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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 { a n } is a convergent sequence of real or complex numbers. Then the sequence { a n } 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 N . By the backward triangle inequality, this implies that | | a n | - | L | | < 1 for all n N , which implies that | a n | | L | + 1 for all n 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 | < M for all n . Hence, the sequence { a n } 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 { a n } is a sequence of real numbers that converges to a number a , and assume that a n c for all n . Prove that a 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 { a n } is a sequence of real numbers for which lim a n = a , and if a 0 , then prove that a n 0 for all large enough n . Show in fact that there exists an N such that | a n | > | a | / 2 for all n N . HINT: Make use of the positive number ϵ = | a | / 2 .
  1. If { a n } is a sequence of positive real numbers for which lim a n = a > 0 , prove that lim a n = a . HINT: Multiply the expression a n - a above and below by a n + a .
  2. If { a n } is a sequence of complex numbers, and lim a n = a , prove that lim | a n | = | a | . HINT: Use the backward triangle inequality.

Suppose { a n } 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 a n M 2 for all n . Prove that M 1 L 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 { a n } be a sequence of complex numbers.

  1. Suppose that, for each n , | a n | < 1 / n . Prove that 0 = lim a n .
  2. Suppose { b n } 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 { a n } is a sequence of real numbers and that { b n } and { c n } are two sequences ofreal numbers for which b n a n c n for all n . Suppose further that lim b n = lim c n = L . Then the sequence { a n } also converges to L .

We examine the quantity | a n - L , | employ some add and subtract tricks, and make the following computations:

r c l | a n - L | | a n - b n + b n - L | | a n - b n | + | b n - L | = a n - b n + | b n - L | c n - b n + | b n - L | = | c n - b n | + | b n - L | | c n - L | + | L - b n | + | b n - L | .

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 N 1 , and let N 2 be a positive integer so that | b n - L | < ϵ / 3 if n N 2 . Then set N = max ( N 1 , N 2 ) . Clearly, if n 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 { a n } and { b n } be two sequences of complex numbers with a = lim a n and b = lim b n . Then

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

Part (1) is exactly the same as [link] . Let us prove part (2).

By [link] , both sequences { a n } and { b n } are bounded. Therefore, let M be a number such that | a n | M and | b n | M for all n . Now, let ϵ > 0 be given. There exists an N 1 such that | a n - a | < ϵ / ( 2 M ) whenever n N 1 , and there exists an N 2 such that | b n - b | < ϵ / ( 2 M ) whenever n N 2 . Let N be the maximum of N 1 and N 2 . Here comes the add and subtract trick again.

r c l | a n b n - a b | = | a n b n - a b n + a b n - a b | | a n b n - a b n | + | a b n - a b | = | a n - a | | b n | + | a | | b - b n | | a n - a | M + M | b n - b | < ϵ

if n N , which shows that lim ( a n b n ) = a b .

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 | < ( ϵ | b | 2 ) / ( 4 M ) whenever n N 1 ; there also exists an N 2 such that | b n - b | < ( ϵ | b | 2 ) / ( 4 M ) whenever n N 2 ; and there exists an N 3 such that | b n | > | b | / 2 whenever n N 3 . (See [link] .) Let N be the maximum of the three numbers N 1 , N 2 and N 3 . Then:

r c l | a n b n - a b | = | a n b - b n a b n b | = | a n b - b n a | 1 | b n b | < | a n b - b n a | 1 | b | 2 / 2 ( | a n - a | | b | + | a | | b n - b | ) 2 | b | 2 < ( M | a n - a | + M | b n - b | ) 2 | b | 2 < ϵ

if n 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 / 4 M 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

| a n b n - a b | ( M | a n - a | + M | b n - b | ) 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. { n 1 / n 2 } .
  2. { ( n 2 ) 1 / n } .
  3. { ( 1 + n ) 1 / n } .
  4. { ( 1 + n 2 ) 1 / n 3 } .
  5. { ( 1 + 1 / n ) 2 / n } .
  6. { ( 1 + 1 / n ) 2 n } .
  7. { ( 1 + 1 / n ) n 2 } .
  8. { ( 1 - 1 / n ) n } . HINT: Note that
    1 - 1 / n = n - 1 n = 1 n n - 1 = 1 n - 1 + 1 n - 1 = 1 1 + 1 n - 1 .
  9. { ( 1 - 1 / ( 2 n ) ) 3 n } .
  10. { ( n ! ) 1 / n } .

Questions & Answers

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Damian Reply
research.net
kanaga
Introduction about quantum dots in nanotechnology
Praveena Reply
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Anassong Reply
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
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Damian Reply
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Anassong
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s. Reply
there is no specific books for beginners but there is book called principle of nanotechnology
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Devang Reply
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s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
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Abhijith Reply
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
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s. Reply
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
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SUYASH Reply
for screen printed electrodes ?
SUYASH
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s. Reply
of graphene you mean?
Ebrahim
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Ebrahim
in general
s.
Graphene has a hexagonal structure
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China
Cied
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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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