Let’s Get Real… Analysis (Part 7): Limit Laws for Sequences

We’re continuing in our Let’s Get Real… Analysis series mission to gather more tools for analyzing sequences. Today, we will learn what I call, our limit laws. These will likely be intuitive, but they give us an opportunity to practice writing proofs, which is very useful!

I will admit, however, that today’s content is not the most interesting math(s) that we can do. There are useful techniques used in these proofs that you should see in action at least once, though! So, I hope you humor me and take a look at some of them!

Limit Laws:

Let $(a_n)$ and $(b_n)$ be sequences such that $a_n \rightarrow \alpha$ and $b_n \rightarrow \beta.$ Then, they satisfy the following properties that we will call the Limit Laws of sequences or simply our limit laws,

$a_n$ is bounded above and below, that is there exist $A\in \mathbb{R}$ such that $|a_n|\leq A$ for all $n$ .
If $a_n < b_n$ for all $n$ then $\alpha \leq \beta,$ (similarly, if $a_n \leq b_n$ for all $n$ then $\alpha \leq \beta,$ )
Let $c \in \mathbb{R},$ then $c\cdot \alpha \rightarrow c\cdot \alpha,$
$a_n \pm b_n \rightarrow \alpha \pm \beta,$
$a_n \cdot b_n \rightarrow \alpha \cdot \beta,$
If $b_n \neq 0$ for all $n$ and $\beta \neq 0$ then $\frac{a_n}{b_n} \rightarrow \frac{\alpha}{\beta},$
If $a_n \geq 0$ for all $n,$ then $\lim{\sqrt{a_n}} = \sqrt{\lim{a_n}}.$

Proof: Let $(a_n)$ and $(b_n)$ be sequences that converge to $\alpha$ and $\beta$ respectively.

(1) $a_n$ is bounded above and below, that is there exist $A\in \mathbb{R}$ such that $|a_n|\leq A$ for all $n$ .

Proof of (1): Since, $a_n \longrightarrow \alpha$ there is some $N \in \mathbb{N}$ such that $|a_n -\alpha| <1$ for all $n>N$ . It’s then clear that $a_n$ is bounded in the following way $\alpha-1 N$ . However, this says nothing about the first $N$ elements of the sequence $a_n$ . To handle those, we note that the set ${ |a_1|,|a_2|,\cdots, |a_N|}$ is finite and thus has a maximum element and is therefore bounded! We just need to take the maximum of the bounds that we’ve found.

Let $A = \max{\big\{|1+\alpha|\,,\,|1-\alpha|\,,\, |a_1|\,,\,|a_2|\,,\,\cdots\,,\, |a_N| \big\}}$ . Then, for all $n$ we can see that $|a_n|\leq A$ using what we discussed in the last paragraph.

$\square_{\mathrm{Part}\;(1)}$

(2) If $a_n < b_n$ for all $n$ then $\alpha \leq \beta.$

Proof of (2)

Let $\varepsilon>0$ and $a_n < b_n$ for all $n.$ Since we know that $a_n \longrightarrow \alpha$ and $b_n \longrightarrow \beta,$ there exists two natural numbers $N_a\,,\,N_b \in \mathbb{N}$ such that,

$|a_n - \alpha| < \frac{\varepsilon}{2}\;\;\mathrm{for}\;\;\mathrm{all}\;n>N_a$

and

$|b_n - \beta| < \frac{\varepsilon}{2}\;\;\mathrm{for}\;\;\mathrm{all}\;n>N_b.$

Let $N = \max{{N_a\,,\,N_b}}.$ (We will use this trick often.) Then, for all $n > N$ we have both of the above inequalities satisfied. In particular, we have,

$\beta - \alpha > \beta - \alpha - (b_n- a_n)\\\textit{ }\qquad\,= -(b_n - \beta) + (a_n- \alpha) \\ \textit{ }\qquad\,= -\frac{\varepsilon}{2} -\frac{\varepsilon}{2}\\\textit{ }\qquad\, = -\varepsilon \nonumber$

Which implies that $\beta - \alpha \geq 0$ . “Why?” you ask. If it weren’t the case, then $\beta-\alpha < 0$ would imply that there is some real number $x$ larger than $\beta - \alpha$ and smaller than $0$ , that is $\beta - \alpha <x = -|x|<0$ . Since $\varepsilon$ is arbitrary, we could let $\varepsilon = |x|$ and get a contradiction; therefore, it must be that $\beta - \alpha \geq 0$ .

