Intuition is a dangerous weapon
Nearly every mathematician had very sharp and concise, yet sometimes dangerous intuition, which led mathematics into the very flawy statements. The theorems and even axioms in a system should be precise internally. I had talked about that topic in the page Gödel’s Proof. The definition (Definition 1.14) in section Analysis I Part 4 stated after the concept of convergence had been used without proof, relying only on intuition. Hopefully, it did not lead mathematics to flawy theorems, statements and etc. Why do we need such a solid definition of convergence? Well, it is because we need to prove them in general terms. Let us continue with boundedness and then move to the algebraic properties of limit?
Our very intuition says that if there is a thing in front/ behind of other thing then we can say the second thing is bounded by the first thing. It is hard to depict when it is written. It is way easier when it is descripted by the language of nature, mathematics. I did this little trick because I wanted to make sure that writing mathematics in daily language not always gives us the clearest understanding of concept.
$\textbf{Definition 1.15.:} $(Boundedness of a Sequence) A sequence $(x_n)$ is bounded if there exists a number $M > 0$ such that $|x_n| \leq M$ for all $n \in \mathbb{N}$
It is not surprising that if one could understand this definition at first glance. Moreover, understanding the definition is pretty trivial because its purpose is to be trivial in first place. This definition reads the following theorem.
Not only bounded but also logical
$\textbf{Theorem 1.2.:} $ Every convergent sequence is bounded.
$\textbf{Proof:} $ Let $(x_n)$ ve a convergent sequence. Then by definition given any $\epsilon$ there exist a number $N(\epsilon)$ such that $|x_n - L| < \epsilon$ whenever $n < N$. See that, $-\epsilon < x_n - L < \epsilon$ and $L - \epsilon < x_n < L + \epsilon$. Let us take $\epsilon = 1$. Then, $L-1 < x_n < L+1$.
Observe that we do not know whether $L$ is positive or not. We can rearrange the last inequality. $-(-L+\epsilon)< x_n < L+\epsilon$.
The ambiguity of $L$ allows us to write $-(L+\epsilon) < x_n < L+\epsilon$, which exactly the same thing as writing $|x_n| < |L|+\epsilon$, for all $n < N$.
But it is still worth to worry for $N$ because we did not choose any $N$ yet.
Now let us think sequences as two parts one of them is tail and the other is head. The head consist of finite elements, namely $\{x_1, x_2, \dots, x_{N-1}\}$ and the tail is $\{x_N, x_{N+1}, \dots\}$. In the light of this informations, we can freely say that the tail is bounded by $|x_n| < |L| + 1$.
However, this bound might not be large enough for the unruly terms in the head. Remember, the tail is well-behaved in some sense. The head is more angry and aggresive part.
So we generally choose $\max$ from the set that consists of the head and the bound for the tail. Namely, $M=\max\{\underbrace{|x_1|, |x_2|, \dots, |x_{N-1}|}_{the\ head}, |L|+1\}$.
We are now sure that for every elements in a convergent sequence there exists a appropriate bound.
As a comprehensive example take the sequence $|x_n| = \{100, -50, 200, \frac{1}{4}, \frac{1}{5}, \dots\}$. Using the intuition we have we presume that the sequence $(x_n)$ converges to the $L=0$. Choose $\epsilon = 1$ and $N=4$ since after $4$th term all terms are less than $1$. For the tail of this sequence the above proof gives us a bound $|x_n|<|L|+1$, which equals to $|x_n| < 1$ for all $n \geq 4$.
However, the last part of the problem implies that choosing $1$ as a bound is not enough, we should choose it wisely. A simple accident of choice would lead us to the fatal logical errors. For instance, we do not know whether the bound $1$ work for the head of the sequence $(x_1, x_2, x_3)$? Let us check the validity.
- Is $|x_1| < 1$? No. |100| is not less than 1.
- Is $|x_2| < 1$? No. |-50| is not less than 1.
- Is $|x_3| < 1$? No. |200| is not less than 1.
Therefore, if we had considered only the tail, we would have reached the wrong conclusion about boundedness. Using the proof,
$M = \max \{|100|,|-50|,|200|,1 \}$.
