# 2.4 Properties of convergent sequences

 Page 1 / 1
A discussion of the properties of convergent sequences, such as when they form a bounded set, the squeeze theorem, and the properties of convergence when sequences are combined algebraically.

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 $ϵ$ 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 $||{a}_{n}|-|L||<1$ 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 $|L|+1.$

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 $|L|+1.$ Let $M$ be the larger of the two numbers $K$ and $|L|+1.$ Then $|{a}_{n}| 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.

1. Suppose $\left\{{a}_{n}\right\}$ is a sequence of real numbers that converges to a number $a,$ and assume that ${a}_{n}\ge c$ for all $n.$ Prove that $a\ge c.$ HINT: Suppose not, and let $ϵ$ be the positive number $c-a.$ Let $N$ be a natural number corresponding to this choice of $ϵ,$ and derive a contradiction.
2. If $\left\{{a}_{n}\right\}$ is a sequence of real numbers for which $lim{a}_{n}=a,$ and if $a\ne 0,$ then prove that ${a}_{n}\ne 0$ for all large enough $n.$ Show in fact that there exists an $N$ such that $|{a}_{n}|>|a|/2$ for all $n\ge N.$ HINT: Make use of the positive number $ϵ=|a|/2.$
1. If $\left\{{a}_{n}\right\}$ is a sequence of positive real numbers for which $lim{a}_{n}=a>0,$ prove that $lim\sqrt{{a}_{n}}=\sqrt{a}.$ HINT: Multiply the expression $\sqrt{{a}_{n}}-\sqrt{a}$ above and below by $\sqrt{{a}_{n}}+\sqrt{a}.$
2. If $\left\{{a}_{n}\right\}$ is a sequence of complex numbers, and $lim{a}_{n}=a,$ prove that $lim|{a}_{n}|=|a|.$ HINT: Use the backward triangle inequality.

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.

1. Suppose that, for each $n,$ $|{a}_{n}|<1/n.$ Prove that $0=lim{a}_{n}.$
2. Suppose $\left\{{b}_{n}\right\}$ is a sequence that converges to $0,$ and suppose that, for each $n,$ $|{a}_{n}|<|{b}_{n}|.$ Prove that $0=lim{a}_{n}.$

The next result is perhaps the most powerful techniquewe have for showing that a given sequence converges to a given number.

## Squeeze theorem

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:

$\begin{array}{ccc}\hfill rcl|{a}_{n}-L|& \le & |{a}_{n}-{b}_{n}+{b}_{n}-L|\hfill \\ & \le & |{a}_{n}-{b}_{n}|+|{b}_{n}-L|\hfill \\ & =& {a}_{n}-{b}_{n}+|{b}_{n}-L|\hfill \\ & \le & {c}_{n}-{b}_{n}+|{b}_{n}-L|\hfill \\ & =& |{c}_{n}-{b}_{n}|+|{b}_{n}-L|\hfill \\ & \le & |{c}_{n}-L|+|L-{b}_{n}|+|{b}_{n}-L|.\hfill \end{array}$

So, we can make $|{a}_{n}-L|<ϵ$ by making $|{c}_{n}-L|<ϵ/3$ and $|{b}_{n}-L|<ϵ/3.$ So, let ${N}_{1}$ be a positive integer such that $|{c}_{n}-L|<ϵ/3$ if $n\ge {N}_{1},$ and let ${N}_{2}$ be a positive integer so that $|{b}_{n}-L|<ϵ/3$ if $n\ge {N}_{2}.$ Then set $N=max\left({N}_{1},{N}_{2}\right).$ Clearly, if $n\ge N,$ then both inequalities $|{c}_{n}-L|<ϵ/3$ and $|{b}_{n}-L|<ϵ/3,$ and hence $|{a}_{n}-L|<ϵ.$ 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

1. The sequence $\left\{{a}_{n}+{b}_{n}\right\}$ converges, and
$lim\left({a}_{n}+{b}_{n}\right)=lim{a}_{n}+lim{b}_{n}=a+b.$
2. The sequence $\left\{{a}_{n}{b}_{n}\right\}$ is convergent, and
$lim\left({a}_{n}{b}_{n}\right)=lim{a}_{n}lim{b}_{n}=ab.$
3. If all the ${b}_{n}$ 's as well as $b$ are nonzero, then the sequence $\left\{{a}_{n}/{b}_{n}\right\}$ is convergent, and
$lim\left(\frac{{a}_{n}}{{b}_{n}}=\frac{lim{a}_{n}}{lim{b}_{n}}=\frac{a}{b}.$

