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2.1 Convergent sequences

We have already considered convergent sequences for rational numbers. Without proper definition of “differences of real numbers” we did not have other choice. Now however, we can talk about adding and subtracting real numbers without any worry. Convergence is the key notion of real analysis, so let us repeat its definition for real numbers.

Definition 2.1.1 (Plain English definition of convergence of real numbers)

A sequence of real numbers $\{x_{n}\}^{\infty}_{n=1}$
converges to the real number $x$ , if after a while the elements of the sequence are very close to the number $x$ .

Definition 2.1.2 (Convergence of real numbers)

A sequence of real numbers $\{x_{n}\}^{\infty}_{n=1}$
converges to the real number $x$ , $x_{n}\to x$ , if for any $\varepsilon>0$ there exists $N>0$ such that $|x_{n}-x|\leq\varepsilon$ provided that $n\geq N$ . The number $x$ is called the limit of the sequence $\{x_{n}\}^{\infty}_{n=1}$ . Sometimes we will say that $\{x_{n}\}^{\infty}_{n=1}$ “tends” to $x$ .

Sometimes we will use the following notation for the convergence of $\{x_{n}\}^{\infty}_{n=1}$ :

\lim_{n\to\infty}x_{n}=x\,.

The following simple observation is crucial.

Lemma 2.1.1

If $\lim_{n\to\infty}x_{n}=x$ and $\lim_{n\to\infty}x_{n}=y$ , then $x=y$ .

Proof: Suppose that $\lim_{n\to\infty}x_{n}=x$ and $\lim_{n\to\infty}x_{n}=y$ , but $x\neq y$ . By definition, there exists $N>0$ such that if $n\geq N$ , then both $|x_{n}-x|<\frac{|x-y|}{2}$ and $|x_{n}-y|<\frac{|x-y|}{2}$ . However, this leads to a contradiction, since $|x-y|<|x_{n}-x|+|x_{n}-y|$ . $\square$

We can define Cauchy-sequences of real numbers as well.

Definition 2.1.3 (Cauchy-sequences of real numbers)

A sequence of real numbers $\{x_{n}\}^{\infty}_{n=1}$ is a Cauchy-sequence if for all $\varepsilon>0$ there exists $N>0$ such that $|x_{n}-x_{m}|\leq\varepsilon$ provided that $n,m\geq N$ .

As in the case of rationals we have the following results:

Proposition 2.1.1 (CONVERGENCE IMPLIES CAUCHY)

If for a real sequence $\{x_{n}\}^{\infty}_{n=1}$ $\lim_{n\to\infty}x_{n}=x\,$ , then $\{x_{n}\}^{\infty}_{n=1}$ is a Cauchy-sequence.

Definition 2.1.4

A sequence $\{x_{n}\}^{\infty}_{n=1}$ of real numbers is bounded, if there exists a positive number $M>0$ such that for any $n\geq 1$ , $|x_{n}|<M$ (or, if you wish, $-M<x_{n}<M$ ).

Example 2.1.1

$1,0,1,0,1,\dots$ is a bounded (but not convergent) sequence. The sequence $\{x_{n}=n\}^{\infty}_{n=1}$ is not bounded.

Question 2.1.1

Let $\{x_{n}\}^{\infty}_{n=1}$ and $\{y_{n}\}^{\infty}_{n=1}$ be bounded sequences. Then, both $\{x_{n}+y_{n}\}^{\infty}_{n=1}$ , $\{x_{n}y_{n}\}^{\infty}_{n=1}$ and $\{10^{100000}x_{n}\}^{\infty}_{n=1}$ are bounded sequences.

Solution There exist an $M$ and $N$ such that $|x_{n}|<M$ , $|y_{n}|<N$ . Hence, for any $n\geq 1$ $|x_{n}+y_{n}|<M+N$ , $|x_{n}y_{n}|<MN$ and $|10^{100000}x_{n}|<10^{100000}M$ . $\square$

Nitpicking 2.1.1

Is the sign $<$ important in the definition of boundedness? Can we substitute it with a $\leq$ sign?

Proposition 2.1.2 (CAUCHY IMPLIES BOUNDED)

A Cauchy-sequence $\{x_{n}\}^{\infty}_{n=1}$ of real numbers is bounded (that is there exists some integer $M$ such that for any $n\geq 1$ , $|x_{n}|\leq M$ ).

