If we need to give a concrete sequence, we often give each term as a formula in terms of $n$. For instance, $\left\{ \dfrac{1}{n} \right\}_{n=1}^\infty$, or simply $\left\{ \dfrac{1}{n} \right\}$, stands for the sequence $$\ds 1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \ldots $$ This is a bounded sequence.

Claim: The sequence $\left\{ \dfrac{1}{n} \right\}$ is convergent and \begin{equation*} \lim_{n\to \infty} \frac{1}{n} = 0 . \end{equation*}

Proof. Given an $\vre > 0$, we find an $N \in \N$ such that $0 \lt \dfrac{1}{N} \lt \vre$ (thanks to the Archimedean property 😎).

Then for all $n \geq N$,

Claim: The sequence $\{ {(-1)}^n \}$ is divergent.

Proof. If there were a limit $L$, then for $\vre = \dfrac{1}{2}$ we expect an $N$ that satisfies the definition.

Suppose such an $N$ exists. Then for an even $n \geq N$ we compute

Claim: The sequence $\{ {(-1)}^n \}$ is divergent.

Proof. If there were a limit $L$, then for $\vre = \dfrac{1}{2}$ we expect an $N$ that satisfies the definition.

Suppose such an $N$ exists. Then for an even $n \geq N$ we compute

But $\,\ds 2 = \abs{1 - L - (-1 -L)}$ $\ds \leq \abs{1 - L} + \abs{-1 -L}$

which is a contradiction. ❗

Proof. Suppose $\{ x_n \}$ converges to $L$. Thus there exists an $N \in \N$ such that for all $n \geq N$, we have $\abs{x_n - L} \lt 1$. Let $C_1 := \abs{L}+1$ and note that for $n \geq N$,

The set $\{ \abs{x_1}, \abs{x_2}, \ldots, \abs{x_{N-1}} \}$ is a finite set and hence let

Let $B := \max \{ C_1, C_2 \}$. Then for all $n \in \N$, $\abs{x_n} \leq B. \; \blacksquare$

Definition 2.2.1 A sequence $\{ x_n \}$ is called:

1. monotone increasing if $x_n \leq x_{n+1}$ for all $n \in \N$.
2. strictly monotone increasing if $x_n \lt x_{n+1}$ for all $n \in \N$.
3. monotone decreasing if $x_n \geq x_{n+1}$ for all $n \in \N$.
4. strictly monotone decreasing if $x_n > x_{n+1}$ for all $n \in \N$.

Theorem 2.2.1. A monotone sequence $\{ x_n \}$ is bounded if and only if it is convergent.

Proof. Let $\{ x_n \}$ be monotone and bounded. To prove that $\{ x_n \}$ converges using the definition of convergence, we need a candidate for the limit. Let's assume that the sequence is increasing (the decreasing case is handled similarly), and consider the set of points $\{x_n : n\in \N\}.$ By assumption, this set is bounded, so define $$s := \sup\, \{x_n : n \in \N\}.$$ It seems reasonable to claim that $\lim \{x_n\} = s.$

Theorem 2.2.1. A monotone sequence $\{ x_n \}$ is bounded if and only if it is convergent.

Proof. It seems reasonable to claim that $\lim \{x_n\} = s$. To prove this, let $\vre >0.$ Because $s$ is the least upper bound of $\{x_n : n \in \N\},$ $s-\vre$ is not an upper bound, so there exists a point in the sequence $x_N$ such that $s-\vre \lt x_N.$ Now, the fact that $\{x_n \}$ is increasing implies that if $n\geq N,$ then $x_N\leq x_n.$ Thus we have

Proof. The sequence is bounded below as $\dfrac{1}{\sqrt{n}} > 0$ for all $n \in \N$. Let us show that it is monotone decreasing. We start with $\sqrt{n+1} \geq \sqrt{n}$ (why is that true?). From this inequality we obtain \begin{equation*} \frac{1}{\sqrt{n+1}} \leq \frac{1}{\sqrt{n}} . \end{equation*}

So the sequence is monotone decreasing and bounded below (hence bounded).

