2.2 Existence of certain fundamental limits

Analysis of functions of a single Page 1 / 1

Nondecreasing, nonincreasing, eventually nondecreasing, and eventually nonincreasing sequences are defined. Convergence is then established for these sequences, and some practice exercises are included.

We have, in the preceding exercises, seen that certain specific sequences converge. It's time to develop some general theory, something that will apply to lots of sequences,and something that will help us actually evaluate limits of certain sequences.

A sequence ${a_{n}}$ of real numbers is called nondecreasing if $a_{n} \leq a_{n + 1}$ for all $n,$ and it is called nonincreasing if $a_{n} \geq a_{n + 1}$ for all $n .$ It is called strictly increasing if $a_{n} < a_{n + 1}$ for all $n,$ and strictly decreasing if $a_{n} > a_{n + 1}$ for all $n .$

A sequence ${a_{n}}$ of real numbers is called eventually nondecreasing if there exists a natural number $N$ such that $a_{n} \leq a_{n + 1}$ for all $n \geq N,$ and it is called eventually nonincreasing if there exists a natural number $N$ such that $a_{n} \geq a_{n + 1}$ for all $n \geq N .$ We make analogous definitions of “eventually strictly increasing” and “eventually strictly decreasing.”

It is ordinarily very difficult to tell whether a given sequence converges or not; and even if we know in theory that a sequence converges, it isstill frequently difficult to tell what the limit is. The next theorem is therefore very useful.It is also very fundamental, for it makes explicit use of the existence of a least upper bound.

Let ${a_{n}}$ be a nondecreasing sequence of real numbers. Suppose that the set $S$ of elements of the sequence ${a_{n}}$ is bounded above. Then the sequence ${a_{n}}$ is convergent, and the limit $L$ is given by $L = sup S = sup a_{n} .$

Analogously, if ${a_{n}}$ is a nonincreasing sequence that is bounded below, then ${a_{n}}$ converges to $inf a_{n} .$

We prove the first statement. The second is done analogously, and we leave it to an exercise.Write $L$ for the supremum $sup a_{n} .$ Let $ϵ$ be a positive number. By Theorem 1.5, there exists an integer $N$ such that $a_{N} > L - ϵ,$ which implies that $L - a_{N} < ϵ .$ Since ${a_{n}}$ is nondecreasing, we then have that $a_{n} \geq a_{N} > L - ϵ$ for all $n \geq N .$ Since $L$ is an upper bound for the entire sequence, we know that $L \geq a_{n}$ for every $n,$ and so we have that

| L - a_{n} | = L - a_{n} \leq L - a_{N} < ϵ

for all $n \geq N .$ This completes the proof of the first assertion.

Prove the second assertion of the preceding theorem.
Show that [link] holds for sequences that are eventually nondecreasing or eventually nonincreasing.(Re-read the remark following the definition of the limit of a sequence.)

The next exercise again demonstrates the “denseness” of the rational and irrational numbers in the set $R$ of all real numbers.

Let $x$ be a real number. Prove that there exists a sequence ${r_{n}}$ of rational numbers such that $x = lim r_{n} .$ In fact, show that the sequence ${r_{n}}$ can be chosen to be nondecreasing. HINT: For example, for each $n,$ use [link] to choose a rational number $r_{n}$ between $x - 1 / n$ and $x .$
Let $x$ be a real number. Prove that there exists a sequence ${{r^{'}}_{n}}$ of irrational numbers such that $x = lim r_{n}^{'} .$
Let $z = x + i y$ be a complex number. Prove that there exists a sequence ${α_{n}} = {β_{n} + i γ_{n}}$ of complex numbers that converges to $z,$ such that each $β_{n}$ and each $γ_{n}$ is a rational number.

Suppose ${a_{n}}$ and ${b_{n}}$ are two convergent sequences, and suppose that $lim a_{n} = a$ and $lim b_{n} = b .$ Prove that the sequence ${a_{n} + b_{n}}$ is convergent and that

lim (a_{n} + b_{n}) = a + b .