$\square_{\mathrm{Part}\;(2)}$

(3) Let $c \in \mathbb{R},$ then $c\cdot \alpha \rightarrow c\cdot \alpha,$

Proof (3): Let $\varepsilon>0$ . We have two cases to consider, the first is when $c=0$ . In this case, $c\cdot a_n = 0$ for all $n$ , which indeed converges to $0$ . The second case, when $c\neq 0$ , is where the real action happens. In this second case we use that $a_n \longrightarrow \alpha$ , to deduce that there is an $N \in \mathbb{N}$ such that,

$|a_n - \alpha| < \frac{\varepsilon}{|c|}$

Then, for this same $N$ , we have

$|c\cdot a_n - c\cdot\alpha| \;=\; |c|\cdot |a_n - \alpha| \;<\; |c|\cdot\frac{\varepsilon}{|c|} \;=\; \varepsilon.$

Concluding the proof of part (3).

$\square_{\mathrm{Part}\;(3)}$

(4) $a_n \pm b_n \rightarrow \alpha \pm \beta,$

Proof (4): Let $\varepsilon>0$ . Guess what? Yet again, we will use that there are two natural numbers $N_a\,,\,N_b \in \mathbb{N}$ such that,

$|a_n - \alpha| < \frac{\varepsilon}{2}\;\;\mathrm{for}\;\;\mathrm{all}\;n>N_a$

and

$|b_n - \beta| < \frac{\varepsilon}{2}\;\;\mathrm{for}\;\;\mathrm{all}\;n>N_b.$

Again, let $N = \max{{N_a\,,\,N_b}}.$ Therefore, for all $n>N$ we have,

$\Big| a_n + b_n - (\alpha + \beta)\Big| = \Big|(a_n - \alpha) - (b_n - \beta) \Big| \\\textit{ }\qquad\qquad\qquad\qquad\leq \Big|a_n - \alpha\Big| + \Big|b_n - \beta \Big| \qquad \qquad\mathrm{by}\;\mathrm{the}\;\mathrm{triangle}\;\mathrm{inequality}\\ \textit{ }\qquad\qquad\qquad\qquad\leq \frac{\varepsilon}{2}+\frac{\varepsilon}{2} \\\textit{ }\qquad\qquad\qquad\qquad= \varepsilon$

(check this out for those who are unfamiliar with the triangle inequality) Completing the proof for the addition case. I leave it as a challenge to prove the subtraction case.

$\square_{\mathrm{Part}\;(4)}$

(5) $a_n \cdot b_n \rightarrow \alpha \cdot \beta,$

Scratch Work for (5): This proof is more challenging than the former four, so let’s do some scratch work for this one. We will have to use a trick that you have used before: add 0 to an equation in a helpful way. The issue is, what is the most helpful way to do so? Well, we want to show there is some $N \in \mathbb{N}$ such that,

$|a_n \cdot b_n - \alpha\cdot\beta| < \varepsilon,$

for all $n>N$ . And we know that we can make $|a_n- \alpha|$ and $| b_n - \beta|$ as small as we need. Let’s try adding and subtracting $\alpha \cdot b_n$ ,

Proof (5): Let $\varepsilon>0$ . Since $b_n$ converges, we know that $b_n$ is bounded by some $B\in \mathbb{R}\setminus{0}$ by part (1). Thus, there exist two natural numbers $N_a\,,\,N_b \in \mathbb{N}$ such that,

$|a_n - \alpha| < \frac{\varepsilon}{2B}\;\;\mathrm{for}\;\;\mathrm{all}\;n>N_a$

and

$|b_n - \beta| < \frac{\varepsilon}{2(|\alpha| +1)}\;\; \mathrm{for}\;\;\mathrm{all}\;n>N_b.$

We added 1 to the denominator because it’s possible that $\alpha = 0$ . Now, again, let $N = \max{{N_a\,,\,N_b}}.$ Therefore, for all $n>N$ we have,