In order to get plenty of examples of convergent sequences we need the following proposition.

Proposition 2.1.3

Let $\lim_{n\to\infty}x_{n}=x$ , $\lim_{n\to\infty}y_{n}=y$ . Then:

1.

$\lim_{n\to\infty}(x_{n}+y_{n})=x+y\,.$ [sum rule]
2.

$\lim_{n\to\infty}(x_{n}-y_{n})=x-y\,.$ [subtraction rule]
3.

$\lim_{n\to\infty}x_{n}y_{n}=xy\,.$ [product rule]
4.

If $y\neq 0$ and for all $n\geq 1$ $y_{n}\neq 0$ , then $\lim_{n\to\infty}\frac{x_{n}}{y_{n}}=\frac{x}{y}\,.$ [quotient rule]

Proof: In order to show 1. we need to see that for any $\varepsilon>0$ there exists $N>0$ such that if $n\geq N$ then $|(x_{n}+y_{n})-(x+y)|\leq\varepsilon$ . We know that there exists an integer $P>0$ such that if $n\geq P$ , then $|x_{n}-x|<\varepsilon/2$ and there exists an integer $R>0$ such that if $n\geq R$ , then $|y_{n}-y|<\varepsilon/2$ . Let $N$ be the maximum of $P$ and $R$ . If $n\geq N$ , then $|x_{n}-x|<\varepsilon/2$ and $|y_{n}-y|<\varepsilon/2$ , so $|(x_{n}+y_{n})-(x-y)|\leq|x_{n}-x|+|y_{n}-y|\leq\varepsilon\,$ (by the Triangle Inequality). Statement 2. can be seen in the same way. In order to show 3. we need to see that for any $\varepsilon>0$ there exists $N>0$ such that if $n\geq N$ then $|(x_{n}y_{n})-(xy)|\leq\varepsilon$ . Since, $\{x_{n}\}^{\infty}_{n=1}$ and $\{y_{n}\}^{\infty}_{n=1}$ are convergent they are bounded. Fix $\varepsilon$ . Then, there exists a positive number $M>0$ so that for all $n$ , $|x_{n}|\leq M,|y_{n}|\leq M$ . (We have a bound for $x_{n}$ and a bound for $y_{n}$ and then we should pick the maximum as a joint bound, we will do it automatically in the future, since it is so obvious). By definition, we have an (joint) $N$ (well, we are in the future now…) so that if $n\geq N$ , then

|x_{n}-x|\leq\frac{\varepsilon}{2M}\,.

|y_{n}-y|\leq\frac{\varepsilon}{2M}\,.

We have that $x_{n}y_{n}-xy=(x_{n}-x)y_{n}+x(y_{n}-y)$ , so by the Triangle Inequality, $|x_{n}y_{n}-xy|\leq|x_{n}-x||y_{n}|+|x||y_{n}-y|\leq\varepsilon\,.$ Hence Statement 3. follows. Now we prove Statement 4. By Statement 3., we can suppose that all $x_{n}$ equal to $1$ . We need to show that for any $\varepsilon>0$ there exists $N>0$ such that if $n\geq N$ , then $|\frac{1}{y_{n}}-\frac{1}{y}|\leq\varepsilon$ . We use the fact that

\left|\frac{1}{y_{n}}-\frac{1}{y}\right|=\frac{|y_{n}-y|}{|y_{n}||y|}\,.

(2.1)

Since $y_{n}\to y$ and $y\neq 0$ , there exists an $P>0$ such that if $n\geq P$ , then

|y_{n}-y|<\frac{1}{2}|y|\,.

(2.2)

So, if $n\geq P$ , then $|y_{n}|\geq\frac{1}{2}|y|\,.$ Hence if $n\geq N$ , then

\frac{|y_{n}-y|}{|y_{n}||y|}\leq\frac{|y_{n}-y|}{\frac{1}{2}|y|^{2}}\,.

(2.3)

Now let $\varepsilon>0$ . Again, since $y_{n}\to y$ and $y\neq 0$ we have an $M>0$ such that for any $n\geq M$ :

|y_{n}-y|\leq\frac{1}{2}\varepsilon|y|^{2}\,.

