} + y_{n})$ must be unbounded, and thus diverge. Similarly, the Limit Laws (Algebraic Limit Theorem) says that if $lim x_{n}$ exists and $lim x_{n} + y_{n}$ by the assumption, then:

lim x_{n} + y_{n} - lim x_{n} = lim y_{n}

must also exist, which is a contradiction.

c. Have $b_{n} = (\frac{1}{n})$ . Then clearly $b_{n} \neq 0$ and similarly we know that $\frac{1}{b_{n}} = (n)$ diverges.

d. Impossible. $| b_{n} |$ is convergent, so it is bounded. Thus $| b_{n} | \leq M_{1}$ and $| a_{n} - b_{n} | \leq M_{2}$ , so:

| a_{n} | = | a_{n} - b_{n} + b_{n} | \leq | a_{n} - b_{n} | + | b_{n} | \leq M_{1} + M_{2}

must bound $(a_{n})$ via the Triangle Inequality.

e. Have $(a_{n}) = (0)$ and $(b_{n}) = (n)$ .
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2.3.9

Question

a. Let $(a_{n})$ be a bounded (but not necessarily convergent) sequence. Assume $lim b_{n} = 0$ . Show that $lim a_{n} b_{n} = 0$ . Why can't we use the Limit Laws (Algebraic Limit Theorem)?
b. Can we conclude anything about the convergence of $(a_{n} b_{n})$ if we assume that $(b_{n})$ converges to some nonzero limit $b$ ?
c. Use (a) to prove the Limit Laws (Algebraic Limit Theorem) (iii) for the case when $a = 0$ .

a.
Proof
We can't use the ALT since we don't know if $(a_{n})$ is convergent (boundedness doesn't mean convergence). We'll have to show this the old fashioned way.

Let $ε > 0$ . We want to show:

| a_{n} b_{n} | < ε

for all $n \geq N$ . Notice that since $b_{n} \to 0$ then $\forall ε > 0 \exists N_{b} \in N \forall n \geq N_{b} (| b_{n} | < ε)$ . Now similarly since $(a_{n})$ is bounded, then $| a_{n} | \leq M$ for some $\exists M > 0$ .

Notice then that if we used $ε$ here then:

| a_{n} b_{n} | = | a_{n} | | b_{n} | < M ε

for $n \geq N_{b}$ . We should then choose $\frac{1}{M} ε$ as a result. Let's do that.

Notice that $\frac{ε}{M} > 0$ , so using our $b_{n} \to 0$ given, then $\exists N_{b} \in N$ where for all $n \geq N_{b}$ :

| b_{n} | < \frac{ε}{M}

Multiplying both sides by $| a_{n} |$ :

| a_{n} b_{n} | = | a_{n} | | b_{n} | < ε \frac{| a_{n} |}{M} \leq ε

Since we have $| a_{n} | \leq M \Leftrightarrow \frac{| a_{n} |}{M} \leq 1$ , so multiplying by $ε$ gives the right inequality. Choosing $N = N_{b}$ works here.

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b. We cannot assume $(a_{n} b_{n})$ converges if $lim b_{n} \to a$ where $a \neq 0$ . This is because, for a counter example, $(b_{n}) = 1$ and $(a_{n}) = (- 1)^{n}$ has $(a_{n} b_{n})$ doesn't converge, while $(a_{n})$ is still bonded and $(b_{n})$ convergent.

c. Using (a), then $lim a_{n} b_{n} = a b$ when $a = 0$ since then when $b_{n} \to b$ then $0 b = a b = lim a_{n} b_{n}$ as desired.

2.3.10

Question

Consider the following list of conjectures. Provide a short proof for those that are true, and a counterexample for any that are false.
a. If $lim (a_{n} - b_{n}) = 0$ then $lim a_{n} = lim b_{n}$ .
b. If $(b_{n}) \to b$ then $| b_{n} | \to | b |$
c. If $(a_{n}) \to a$ and $(b_{n} - a_{n}) \to 0$ then $(b_{n}) \to a$ .
d. If $(a_{n}) \to 0$ and $| b_{n} - b | \leq a_{n}$ for all $n \in N$ then $(b_{n}) \to b$ .

Proof
a. False via $(a_{n}) = (b_{n}) = (n)$ , where clearly the limits don't exist.

b. True. If $(b_{n}) \to b$ then $(b_{n}^{2}) \to b^{2}$ (apply the multiplication law). Then via 23 Algebraic and Order Limit Theorems Practice#2.3.1 (Finish) we have it that $(\sqrt{b_{n}^{2}}) \equiv (| b_{n} |) \to \sqrt{b^{2}} = | b |$ .

c. True. Since $(a_{n}) \to a$ and $(b_{n} - a_{n}) \to 0$ then $(a_{n} + (b_{n} - a_{n})) = (b_{n}) \to a + 0 = a$ .
d.
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