Home page for accesible maths 2 Sequences 2.3 The Least Upper Bound Principle. Maximum vs. Supremum 2.5 Increasing and decreasing sequences

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2.4 Bounded and unbounded sequences

Proposition 2.4.1

Let $\{x_{n}\}^{\infty}_{n=1}$ be a bounded sequence and $\{y_{n}\}^{\infty}_{n=1}$ be a sequence converging to zero. Then the sequence $\{x_{n}y_{n}\}^{\infty}_{n=1}$ is converging to the zero as well.

Proof: We need to prove that for any $\varepsilon>0$ there exists $N>0$ such that if $n\geq N$ then $|x_{n}y_{n}|\leq\varepsilon$ . There exists a number $M>0$ such that for all $n\geq 1$ , $|x_{n}|\leq M$ . Also, there exists some $N\geq 0$ such that if $n\geq N$ , then $|y_{n}|\leq\frac{\varepsilon}{M}$ . That is, if $n\geq N$ , then $|x_{n}y_{n}|\leq\varepsilon$ . $\square$

Question 2.4.1

Calculate the limit of the sequence $\{(1-\frac{1}{(1+\frac{1}{n^{2}})})sin(n)\}^{\infty}_{n=1}$ .

Solution: By the Quotient Rule, the Product Rule, the Subtraction Rule and the Sum Rule $\lim_{n\to\infty}1-\frac{1}{(1+\frac{1}{n^{2}})}=0\,.$ . Also, the sequence $\{sin(n)\}^{\infty}_{n=1}$ is clearly bounded. Hence by Proposition 2.4, $\lim_{n\to\infty}(1-\frac{1}{(1+\frac{1}{n^{2}})})sin(n)=0\,.$

We call ALL sequences that are not bounded unbounded sequences.

Example 2.4.1

$\{n^{9}\}^{\infty}_{n=1}$ is an unbounded sequence. $0,1,0,2,0,3,0,4,....$ is also an unbounded sequence.

The most important unbounded sequences are those sequences that are tending to infinity.

Definition 2.4.1 (Plain English Definition)

A sequence of positive real numbers
$\{x_{n}\}^{\infty}_{n=1}$ are tending to infinity if after a while they are becoming really large.

Definition 2.4.2 (More Mathematical Definition)

A sequence of positive real numbers $\{x_{n}\}^{\infty}_{n=1}$ is tending to infinity if for any $K>0$ , there exists some $N>0$ such that if $n\geq N$ then $x_{n}>K$ .

Question 2.4.2

Negate the statement: $\{x_{n}\}^{\infty}_{n=1}$ is tending to infinity.

Solution: It is not true that some statement holds for ALL $K>0$ . Hence, there exists some $K>0$ for which the statement does not hold. The statement for this $K$ is: There exists some $N>0$ so that if $n\geq N$ then $x_{n}>K$ . So, we need to negate this statement. It is not true that there exists some $N>0$ such that some statement is true. Hence for ALL $N>0$ there is some $n\neq N$ for which the statement does not hold. That is, there exists some $n\geq N$ such that $x_{n}\leq K$ . So, the negation of tending to infinity is: There exists some $K$ such that for every $N$ there exists some $n\geq N$ such that $x_{n}\leq K$ . In other words: $\{x_{n}\}^{\infty}_{n=1}$ has a bounded subsequence.

Proposition 2.4.2

A sequence of positive real numbers $\{x_{n}\}^{\infty}_{n=1}$ tends to infinity if and only if $\{\frac{1}{x_{n}}\}^{\infty}_{n=1}$ tends to zero.

Proof: This is easy enough to be left for the reader. $\square$

Proposition 2.4.3

If $\{x_{n}\}^{\infty}_{n=1}$ and $\{y_{n}\}^{\infty}_{n=1}$ are both tending to infinity then $\{x_{n}y_{n}\}^{\infty}_{n=1}$ , $\{x_{n}+y_{n}\}^{\infty}_{n=1}$ are tending to infinity.

Proof: This is even easier than the previous one.

Example 2.4.2

The following sequences are tending to infinity:

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$\{n^{10}\}^{\infty}_{n=1}$
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$\{log(n)\}^{\infty}_{n=1}$
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$\{2^{n}\}^{\infty}_{n=1}$
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$\{n!\}^{\infty}_{n=1}$
•

If $\{f(n)\}^{\infty}_{n=1}$ (see the notation, it is completely OK) tends to infinity having only integer values and $\{x_{n}\}^{\infty}_{n=1}$ tends to infinity, then $\{x_{f(n)}\}^{\infty}_{n=1}$ tends to infinity as well.

Question 2.4.3

Show that if the sequence $\{x_{n}\}^{\infty}_{n=1}$ tends to infinity, then the sequence $\{\sqrt{x_{n}}\}^{\infty}_{n=1}$ still tends to infinity. (we can write $\sqrt{x_{n}}\to\infty$ , but it is not a good idea to write $\lim_{n\to\infty}\sqrt{x_{n}}=\infty\,.$ )

Solution: For every $K>0$ there exists some $N$ such that if $n\geq N$ then $x_{n}\geq K^{2}$ . Hence if $n\geq N$ then $\sqrt{x_{n}}\geq K$ , proving that $\{\sqrt{x_{n}}\}^{\infty}_{n=1}$ tends to infinity.