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 $ϵ>0$ be given. There exists an ${N}_{1}$ such that $|{a}_{n}-a|<ϵ/\left(2M\right)$ whenever $n\ge {N}_{1},$ and there exists an ${N}_{2}$ such that $|{b}_{n}-b|<ϵ/\left(2M\right)$ whenever $n\ge {N}_{2}.$ Let $N$ be the maximum of ${N}_{1}$ and ${N}_{2}.$ Here comes the add and subtract trick again.

$\begin{array}{ccc}\hfill rcl|{a}_{n}{b}_{n}-ab|& =& |{a}_{n}{b}_{n}-a{b}_{n}+a{b}_{n}-ab|\hfill \\ & \le & |{a}_{n}{b}_{n}-a{b}_{n}|+|a{b}_{n}-ab|\hfill \\ & =& |{a}_{n}-a||{b}_{n}|+|a||b-{b}_{n}|\hfill \\ & \le & |{a}_{n}-a|M+M|{b}_{n}-b|\hfill \\ & <& ϵ\hfill \end{array}$

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 $ϵ>0$ be given. There exists an ${N}_{1}$ such that $|{a}_{n}-{a|<\left(ϵ|b|}^{2}\right)/\left(4M\right)$ whenever $n\ge {N}_{1};$ there also exists an ${N}_{2}$ such that $|{b}_{n}-{b|<\left(ϵ|b|}^{2}\right)/\left(4M\right)$ 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:

$\begin{array}{ccc}\hfill rcl|\frac{{a}_{n}}{{b}_{n}}-\frac{a}{b}|& =& |\frac{{a}_{n}b-{b}_{n}a}{{b}_{n}b}|\hfill \\ & =& |{a}_{n}b-{b}_{n}a|\frac{1}{|{b}_{n}b|}\hfill \\ & <& |{a}_{n}b-{b}_{n}a|\frac{1}{{|b|}^{2}/2}\hfill \\ & \le & \left(|{a}_{n}-a||b|+|a||{b}_{n}-b|\right)\frac{2}{{|b|}^{2}}\hfill \\ & <& \left(M|{a}_{n}-a|+M|{b}_{n}-b|\right)\frac{2}{{|b|}^{2}}\hfill \\ & <& ϵ\hfill \end{array}$

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 ${ϵ|b|}^{2}/4M$ 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

$|\frac{{a}_{n}}{{b}_{n}}-\frac{a}{b}|\le \left(M|{a}_{n}-a|+M|{b}_{n}-b|\right)\frac{2}{{|b|}^{2}},$

and one can then see exactly how small to make $|{a}_{n}-a|$ and $|{b}_{n}-b|$ so that $|{a}_{n}/{b}_{n}-a/b|<ϵ.$ 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] .

1. $\left\{{n}^{1/{n}^{2}}\right\}.$
2. $\left\{{\left({n}^{2}\right)}^{1/n}\right\}.$
3. $\left\{{\left(1+n\right)}^{1/n}\right\}.$
4. $\left\{{\left(1+{n}^{2}\right)}^{1/{n}^{3}}\right\}.$
5. $\left\{{\left(1+1/n\right)}^{2/n}\right\}.$
6. $\left\{{\left(1+1/n\right)}^{2n}\right\}.$
7. $\left\{{\left(1+1/n\right)}^{{n}^{2}}\right\}.$
8. $\left\{{\left(1-1/n\right)}^{n}\right\}.$ HINT: Note that
$1-1/n=\frac{n-1}{n}=\frac{1}{\frac{n}{n-1}}=\frac{1}{\frac{n-1+1}{n-1}}=\frac{1}{1+\frac{1}{n-1}}.$
9. $\left\{{\left(1-1/\left(2n\right)\right)}^{3n}\right\}.$
10. $\left\{{\left(n!\right)}^{1/n}\right\}.$

how can chip be made from sand
is this allso about nanoscale material
Almas
are nano particles real
yeah
Joseph
Hello, if I study Physics teacher in bachelor, can I study Nanotechnology in master?
no can't
Lohitha
where is the latest information on a no technology how can I find it
William
currently
William
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
has a lot of application modern world
Kamaluddeen
yes
narayan
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!