Redefining the Limits of Sequences in University Mathematics 📂Analysis

Redefining the Limits of Sequences in University Mathematics

Definitions¹ ²

$\mathbb{N}$ represents the set of natural numbers, and $\mathbb{R}$ represents the set of real numbers.

A function with a domain of $\mathbb{N}$ is called a sequence.
For a sequence of natural numbers $\left\{ n_{k} \right\}_{ k \in \mathbb{N}}$ , $\left\{ x_{n_{k}} \right\}_{ k \in \mathbb{N}}$ is called a subsequence of $\left\{ x_{n} \right\}_{ n \in \mathbb{N}}$ .
If for every $x \in \left\{ x_{n} \right\}_{ n \in \mathbb{N}}$ there exists $M \in \mathbb{R}$ such that $x \le M$ is satisfied, then $\left\{ x_{n} \right\}_{ n \in \mathbb{N}}$ is bounded above, and if there exists $m \in \mathbb{R}$ such that $m \le x$ is satisfied, then it is bounded below; if it is bounded above and below, it is simply bounded.
Let’s say $\left\{ x_{n } \right\}_{n = 1}^{\infty}$ is a sequence of real numbers. If for every $\varepsilon > 0$ there exists $N \in \mathbb{N}$ such that $n \ge N \implies | x_{n} - a | < \varepsilon$ is satisfied, then $\left\{ x_{n } \right\}$ is said to converge to $a \in \mathbb{R}$ , denoted $\lim \limits_{n\to \ \infty}x_{n}=a$ .
If $\left\{ x_{n } \right\}$ does not converge, it is said to diverge.
- If for every $M \in \mathbb{R}$ there exists $N \in \mathbb{N}$ such that $n \ge N \implies x_{n} > M$ is satisfied, $\lim \limits_{n\to \infty} x_{n} = +\infty \quad \text{ or } \quad x_{n} \to +\infty$ is denoted.
- If for every $M \in \mathbb{R}$ there exists $N \in \mathbb{N}$ such that $n \ge N \implies x_{n} < M$ is satisfied, $\lim \limits_{n\to \infty} x_{n} = -\infty \quad \text{ or } \quad x_{n} \to -\infty$ is denoted.

Explanation

When you first encounter the definitions of convergence and divergence in university, it seems like an arrow going nowhere with confusing terms such as $\varepsilon$ , $M$ , and $N$ popping up. Honestly, you might not want to learn it. From the perspective of an undergrad, not knowing this new definition of limits doesn’t mean they don’t understand the concept of limits at all, especially because it feels like if they can just somehow get through the midterms, they’ll never have to deal with it again. Of course, this is a foolish thought.

Looking back at high school, teachers did use phrases like “grows infinitely large” or “sent to infinity” when talking about $n \to \infty$ , but somehow, they seemed overly cautious about treating sequences as “something moving”. It’s because they are educated people.

The reason for using strict definitions instead of intuition is that strict definitions are, in fact, easier. While intuition works quickly for ‘simple sequences’ that might appear in the SATs, there’s a deficiency when dealing with ‘complex sequences’, which is why strict definitions were introduced. Historically, despite British mathematics significantly outpacing continental mathematics since Newton, the insistence on intuitionism led to the loss of academic dominance to the continent.

The biggest stumbling block in learning about sequence convergence isn’t the inherent difficulty but the annoyance from ‘having to’ ‘complicatedly’ ’re-learn’ something that seems easily approachable. For instance, taking an unnecessarily roundabout way to solve an easy problem like $\displaystyle \lim_{n \to \infty} {{n + 3} \over {2n}} = {{1} \over {2}}$ feels this way.

The problem is not so much that learning is difficult, but rather not wanting to learn, and unfortunately, teaching sequences worth the difficulty of understanding to students who struggle with it is a very difficult task. To aid in understanding and empathy, the following two theorems are introduced.

Theorems

Let $\left\{ w_{n} \right\}, \left\{ x_{n} \right\}, \left\{ y_{n} \right\}$ be a sequence of real numbers, and let’s say it is $a \in \mathbb{R}$ .

(a) The Sandwich Theorem:
$\displaystyle \lim_{n \to \infty} x_{n} = \lim_{n \to \infty} y_{n} = a$
If this holds and
$n \ge N_{0} \implies x_{n} \le w_{n} \le y_{n}$
a $N_{0} \in \mathbb{N}$ exists that satisfies
$\displaystyle \lim_{n \to \infty} w_{n} = a$
(b) The Comparison Theorem:
$n \ge N_{0} \implies x_{n} \le y_{n}$
If a $N_{0} \in \mathbb{N}$ exists that satisfies
$\displaystyle \lim_{n \to \infty} x_{n} \le \lim_{n \to \infty} y_{n}$

Of course, both the Sandwich Theorem and the Comparison Theorem seem to obviously hold from an intuitive standpoint. They aren’t particularly hard facts. But how exactly will you, who resists the new definitions of convergence, prove these facts?

