<< Chapter < Page | Chapter >> Page > |
Often, our goal is to show that a given sequence is convergent. However, as we study convergent sequences, we would liketo establish various properties that they have in common. The first theorem of this section is just such a result.
Suppose $\left\{{a}_{n}\right\}$ is a convergent sequence of real or complex numbers. Then the sequence $\left\{{a}_{n}\right\}$ forms a bounded set.
Write $L=lim{a}_{n}.$ Let $\u03f5$ be the positive number $1.$ Then, there exists a natural number $N$ such that $|{a}_{n}-L|<1$ for all $n\ge N.$ By the backward triangle inequality, this implies that $\left|\right|{a}_{n}|-|L\left|\right|<1$ for all $n\ge N,$ which implies that $|{a}_{n}|\le |L|+1$ for all $n\ge N.$ This shows that at least the tail of the sequence is bounded by the constant $\left|L\right|+1.$
Next, let $K$ be a number larger than the finitely many numbers $|{a}_{1}|,...,|{a}_{N-1}|.$ Then, for any $n,$ $|{a}_{n}|$ is either less than $K$ or $\left|L\right|+1.$ Let $M$ be the larger of the two numbers $K$ and $\left|L\right|+1.$ Then $|{a}_{n}|<M$ for all $n.$ Hence, the sequence $\left\{{a}_{n}\right\}$ is bounded.
Note that the preceding theorem is a partial converse to [link] ; i.e., a convergent sequence is necessarily bounded. Of course, not every convergent sequence must beeither nondecreasing or nonincreasing, so that a full converse to [link] is not true. For instance, take $z=-1/2$ in part (1) of [link] . It converges to 0 all right, but it is neither nondecreasing nor nonincreasing.
Suppose $\left\{{a}_{n}\right\}$ is a sequence of real numbers and that $L=lim{a}_{n}.$ Let ${M}_{1}$ and ${M}_{2}$ be real numbers such that ${M}_{1}\le {a}_{n}\le {M}_{2}$ for all $n.$ Prove that ${M}_{1}\le L\le {M}_{2}.$
HINT: Suppose, for instance, that $L>{M}_{2}.$ Make use of the positive number $L-{M}_{2}$ to derive a contradiction.
We are often able to show that a sequence converges by comparing it to another sequence that we already know converges.The following exercise demonstrates some of these techniques.
Let $\left\{{a}_{n}\right\}$ be a sequence of complex numbers.
The next result is perhaps the most powerful techniquewe have for showing that a given sequence converges to a given number.
Suppose that $\left\{{a}_{n}\right\}$ is a sequence of real numbers and that $\left\{{b}_{n}\right\}$ and $\left\{{c}_{n}\right\}$ are two sequences ofreal numbers for which ${b}_{n}\le {a}_{n}\le {c}_{n}$ for all $n.$ Suppose further that $lim{b}_{n}=lim{c}_{n}=L.$ Then the sequence $\left\{{a}_{n}\right\}$ also converges to $L.$
We examine the quantity $|{a}_{n}-L,|$ employ some add and subtract tricks, and make the following computations:
So, we can make $|{a}_{n}-L|<\u03f5$ by making $|{c}_{n}-L|<\u03f5/3$ and $|{b}_{n}-L|<\u03f5/3.$ So, let ${N}_{1}$ be a positive integer such that $|{c}_{n}-L|<\u03f5/3$ if $n\ge {N}_{1},$ and let ${N}_{2}$ be a positive integer so that $|{b}_{n}-L|<\u03f5/3$ if $n\ge {N}_{2}.$ Then set $N=max({N}_{1},{N}_{2}).$ Clearly, if $n\ge N,$ then both inequalities $|{c}_{n}-L|<\u03f5/3$ and $|{b}_{n}-L|<\u03f5/3,$ and hence $|{a}_{n}-L|<\u03f5.$ This finishes the proof.
The next result establishes what are frequently called the “limit theorems.” Basically, these results show how convergenceinteracts with algebraic operations.
Let $\left\{{a}_{n}\right\}$ and $\left\{{b}_{n}\right\}$ be two sequences of complex numbers with $a=lim{a}_{n}$ and $b=lim{b}_{n}.$ Then
Part (1) is exactly the same as [link] . Let us prove part (2).
By [link] , both sequences $\left\{{a}_{n}\right\}$ and $\left\{{b}_{n}\right\}$ are bounded. Therefore, let $M$ be a number such that $|{a}_{n}|\le M$ and $|{b}_{n}|\le M$ for all $n.$ Now, let $\u03f5>0$ be given. There exists an ${N}_{1}$ such that $|{a}_{n}-a|<\u03f5/\left(2M\right)$ whenever $n\ge {N}_{1},$ and there exists an ${N}_{2}$ such that $|{b}_{n}-b|<\u03f5/\left(2M\right)$ whenever $n\ge {N}_{2}.$ Let $N$ be the maximum of ${N}_{1}$ and ${N}_{2}.$ Here comes the add and subtract trick again.
if $n\ge N,$ which shows that $lim\left({a}_{n}{b}_{n}\right)=ab.$
To prove part (3), let $M$ be as in the previous paragraph, and let $\u03f5>0$ be given. There exists an ${N}_{1}$ such that $|{a}_{n}-{a|<(\u03f5\left|b\right|}^{2})/\left(4M\right)$ whenever $n\ge {N}_{1};$ there also exists an ${N}_{2}$ such that $|{b}_{n}-{b|<(\u03f5\left|b\right|}^{2})/\left(4M\right)$ whenever $n\ge {N}_{2};$ and there exists an ${N}_{3}$ such that $|{b}_{n}|>|b|/2$ whenever $n\ge {N}_{3}.$ (See [link] .) Let $N$ be the maximum of the three numbers ${N}_{1},{N}_{2}$ and ${N}_{3}.$ Then:
if $n\ge N.$ This completes the proof.
REMARK The proof of part (3) of the preceding theorem may look mysterious. Where, for instance, does this number ${\u03f5\left|b\right|}^{2}/4M$ come from? The answer is that one begins such a proof by examining the quantity $|{a}_{n}/{b}_{n}-a/b|$ to see if by some algebraic manipulation one can discover how to control its size by using the quantities $|{a}_{n}-a|$ and $|{b}_{n}-b|.$ The assumption that $a=lim{a}_{n}$ and $b=lim{b}_{n}$ mean exactly that the quantities $|{a}_{n}-a|$ and $|{b}_{n}-b|$ can be controlled by requiring $n$ to be large enough. The algebraic computation in the proof above shows that
and one can then see exactly how small to make $|{a}_{n}-a|$ and $|{b}_{n}-b|$ so that $|{a}_{n}/{b}_{n}-a/b|<\u03f5.$ Indeed, this is the way most limit proofs work.
If possible, determine the limits of the following sequences by using [link] , [link] , [link] , and the squeeze theorem [link] .
Notification Switch
Would you like to follow the 'Analysis of functions of a single variable' conversation and receive update notifications?