In this case we can freely say the bound for the sequence is $200$. Every single term, from $x_1$ to infinity, has an absolute values less than or equal to $200$.
Some operations on limit
After clearing the gaps in our minds, we can move on to the topic, which shows how sequences behave well with respect to the operations of addition, multiplication and division.
$\textbf{Theorem 1.3.:} $(Algebraic Limit Theorem) Let $\lim a_n = a$, and $\lim b_n = b$. Then,
- $\lim (c a_n) = ca$ for all $c \in \mathbb{R}$
- $\lim (a_n + b_n) = a + b$
- $\lim (a_n b_n) = ab$
- $\lim (a_n / b_n) = a/b$ provided $b \neq 0$
$\textbf{Proof:} $
(i.) For $c=0$ it is trivial, since $\lim(0 a_n) = \lim(0) = 0$. For $c \neq 0$ we want to show that given any $\epsilon > 0$ there exists $N = N(\epsilon)$ such that $|c a_n - ca| < \epsilon$ whenever $n \geq N$.
The term $|c a_n -ca|$ equals to the $|c||a_n - a|$. We already know that $(a_n) \rightarrow a$. This means that we can choose $N$ such that $|a_n - a| < \dfrac{\epsilon}{|c|}$ whenever $n \geq N$. All in all, we observe that $|ca_n - ca| = |c||a_n - a| < |c| \dfrac{\epsilon}{|c|} = \epsilon$ whenever $n \geq N$
(ii.) We use another algebraic manipulation called triangle inequality. Let us write the definition of the limit corresponding to the given term in the question. Given any $\epsilon > 0$ there exists $N = N(\epsilon)$ such that $|(a_n + b_n) - (a + b)| < \epsilon$ whenever $n \geq N$.
I like writing everything, including the definitions. It helps me to understand the question and remind me the inventory I have. In this case, we have the term above literally says “apply me the triangle inequality”. $|(a_n + b_n) - (a + b)| = |(a_n - a) + (b_n - b)| \leq |a_n - a| + |b_n - b|$. We all know where we are going from now on.
The hypothesis $(a_n) \rightarrow a$ and $(b_n) \rightarrow b$ imply there exists $N_1$ and $N_2$ correspondingly. Therefore the followings are valid:
- $|a_n-a| < \dfrac{\epsilon}{2}$ whenever $n \geq N_1$
- $|b_n-b| < \dfrac{\epsilon}{2}$ whenever $n \geq N_2$
Remember that in here we showed the importance of choosing the index number $N$. Using the same logic we choose $N = \max \{N_1, N_2\}$. We now sure that if $n \geq N$, then $n \geq N_1$ and $n \geq N_2$. In the light of these information;
$|(a_n + b_n) - (a + b)| \leq |a_n - a| + |b_n - b| < \dfrac{\epsilon}{2} + \dfrac{\epsilon}{2} = \epsilon$ whenever $n \geq N$.
(iii.) Showing this is trickier than others. Observe that,
$|a_n b_n - a b| = |a_n b_n - a b_n + a b_n - ab| \leq |a_n b_n - a b_n| + |a b_n - ab| = |b_n||a_n - a| + |a||b_n - b|$
We add and substract the term $a b_n$. Why that term? Because we simply need it 😊. Let $|a| \neq 0$. Choose $N_1$ such that $|b_n - b| < \dfrac{1}{|a|}\dfrac{\epsilon}{2}$ whenever $n \geq N_1$.
We should not worry about the term $|b_n|$ on the left hand-side $(|b_n||a_n-a|)$ thanks to Theorem 1.2. We directly conclude that there exist $M > 0$ such that $|b_n| < M$ for all $n \in \mathbb{N}$. Using these ideas we choose $N_2$ so the definition is satisfied;
$|a_n - a| < \dfrac{1}{M}\dfrac{\epsilon}{2}$ whenever $n \geq N_2$.
Same trick applies here. Choose $N = \max\{N_1, N_2\}$. Then,
$|a_n b_n - ab| \leq |a_n b_n - a b_n| + |a b_n - a b| = |b_n||a_n - a| + |a||b_n-b| \leq M|a_n - a| + |a||b_n-b| < M \left(\dfrac{\epsilon}{2M}\right) + |a|\left(\dfrac{\epsilon}{2|a|}\right) = \epsilon$.