Question 2.4.4

Let $\{x_{n}\}^{\infty}_{n=1}$ be a sequence of positive numbers tending to infinity. Let $\{y_{n}\}^{\infty}_{n=1}$ be a sequence of positive numbers tending to $C>0$ . Then $x_{n}y_{n}\to\infty$ .

Solution: Fix $K>0$ and let $N>0$ be such a number that if $n\geq N$ then $x_{n}>\frac{2K}{C}$ , also $y_{n}\geq\frac{C}{2}$ . Then if $n\geq N$ $x_{n}y_{n}>K$ . Hence $x_{n}y_{n}\to\infty$ . $\square$

Definition 2.4.3

Suppose that the sequences of positive numbers $\{x_{n}\}^{\infty}_{n=1}$ and $\{y_{n}\}^{\infty}_{n=1}$ tend to infinity. Then we say that $\{x_{n}\}^{\infty}_{n=1}$ beats $\{y_{n}\}^{\infty}_{n=1}$ if $\{\frac{x_{n}}{y_{n}}\}^{\infty}_{n=1}$ still tends to infinity.

The following proposition is still very easy.

Proposition 2.4.4 (The Race of Sequences)

The following rules apply:

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If $\{x_{n}\}^{\infty}_{n=1}$ beats $\{y_{n}\}^{\infty}_{n=1}$ and $\{y_{n}\}^{\infty}_{n=1}$ beats $\{z_{n}\}^{\infty}_{n=1}$ , then $\{x_{n}\}^{\infty}_{n=1}$ beats $\{z_{n}\}^{\infty}_{n=1}$ .
•

If $\{x_{n}\}^{\infty}_{n=1}$ tends to infinity and $\{y_{n}\}^{\infty}_{n=1}$ beats $\{z_{n}\}^{\infty}_{n=1}$ , then $\{x_{n}+y_{n}\}^{\infty}_{n=1}$ beats $\{z_{n}\}^{\infty}_{n=1}$ .
•

If $\{x_{n}\}^{\infty}_{n=1}$ tends to infinity then for any $k>0$ , $\{x^{k}_{n}\}^{\infty}_{n=1}$ beats $\{x_{n}\}^{\infty}_{n=1}$ consequently, $\{x_{n}\}^{\infty}_{n=1}$ beats $\{(x_{n})^{1/k}\}^{\infty}_{n=1}$ .

Proposition 2.4.5

For every $k>0$ and $\varepsilon>0$ , $\{(1+\varepsilon)^{n}\}^{\infty}_{n=1}$ beats $\{n^{k}\}^{\infty}_{n=1}$ but $\{n!\}^{\infty}_{n=1}$ beats $\{K^{n}\}^{\infty}_{n=1}$ for every single $K>0$ .

Proof: First recall that binomial formula:

(1+\varepsilon)^{n}=\sum_{k=0}^{n}{n\choose k}\varepsilon^{k}\,.

Since all the terms are positive, for any $0\leq k\leq n$ , $(1+\varepsilon)^{n}\geq{n\choose k}\varepsilon^{k}\,.$

Lemma 2.4.1

$\frac{{n\choose k+1}}{n^{k}}\to\infty$ .

Proof:

\frac{{n\choose k+1}}{n^{k}}=\frac{n(n-1)(n-2)\dots(n-k)}{k!n^{k}}=\frac{1}{k!% }1\cdot\frac{n-1}{n}\cdot\frac{n-2}{n}\cdot\frac{n-k+1}{n}\cdot(n-k)\,.

Thus the lemma follows from Question 2.4.4 (and our usual quotient and sum rules). Consequently, $\frac{(1+\varepsilon)^{n}}{n^{k}}\to\infty$ . $\square$

Now let us prove that $n!$ beats $K^{n}$ for any $K>0$ .

n!=1\cdot 2\dots n>([\frac{n}{3}]+1)\cdot([\frac{n}{3}]+2)\dots n>(\frac{n}{3}% )^{n/2}=(\sqrt{\frac{n}{3}})^{n}\,,

where $[]$ denotes the integer part function. So for large enough $n$ , $n!>K^{n}$ . $\square$

Question 2.4.5

Is there a sequence that beats $n!$ and $(n!)!$ and $((n!)!)!$ and so…?

Solution: For any $i\geq 1$ , let $\overline{x}^{i}=\{x^{i}_{n}\}_{n=1}^{\infty}$ be a sequence tending to infinity. Then there exists a sequence $\overline{y}=\{y_{n}\}^{\infty}_{n=1}$ that beats all of them. We define $y_{n}$ as the maximum of the numbers $\{x^{1}_{n},x^{2}_{n},\dots,x^{n}_{n}\}$ times $n$ . Clearly, the sequence $\{y_{n}\}^{\infty}_{n=1}$ will beat any $\overline{x}^{i}$ since if $i\leq n$ then $y_{n}\geq x^{i}_{n}n$ . $\square$