(2.4)

Let $N$ be the maximum of $P$ and $M$ . Then by (2.3), we have that for $n\geq N$

\frac{|y_{n}-y|}{|y_{n}||y|}\leq\frac{\frac{1}{2}\varepsilon|y|^{2}}{\frac{1}{% 2}|y|^{2}}=\varepsilon\,.

Therefore, by (2.1) if $n\geq N$ , then

\left|\frac{1}{y_{n}}-\frac{1}{y}\right|\leq\varepsilon\,.

Hence Statement 4. follows. $\square$

Question 2.1.2

Calculate $\lim_{n\to\infty}x_{n}$ , where $x_{n}=\frac{12n^{7}-6n^{3}+9}{n^{7}+4n^{6}-11n}$ .

Solution: Dividing by $n^{7}$ we get that

x_{n}=\frac{12-6\frac{1}{n^{4}}+9\frac{1}{n^{7}}}{1+4\frac{1}{n}-11\frac{1}{n^% {6}}}\,.

By Example 1.1.1, $\lim_{n\to\infty}\frac{1}{n}=0.$ So by the product rule and the sum rule

\lim_{n\to\infty}12-6\frac{1}{n^{4}}+9\frac{1}{n^{7}}=12\,.

\lim_{n\to\infty}1+4\frac{1}{n}-11\frac{1}{n^{6}}=1\,.

Hence, by the quotient rule,

\lim_{n\to\infty}\frac{12n^{7}-6n^{3}+9}{n^{7}+4n^{6}-11n}=12\,.

$\square$

Finally, we prove the so-called Sandwich (or Squeeze) Rule.

Proposition 2.1.4 (Sandwich Rule)

Let $x_{n}\to a$ , $y_{n}\to a$ be convergent sequences. Suppose that for any $n\geq 1$ , $x_{n}\leq z_{n}\leq y_{n}$ or $x_{n}\geq z_{n}\geq y_{n}$ Then $z_{n}\to a$ as well.

Proof: For any $\varepsilon>0$ there exists $N>0$ such that if $n\geq N$ then $|x_{n}-a|\leq\varepsilon$ and $|y_{n}-a|\leq\varepsilon$ . But then, $|z_{n}-a|\leq\varepsilon$ as well. Hence, $z_{n}\to a$ . $\square$

Question 2.1.3

Suppose that $x_{n}\to a$ , $y_{n}\to a$ are convergent sequences of positive numbers. Then $\frac{x_{n}+y_{n}}{2}\to a$ and $\sqrt{x_{n}y_{n}}\to a$ .

Solution: If $x_{n}\leq y_{n}$ , then $x_{n}\leq\frac{x_{n}+y_{n}}{2}\leq y_{n}$ and $x_{n}\leq\sqrt{x_{n}y_{n}}\leq y_{n}$ , hence the statement follows. $\square$

Proposition 2.1.5 (The limit is monotone)

Let $\{x_{n}\}_{n=1}^{\infty}$ and $\{y_{n}\}^{\infty}_{n=1}$ be convergent sequences such that for any $n\geq 1$ , $x_{n}\leq y_{n}$ . Then $\lim_{n\to\infty}x_{n}\leq\lim_{n\to\infty}y_{n}\,.$

Proof: Let us prove this result by contradiction. Suppose that $x_{n}\to x$ and $y_{n}\to y$ and suppose that $x>y$ . Let $x-y=a>0$ . By the definition of convergence, there exists $N>0$ such that if $n\geq N$ , then $|x-x_{n}|<\frac{a}{2},|y-y_{n}|<\frac{a}{2}$ . Therefore $x_{n}>y_{n}$ , leading to a contradiction. $\square$

Corollary 2.1.1

Let $\{x_{n}\}_{n=1}^{\infty}$ be a convergent sequence tending to $x$ . Suppose that for some $a<b$ it holds that for all $n\geq 1$ , $a\leq x_{n}\leq b$ . Then $a\leq x\leq b$ .

Proof: Just apply the proposition above for the sequences $\{y_{n}=b\}^{\infty}_{n=1}$ and $\{x_{n}\}^{\infty}_{n=1}$ . $\square$