$\Big|a_n \cdot b_n +\alpha \cdot b_n-\alpha \cdot b_n- \alpha\cdot\beta\Big| = \Big|(a_n +\alpha )\cdot b_n -\alpha \cdot (b_n- \beta)\Big|\\\textit{ }\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;\; \,\leq \Big|(a_n +\alpha )\cdot b_n\Big| -\Big|\alpha \cdot (b_n- \beta)\Big|\\\textit{ } \qquad\qquad\qquad\qquad\qquad\qquad\;\;\;\; \,= \Big|a_n +\alpha \Big|\cdot |b_n| -|\alpha| \cdot\Big| b_n- \beta\Big| \\\textit{ }\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;\; \,\leq \Big|a_n +\alpha \Big|\cdot B -|\alpha| \cdot\Big| b_n- \beta\Big| \\\textit{ }\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;\; \,< \frac{\varepsilon}{2B}\cdot B -|\alpha| \cdot\ \frac{\varepsilon}{2(|\alpha| +1)}\\\textit{ }\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;\; \,<\frac{\varepsilon}{2}+\frac{\varepsilon}{2} \\\textit{ }\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;\; \,= \varepsilon$

Boom! We’re done.

$\square_{\mathrm{Part}\;(5)}$

(6) If $b_n \neq 0$ for all $n$ and $\beta \neq 0$ then $\frac{a_n}{b_n} \rightarrow \frac{\alpha}{\beta},$

Scratch Work for (6): This proof is challenging in a similar way to part (5). Let’s do some scratch work again. We want to show there is some $N \in \mathbb{N}$ such that,

$\Big| \frac{a_n}{b_n} - \frac{\alpha}{\beta} \Big| <\varepsilon$

for all $n>N$ . Let’s try something similar to what we did for part (5). Let’s add and subtract $\alpha/b_n$ ,

In a similar way to part (5), we need to guarantee that $1/|b_n|$ doesn’t get too large, however, we cannot use part (1) since we could have $b_n$ be bounded between $-1<b_n<1$ get still really close to 0 and make $1/|b_n|$ very large. What we need is to show that: If $b_n \longrightarrow \beta \neq 0$ , then there is some positive real number $t$ such that $0<t<|b_n|$ for all $n$ . Once we do this, we can finish the proof of (6) in a way similar to part (5).

Proof (6): We will first show that there is some positive real number $t$ such that $0<t<|b_n|$ for all $n$ , since $\beta \neq 0$ . First, because $b_n \longrightarrow \beta$ , there is some $N_{bt} \in \mathbb{N}$ such that $|b_n - \beta| < |\beta|/2$ . Therefore, $|\beta|/2<b_n < 3|\beta|/2$ . By letting $t = |\beta|/2$ we achieved what we wanted! On to the main event.

Let $\varepsilon>0$ . Yet again, there exist two natural numbers
natural numbers $N_a\,,\,N_b \in \mathbb{N}$ such that,

$|a_n - \alpha| < \frac{\varepsilon t}{2}\;\;\mathrm{for}\;\;\mathrm{all}\;n>N_a$

and

$|b_n - \beta| < \frac{\varepsilon t}{2(|\alpha| +1)}\;\;\mathrm{for}\;\;\mathrm{all}\;n>N_b.$

We added 1 to the denominator because it’s possible that $\alpha = 0$ . Now, again, let $N = \max{{N_a\,,\,N_b\,,\,N_{bt}}}.$ Therefore, for all $n>N$ we have,

$\Bigg| \frac{a_n}{b_n} + \frac{\alpha}{b_n} -\frac{\alpha}{b_n}- \frac{\alpha}{\beta} \Bigg| =\Big|a_n - \alpha\Big| \cdot \Bigg|\frac{1}{b_n}\Bigg| + |\alpha|\cdot \Bigg|\frac{1}{b_n} - \frac{1}{\beta} \Bigg| \\\textit{ }\qquad\qquad\qquad\;\;\;\;\;\;=\Big|a_n - \alpha\Big| \cdot \Bigg|\frac{1}{b_n}\Bigg| + |\alpha|\cdot \Bigg|\frac{b_n - \beta}{b_n}\Bigg|\\\textit{ }\qquad\qquad\qquad\;\;\;\;\;\;< \Big|a_n - \alpha\Big| \cdot \frac{1}{t} + \frac{|\alpha|}{t}\cdot \Big|b_n - \beta\Big|\\\textit{ }\qquad\qquad\qquad\;\;\;\;\;\;<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}\\\textit{ }\qquad\qquad\qquad\;\;\;\;\;\;= \varepsilon \nonumber$

And we’re done, boom!