While these two theorems were confidently used without proof even at the high school level, they were actually hypotheses accepted through common-sense guessing rather than logical deduction. Considering how often human common sense is wrong, the need for strict proof becomes understandable, at least for STEM students.

Following the definitions of convergence, the proofs of these theorems are not difficult, but assuming this is the reader’s first encounter with such reasoning, they are shown in as much detail as possible. Reading the proofs, one might consistently feel that there’s an obsession with the existence of $N$ , and indeed there is. When demonstrating convergence, it’s not crucial to establish an inequality where $| x_{n} - a |$ becomes smaller than $\varepsilon$ ; showing the existence of $N$ satisfying the equation is the priority.

To put it bluntly, when demonstrating the convergence of a sequence, how $\varepsilon$ was found does not matter. According to the definition, as long as $N$ exists, the sequence converges, so one should first focus on the existence of $N$ . Failing to understand this leads to ignoring given $N_{1}$ and $N_{2}$ in the problem and laying out plausible inequalities only to present a logically collapsed argument.

Proofs

(a)

Strategy: Rather than vaguely sending to infinity, it specifically demonstrates the existence of $N$ meeting $n \ge N \implies| w_{n} - a | < \varepsilon$ by breaking down inequality.

Let’s assume $\varepsilon > 0$ .

Given $\displaystyle \lim_{n \to \infty} x_{n} = \lim_{n \to \infty} y_{n} = a$ , therefore

$n \ge N_{1} \implies | x_{n} - a | < \varepsilon$

$n \ge N_{2} \implies | y_{n} - a | < \varepsilon$

a $N_{1} , N_{2} \in \mathbb{N}$ exists satisfying these. Summarizing the necessary parts,

$n \ge N_{1} \implies a - \varepsilon < x_{n}$

$n \ge N_{2} \implies y_{n} < a + \varepsilon$

Meanwhile, assuming a $N_{0} \in \mathbb{N}$ exists that satisfies $n \ge N_{0} \implies x_{n} \le w_{n} \le y_{n}$ , and organizing neatly,

$n \ge N_{1} \implies a - \varepsilon < x_{n}$

$n \ge N_{0} \implies x_{n} \le w_{n} \le y_{n}$

$n \ge N_{2} \implies y_{n} < a + \varepsilon$

The existence of $N_{0}$ , $N_{1}$ , and $N_{2}$ already assures us that an $N := \max \left\{ N_{0} , N_{1} , N_{2} \right\}$ exists. Then, for such $N$ when $n \ge N$ ,

$a - \varepsilon < x_{n} \le w_{n} \le y_{n} < a + \varepsilon$

In other words, for the previously proven existing $N$ ,

$n \ge N \implies | w_{n} - a | < \varepsilon$ According to the definition of convergence, we obtain $\displaystyle \lim_{n \to \infty} w_{n} = a$ .

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(b)

Strategy: The outcome is negated and shown not to satisfy the inequality. This logical progression is difficult to use when perceiving sequences as “something moving.” Rather, it’s better to imagine catching an infinitely growing $n$ and fixing it to use mathematical skill.

$x := \lim_{n \to \infty} x_{n}$

$y := \lim_{n \to \infty} y_{n}$

Assuming $x > y$ , meaning

$n \ge N_{1} \implies | x_{n} - x | < \varepsilon$

$n \ge N_{2} \implies | y_{n} - y | < \varepsilon$

There are natural numbers $N_{1}$ , $N_{2}$ existing that satisfy these, and expanding the absolute value,

$n \ge N_{3} \implies \begin{cases} - \varepsilon + x< x_{n} \\ y_{n} < y + \varepsilon \end{cases}$

$N_{3} > N_{0}$

a $N_{3} : = \max \left\{ N_{0} +1 , N_{1} , N_{2} \right\}$ also exists. Now assuming $\displaystyle \varepsilon := {{ x - y} \over {2}}$ means $\varepsilon > 0$ , and by assumption for all $n \ge N_{3}$ ,

$\begin{align*} y_{n} <& y + \varepsilon \\ =& y + \left( {{x - y} \over {2}} \right) \\ =& x - \left( {{x - y} \over {2}} \right) \\ =& x - \varepsilon \\ <& x_{n} \end{align*}$

However, since $N_{3} > N_{0}$ , a contradiction arises with $n \ge N_{0} \implies x_{n} \le y_{n}$ . Therefore, we get $x \le y$ .

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