2.2 Existence of certain fundamental limits

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 $\left\{a_n\right\}$ of real numbers is called nondecreasing if $a_n\le a_{n+1}$ for all $n,$ and it is called nonincreasing if $a_n\ge a_{n+1}$ 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 $\left\{a_n\right\}$ of real numbers is called eventually nondecreasing if there exists a natural number $N$ such that $a_n\le a_{n+1}$ for all $n\ge N,$ and it is called eventually nonincreasing if there exists a natural number $N$ such that $a_n\ge a_{n+1}$ for all $n\ge N.$ 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 $\left\{a_n\right\}$ be a nondecreasing sequence of real numbers. Suppose that the set $S$ of elements of the sequence $\left\{a_n\right\}$ is bounded above. Then the sequence $\left\{a_n\right\}$ is convergent, and the limit $L$ is given by $L=supS=sup{a}_n.$

Analogously, if $\left\{a_n\right\}$ is a nonincreasing sequence that is bounded below, then $\left\{a_n\right\}$ 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 $\left\{a_n\right\}$ is nondecreasing, we then have that $a_n\ge a_N>L-ϵ$ for all $n\ge N.$ Since $L$ is an upper bound for the entire sequence, we know that $L\ge a_n$ for every $n,$ and so we have that

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

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

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 $N.$

1. Let $z\in C$ satisfy $\left|z\right|<1,$ and define $a_n=z^n.$ then the sequence $\left\{a_n\right\}$ converges to $0.$ We write $lim{z}^n=0.$
2. 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 $lim{b}^{1/n}=1.$
3. Let $b$ be a positive number less than 1. Then $lim{b}^{1/n}=1.$
4. 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 $z\ne 0,$ and let $ϵ>0$ be given. Let $w=1/\left|z\right|,$ 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, $w^n>nh,$ and so $1/w^n<1/\left(nh\right).$ See part (a) of [link] . But then

So, if $N$ is any natural number larger than $1/\left(ϵh\right),$ then

for all $n\ge N.$ 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

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