$\square_{\mathrm{Part}\;(6)}$

(7) If $a_n \geq 0$ for all $n,$ then $\lim{\sqrt{a_n}} = \sqrt{\lim{a_n}}.$

Scratch Work for (7): This one requires another little algebra trick! First, let’s state up front what we want to show, we want to find some $N\in \mathbb{N}$ such that,

$\Big| \sqrt{a_n} - \sqrt{\alpha}\Big| < \varepsilon$

for all $n>N$ . How do we achieve this? My first thought would be to square both sides:

$a_n + \alpha - 2 \sqrt{a_n \alpha}< \varepsilon^2$

But now we have $a_n + \alpha$ , we can only guarantee that $|a_n -\alpha|$ is small. Scratch this idea. What do we need to multiply $|\sqrt{a_n} - \sqrt{\alpha}|$ by to get $|a_n -\alpha|$ ? What about $|\sqrt{a_n} - \sqrt{\alpha}|$ ? We get,

$\Big| \sqrt{a_n} - \sqrt{\alpha}\Big|\cdot\Big|\sqrt{a_n} - \sqrt{\alpha}\Big| = \Big|a_n - \alpha\Big|< \varepsilon \cdot|\sqrt{a_n} - \sqrt{\alpha}|$

That’s it! Now, since $a_n$ is bounded by some $A$ , the same can be said regarding $\sqrt{a_n}$ . That is, $\sqrt{a_n}$ is bounded by $\sqrt{A}$ . Let’s try this out!

Proof (7): Let $\varepsilon>0$ . First, we know that $a_n$ is bounded by some positive $A\in \mathbb{R}$ . This then means that $\sqrt{a_n} - \sqrt{\alpha} \leq \sqrt{A}+ \sqrt{\alpha}$ . Also, we (again) know there is some natural number $N \in \mathbb{N}$ such that

$|a_n - \alpha| < \varepsilon\cdot\Big(\sqrt{A} + \sqrt{\alpha}\Big)$

for all $n>N$ . Therefore,

$\Big| \sqrt{a_n} - \sqrt{\alpha}\Big| < \frac{| \sqrt{a_n} + \sqrt{\alpha}|}{| \sqrt{a_n} + \sqrt{\alpha}|}\Big| \sqrt{a_n} - \sqrt{\alpha}\Big| \\\textit{ }\qquad\qquad\;\;\;\,= \frac{|a_n - \alpha|}{| \sqrt{a_n} + \sqrt{\alpha}|} \\\textit{ }\qquad\qquad\;\;\;\,< \frac{|a_n - \alpha|}{\sqrt{A} + \sqrt{\alpha}} \\\textit{ }\qquad\qquad\;\;\;\,<\varepsilon \nonumber$

And we’re done! Finally, am I right?

$\square_{\mathrm{Part}\;(7)}$

$\square$

Closing Remarks

We’ve covered a lot, and yet nothing surprising. I would argue that every one of the limit laws is intuitive. However, we’ve used a lot of tricks while proving them. Let’s take stock of them:

We used that $a_n \rightarrow \alpha$ and $b_n \rightarrow \beta$ to prove that there are natural numbers $N_a,N_b$ such that when $n$ is greater than them both we can make $|a_n-\alpha|$ and $|b_n - \beta|$ small.
We used the triangle inequality as well.
We added 0, in a useful way, to simplify what we had.
We multiplied by 1, in a useful way.
We used a special case of the following formula:

$a^k-b^k = (a-b)(a^{k-1} + a^{k-2}b^{} + a^{k-3}b^{2} + \cdots + a^2b^{k-3} + a^{1}b^{k-2} + b^{k-1} )$

These are all valuable tools to keep in mind when you get stuck.

Be Kind. Be Curious. Be Compassionate. Be Creative.

And Have Fun!

A Kick in the Discovery