Proof. The Monotone Convergence Theorem (Theo 2.2.1) says that that the sequence is convergent and \begin{equation*} \lim_{n\to \infty} \frac{1}{\sqrt{n}} = \inf \left\{ \frac{1}{\sqrt{n}} : n \in \N \right\} . \end{equation*} We already know that the infimum is greater than or equal to 0, as 0 is a lower bound.

Proof. Take a number $b \geq 0$ such that $b \leq \frac{1}{\sqrt{n}}$ for all $n$. We square both sides to obtain \begin{equation*} b^2 \leq \frac{1}{n} \quad \text{for all } \; n \in \N. \end{equation*} This implies that $b^2 \leq 0$. As $b^2 \geq 0$ as well, we have $b^2 = 0$ and so $b = 0$. Hence, $b=0$ is the greatest lower bound, and $$\lim \dfrac{1}{\sqrt{n}} = 0.$$

A warning: Before using the Monotone Convergence Theorem (Theo 2.2.1) to conclude a sequence converges we must show that a monotone sequence is bounded. The sequence $$\left\{ 1 + \frac{1}{2} + \cdots + \frac{1}{n} \right\}$$ is a monotone increasing sequence that grows very slowly. We will see, once we get to series, that this sequence has no upper bound and so does not converge.

It is not at all obvious that this sequence has no upper bound. 😥

Definition 2.2.2. Let $\{ x_n \}$ be a sequence. Let $\{ n_i \}$ be a strictly increasing sequence of natural numbers, that is, $n_i \lt n_{i+1}$ for all $i$ (in other words $n_1 \lt n_2 \lt n_3 \lt \cdots$). The sequence \begin{equation*} \{ x_{n_i} \}_{i=1}^\infty \end{equation*} is called a subsequence of $\{ x_n \}.$

Theorem 2.2.2. If $\{ x_n \}$ is a convergent sequence, then every subsequence $\{ x_{n_i} \}$ is also convergent, and \begin{equation*} \lim_{n\to \infty} x_n = \lim_{i\to \infty} x_{n_i} . \end{equation*}

Proof. Suppose $\lim_{n\to \infty} x_n = L.$ So for every $\vre > 0$ there is an $N \in \N$ such that for all $n \geq N,$ \begin{equation*} \abs{x_n - L} \lt \vre . \end{equation*} It is not hard to prove (do it! 📝) by Induction that $n_i \geq i.$ Hence $i \geq N$ implies $n_i \geq N.$ Thus, for all $i \geq N,$

Let $L := \lim\, a_n = \lim\, b_n.$ Let $\vre > 0$ be given. Find an $N_1$ such that for all $n \geq N_1,$ we have that $\abs{a_n-L} \lt \vre$, and an $N_2$ such that for all $n \geq N_2,$ we have $\abs{b_n-L} \lt \vre.$ Set $N := \max \{N_1, N_2 \}.$ Suppose $n \geq N.$ In particular, $L - a_n \lt \vre,$ or $L - \vre \lt a_n.$ Similarly, $b_n \lt L + \vre.$ Putting everything together, we find \begin{equation*} L - \vre \lt a_n \leq x_n \leq b_n \lt L + \vre . \end{equation*}

In other words, $-\vre \lt x_n-L \lt \vre$ or $\abs{x_n-L} \lt \vre$. So $\{x_n\}$ converges to $L.$ $\blacksquare$

Consider the sequence $\left\{ \dfrac{1}{n\sqrt{n}} \right \}.$ Since $\sqrt{n} \geq 1$ for all $n \in \N,$ we have \begin{equation*} 0 \leq \frac{1}{n\sqrt{n}} \leq \frac{1}{n} \text{ for all } n \in \N. \end{equation*} We already know $\lim \dfrac{1}{n} = 0.$ Hence, using the constant sequence $\{ 0 \}$ and the sequence $\left\{ \frac{1}{n} \right\}$ in the squeeze lemma, \begin{equation*} \text{we conclude } \;\lim_{n\to\infty} \frac{1}{n\sqrt{n}} = 0 . \end{equation*}