(iv.) This will follow from (iii.). Use the following: $(b_n) \rightarrow b$ implies $\left(\dfrac{1}{b_n}\right) \rightarrow \dfrac{1}{b}$.
Ordering on the limits
As you might guess, ordering is also well-behaved propert of limit.
$\textbf{Theorem 1.4.:} $(Order Limit Theorem) Assume $\lim a_n = a$ and $\lim b_n = b$
- If $a_n \geq 0$ for all $n \in \mathbb{N}$, then $a \geq 0$
- If $a_n \leq b_n$ for all $n \in \mathbb{N}$, then $a \leq b$
- If there exists $c \in \mathbb{R}$ for which $c \leq b_n$ for all $n \in \mathbb{N}$, then $c \leq b$
$\textbf{Proof:} $
(i.) Let us assume $a < 0$. Using the definition of the limit we can write: given any $\epsilon > 0$, there exists $N = N(\epsilon)$ such that $|a_n - a| < \epsilon$ whenever $n \geq N$. Choose $\epsilon = |a|$.
Then $|a_N - a| < |a| \rightarrow -|a| < a_N - a < |a|$ and $a - |a| < a_N < a + |a|$. We get $a - (-a) < a_N < a - a = 0$. A contradiction since $a_N < 0$. Therefore, $a \geq 0$
(ii.) By using (ii.) from Theorem 1.3. implies $b_n - a_n$ converges to $b-a$. Assume $b-a < 0$. Take for example $\epsilon = |b-a|$. Then using exact same procedure as in above:
$|b_n - a_n - (b-a)| < |b-a| \rightarrow -|b-a| < b_N - a_N - (b - a) < |b-a|$. We end up with same result: $(b-a) - |b-a| < b_N - a_N < (b-a) + |b-a| \rightarrow b_N - a_N < 0 \rightarrow b_N < a_N$, which contradicts with the hypothesis.
(iii.) Let $c \in \mathbb{R}$ that satisfies $c \leq b_n$ for all $n \in \mathbb{N}$. Assume c > b. Consider $c$ as the limit of a sequence $(c_n)$. Then, apply (ii.). We have $c > b_n$ for all $n \in \mathbb{N}$, which is a fatal contradiction. Thus $c \leq b$.
$\textbf{Theorem 1.5.:} $(Squeeze - Sandwich Theorem) If $a_n \leq b_n \leq c_n$ for all $n \geq N$ and if $\lim a_n = \lim c_n = L$, then $\lim b_n = L$.
$\textbf{Proof:} $ Using (ii.) from Theorem 1.4. we have the following inequalities: $\lim a_n < \lim b_n < \lim c_n \rightarrow L < \lim b_n < L$. Only way out is $\lim b_n = L$.
$\textbf{Question:} $ Show that $\lim_{n \rightarrow \infty} \dfrac{sin(n)}{n}=0$
$\textbf{Solution:} $ We know that $|\sin(n)| < 1$ from our very information of trigonometry. We also know that $\lim_{n \rightarrow \infty} \dfrac{1}{n} = 0$ and $\lim_{n \rightarrow \infty} -\dfrac{1}{n} = 0$. Using this, we have $-1 < \sin(n) < 1$ for all $n \in \mathbb{R}$. If we divide the inequality by $n$ we get: $-\dfrac{1}{n} < \dfrac{\sin(n)}{n} < \dfrac{1}{n}$. Adding limit to the every sides, $\lim_{n \rightarrow \infty} -\dfrac{1}{n} < \lim_{n \rightarrow \infty} \dfrac{\sin(n)}{n} < \lim_{n \rightarrow \infty} \dfrac{1}{n}$. Now, it is trivial to write $\lim_{n \rightarrow \infty} \dfrac{\sin(n)}{n} = 0$.
In the next section, we will talk about Monotone Convergence Theorem, Subsequences, Bolzano-Weierstrass Theorem and Cauchy Criterion.