HINT: Use an $ϵ / 2$ argument. That is, choose a natural number $N_{1}$ so that $| a_{n} - a | < ϵ / 2$ for all $n \geq N_{1},$ and choose a natural number $N_{2}$ so that $| b_{n} - b | < ϵ / 2$ for all $n \geq N_{2} .$ Then let $N$ be the larger of the two numbers $N_{1}$ and $N_{2} .$

The next theorem establishes the existence of four nontrivial and important limits.This time, the proofs are more tricky. Some clever idea will have to be used before we can tell how to choose the $N .$

Let $z \in C$ satisfy $| z | < 1,$ and define $a_{n} = z^{n} .$ then the sequence ${a_{n}}$ converges to $0 .$ We write $lim z^{n} = 0 .$
Let $b$ be a fixed positive number greater than 1, and define $a_{n} = b^{1 / n} .$ See [link] . Then $lim a_{n} = 1 .$ Again, we write $lim b^{1 / n} = 1 .$
Let $b$ be a positive number less than 1. Then $lim b^{1 / n} = 1 .$
If $a_{n} = n^{1 / n},$ then $lim a_{n} = lim n^{1 / n} = 1 .$

We prove parts (1) and (2) and leave the rest of the proof to the exercise that follows. If $z = 0,$ claim (1) is obvious. Assume then that $z \neq 0,$ and let $ϵ > 0$ be given. Let $w = 1 / | z |,$ and observe that $w > 1 .$ So, we may write $w = 1 + h$ for some positive $h .$ (That step is the clever idea for this argument.) Then, using the Binomial Theorem, $w^{n} > n h,$ and so $1 / w^{n} < 1 / (n h) .$ See part (a) of [link] . But then

| z^{n} - 0 | = | z^{n} {| = | z |}^{n} = {(1 / w)}^{n} = 1 / w^{n} < 1 / (n h) .

So, if $N$ is any natural number larger than $1 / (ϵ h),$ then

| z^{n} - 0 | = | z^{n} {| = | z |}^{n} < \frac{1}{n h} \leq \frac{1}{N h} < ϵ

for all $n \geq N .$ This completes the proof of the first assertion of the theorem.

To see part (2), write $a_{n} = b^{1 / n} = 1 + x_{n},$ i.e., $x_{n} = b^{1 / n} - 1,$ and observe first that $x_{n} > 0 .$ Indeed, since $b > 1,$ it must be that the $n$ th root $b^{1 / n}$ is also $> 1 .$ (Why?) Therefore, $x_{n} = b^{1 / n} - 1 > 0 .$ (Again, writing $b^{1 / n}$ as $1 + x_{n}$ is the clever idea.) Now, $b = {b^{1 / n}}^{n} = {(1 + x_{n})}^{n},$ which, again by the Binomial Theorem, implies that $b > 1 + n x_{n} .$ So, $x_{n} < (b - 1) / n,$ and therefore

| b^{1 / n} - 1 | = b^{1 / n} - 1 = x_{n} < \frac{b - 1}{n} < ϵ

whenever $n > ϵ / (b - 1),$ and this proves part (2).

Prove part (3) of the preceding theorem. HINT: For $b \leq 1,$ use the following algebraic calculation:
$| b^{1 / n} - 1 | = b^{1 / n} {| 1 - (1 / b)}^{1 / n} {| \leq | 1 - (1 / b)}^{1 / n} |,$
and then use part (2) as applied to the positive number $1 / b .$
Prove part (4) of the preceding theorem. Explain why it does not follow directly from part (2).HINT: Write $n^{1 / n} = 1 + h_{n} .$ Observe that $h_{n} > 0 .$ Then use the third term of the binomial theorem in the expansion $n = {(1 + h_{n})}^{n} .$
Construct an alternate proof to part (2) of the preceding theorem as follows: Show that the sequence ${b^{1 / n}}$ is nonincreasing and bounded below by 1. Deduce, from [link] , that the sequence converges to a number $L .$ Now prove that $L$ must be $1 .$

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Source: OpenStax, Analysis of functions of a single variable. OpenStax CNX. Dec 11, 2010 Download for free at http://cnx.org/content/col11249/1.1

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