Properties of limits

We see some basic properties of the limit of a sequence, such as

• The limit of a sequence is unique.
• A convergent sequence is bounded.
• The squeeze theorem.
• Limiting operation is linear.
• The limit of the product of sequences is the product of the limits.
• etc.

Theorem (Uniqueness of limit)

If the sequence $\{a_n\}$ converges, its limit is unique. In other words, if we have $\underset{n\to \mathrm{\infty }}{lim}{a}_n=\alpha$ and $\underset{n\to \mathrm{\infty }}{lim}{a}_n=\beta$, then $\alpha =\beta$.

Proof. Suppose that $\{a_n\}$ converges to $\alpha$ as well as to $\beta$. Take any positive real number $\epsilon$. 

Since $\underset{n\to \mathrm{\infty }}{lim}{a}_n=\alpha$, there exists $N^{\prime }\in \mathbb{N}$ such that for any $n\ge N^{\prime }$, $|a_n-\alpha |<\epsilon$.

Similarly, there exists $N^{″}\in \mathbb{N}$ such that for any $n\ge N^{″}$, $|a_n-\beta |<\epsilon$.

Let $N=max\{N^{\prime },N^{″}\}$. Then, if $n\ge N$,

$$|\alpha -\beta |=|\alpha -a_n+a_n-\beta |\le |\alpha -a_n|+|a_n-\beta |<\epsilon +\epsilon =2\epsilon .$$

Since $\epsilon >0$ is arbitrary, it must be that $\alpha =\beta$. ■

Definition (Bounded sequence)

Given a sequence $\{a_n\}$, we can define a set of real numbers that appear in the sequence: $\{a_n\mid n\in \mathbb{N}\}\subset \mathbb{R}$. If this set is bounded above, bounded below, or bounded, we say that the sequence $\{a_n\}$ is bounded above, bounded below, or bounded, respectively.

Theorem (Convergent sequences are bounded)

Any convergent sequence (i.e., a sequence that converges) is bounded.

Proof. Let $\{a_n\}$ be a converging sequence such that $\underset{n\to \mathrm{\infty }}{lim}{a}_n=\alpha$. Let $\epsilon =1$. Then there exists $N\in \mathbb{N}$ such that for all $n\ge N$, $|a_n-\alpha |<1$. Then, by the triangle inequality,

$$|a_n|-|\alpha |\le |a_n-\alpha |<1$$

so that

$$|a_n|<|\alpha |+1.$$

Now, let us define

$$M=max\{|\alpha |+1,|a_1|,|a_2|,\cdots ,|a_{N-1}|\}.$$

Then, whether $n\ge N$ or $n<N$, we always have $|a_n|\le M$. Hence $\{a_n\}$ is bounded. ■

Remark. The converse of this theorem is not necessarily true. e.g., $a_n=(-1{)}^n$ is bounded, but not convergent. □

Theorem (Squeeze Theorem)

Let $\{a_n\},\{b_n\},\{c_n\}$ be sequences. Suppose that for all $n\in \mathbb{N}$, we have $a_n\le b_n\le c_n$ and $\underset{n\to \mathrm{\infty }}{lim}{a}_n=\underset{n\to \mathrm{\infty }}{lim}{c}_n=\alpha$. Then $\underset{n\to \mathrm{\infty }}{lim}{b}_n=\alpha$.

Proof. Let $\epsilon >0$ be an arbitrary positive real number. Since $a_n\to \alpha$, there exists $N^{\prime }\in \mathbb{N}$ such that for any $n\ge N^{\prime }$,  $|a_n-\alpha |<\epsilon$. In particular, for any $n\ge N^{\prime }$,

$$-\epsilon <a_n-\alpha .$$

Similarly, there exists $N^{″}\in \mathbb{N}$ such that for any $n\ge N^{″}$,

$$c_n-\alpha <\epsilon .$$

Let $N=max\{N^{\prime },N^{″}\}$. Then for any $n\ge N$, we have

$$-\epsilon <a_n-\alpha \le c_n-\alpha <\epsilon .$$

However, since $a_n-\alpha \le b_n-\alpha \le c_n-\alpha$, we have

$$|b_n-\alpha |<\epsilon .$$

Therefore $\underset{n\to \mathrm{\infty }}{lim}{b}_n=\alpha$. ■

Let $\{a_n\}$ be a sequence. Let's pick a subset $\{n_k\}$ of $\mathbb{N}$ which we arrange in increasing order:

$$n_1<n_2<n_3<\cdots n_k<n_{k+1}<\cdots .$$

Then, for $k=1,2,\cdots$ we obtain a sequence

$$a_{n_1},a_{n_2},\cdots ,a_{n_k},a_{n_{k+1}},\cdots$$

This new sequence $\{a_{n_k}\}$ is called a subsequence of $\{a_n\}$.

Theorem (Convergence of subsequence)

Let $\{a_n\}$ be a sequence that converges to $\alpha \in \mathbb{R}$. Then, any subsequence $\{a_{n_k}\}$ also converges to $\alpha$.

Proof. Let $\epsilon >0$ be any positive real number. Since $\underset{n\to \mathrm{\infty }}{lim}{a}_n=\alpha$, there exists $N\in \mathbb{N}$ such that for all $n\ge N$, $|a_n-\alpha |<\epsilon$. 

Now take $K\in \mathbb{N}$ such that $n_K\ge N$. Then, for any $k\ge K$, $n_k\ge N$. Thus $|a_{n_k}-\alpha |<\epsilon$. 

That is, for any $\epsilon >0$, there exists $K\in \mathbb{N}$ such that for all $k\ge K$, $|a_{n_k}-\alpha |<\epsilon$. This means $\underset{k\to \mathrm{\infty }}{lim}{a}_{n_k}=\alpha$. ■

Theorem

Let $\{a_n\}$ be a sequence that converges to $\alpha \in \mathbb{R}$.

1. If there exists a real number $a$ such that $a_n\ge a$ holds for infinitely many $n\in \mathbb{N}$, then $\alpha \ge a$.
2. If there exists a real number $b$ such that $a_n\le b$ holds for infinitely many $n\in \mathbb{N}$, then $\alpha \le b$.

Proof. We only prove part 1. Part 2 is similar.

Let us define a subsequence of $\{a_n\}$ such that $\{a_n\mid a_n\ge a\}$, and re-index this subsequence so that we may assume $a_n\ge a$ for all $n\in \mathbb{N}$ in this subsequence. We prove it by contradiction.

Suppose $\alpha <a$. Let $\epsilon =a-\alpha (>0)$. Since $\underset{n\to \mathrm{\infty }}{lim}{a}_n=\alpha$, there exists $N\in \mathbb{N}$ such that for all $n\ge N$, $|a_n-\alpha |<\epsilon$. But then, $a_n<\alpha +ϵ=a$, which is a contradiction. ■

Theorem

Let $\{a_n\}$ and $\{b_n\}$ be sequences such that $\underset{n\to \mathrm{\infty }}{lim}{a}_n=\alpha$ and $\underset{n\to \mathrm{\infty }}{lim}{b}_n=\beta$.

1. $\underset{n\to \mathrm{\infty }}{lim}k{a}_n=k\alpha$ where $k$ is a constant.
2. $\underset{n\to \mathrm{\infty }}{lim}(a_n+b_n)=\alpha +\beta$.
3. $\underset{n\to \mathrm{\infty }}{lim}{a}_n{b}_n=\alpha \beta$.
4. $\underset{n\to \mathrm{\infty }}{lim}\frac{a_n}{b_n}=\frac{\alpha }{\beta }$ given $\beta \ne 0$.

Proof. Let $\epsilon >0$. There exists $N_1\in \mathbb{N}$ such that for any $n\ge N_1$, $|a_n-\alpha |<\epsilon$.

Also, there exists $N_2\in \mathbb{N}$ such that for any $n\ge N_2$, $|b_n-\beta |<\epsilon$. Let $N=max\{N_1,N_2\}$. Then for $n\ge N$, we have both $|a_n-\alpha |<\epsilon$ and $|b_n-\beta |<\epsilon$.

1. This is a special case of Part 3 (where $b_n=k$).
2. By the triangular inequality, $$\begin{array}{rcl}|(a_n+b_n)-(\alpha +\beta )|& =& |(a_n-\alpha )+(b_n-\beta )|\\ & \le & |a_n-\alpha |+|b_n-\beta |\\ & <& \epsilon +\epsilon =2\epsilon .\end{array}$$ Note that $2\epsilon$ is also an arbitrary positive real number. Therefore, for any $n\ge N$, we have $$|(a_n+b_n)-(\alpha +\beta )|<2\epsilon .$$ Thus, $$\underset{n\to \mathrm{\infty }}{lim}(a_n+b_n)=\alpha +\beta .$$
3. $\{a_n\}$ and $\{b_n\}$ are convergent, and are hence bounded. Therefore, there exists some real number $M>0$ such that $$|a_n|\le M,|b_n|\le M.$$ By the above theorem, we also have $$|\alpha |\le M,|\beta |\le M.$$ Then, $$\begin{array}{rcl}|a_n{b}_n-\alpha \beta |& =& |a_n{b}_n-a_n\beta +a_n\beta -\alpha \beta |\\ & =& |a_n(b_n-\beta )+(a_n-\alpha )\beta |\\ & \le & |a_n|\cdot |b_n-\beta |+|a_n-\alpha |\cdot |\beta |\\ & <& M\epsilon +\epsilon M=2M\epsilon .\end{array}$$ Note that $2M\epsilon$ is an arbitrary positive real number. Thus, $$\underset{n\to \mathrm{\infty }}{lim}{a}_n{b}_n=\alpha \beta .$$
4. First we show that there exists $N\in \mathbb{N}$ such that for any $n\ge N$, $b_n\ne 0$. Since $\beta \ne 0$, $|\beta |>0$. There exists $N$ such that for any $n\ge N$, $|b_n-\beta |<\frac{1}{2}|\beta |$. By the triangular inequality, $$|\beta |-|b_n|\le |\beta -b_n|<\frac{1}{2}|\beta |$$ so that $$|b_n|>|\beta |-\frac{1}{2}|\beta |=\frac{1}{2}|\beta |>0.$$ Therefore, for $n\ge N$, $b_n\ne 0$. For any $\epsilon >0$ and a sufficiently large $n$, we have $|b_n-\beta |<\epsilon$, and $$|\frac{1}{b_n}-\frac{1}{\beta }|=\frac{|\beta -b_n|}{|b_n\beta |}<\frac{2\epsilon }{{\beta }^2}.$$ Thus, $\{\frac{1}{b_n}\}$ converges to $\frac{1}{\beta }$. Therefore, using Part 3, we have $$\underset{n\to \mathrm{\infty }}{lim}\frac{a_n}{b_n}=\underset{n\to \mathrm{\infty }}{lim}{a}_n\frac{1}{b_n}=\alpha \cdot \frac{1}{\beta }=\frac{\alpha }{\beta }.$$

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