[PDF] [PDF] 2 Sequences

[PDF] Math 3150 Fall 2015 HW2 Solutions

Let (sn) be a sequence that converges (a) Show that (c) Conclude that if all but finitely many sn belong to [a, b], then limsn ∈ [a, b] Then supposing that lim

[PDF] 4 Sequences 41 Convergent sequences • A sequence (s n

(sn) be a sequence, let s be a number, and suppose that sn −s ≤ an for all n ≥ 1, where (an) is a sequence with limit 0 Then limn sn = s Proof: We have 0 

[PDF] Convergent Sequences

Suppose n1 < n2 < n3 < ··· is a strictly increasing sequence of indices, then (snk ) is a subsequence of (sn) We will Now we state some limit theorems Let (sn) be a sequence that converges to s ∈ R Applying the definition to ε = 1, we see

[PDF] Homework 3

(b) Suppose (sn) and (tn) are sequences such that sn ≤ tn for all n and lim tn = 0 (c) lim[ √ 4n2 + n − 2n] = 1 4 8 9 Let (sn) be a sequence that converges

[PDF] Math131A Midterm Solutions July 15, 2013 Math131A Midterm

15 juil 2013 · Solution Let sn = n Suppose (sn) is a convergent sequence such that lim sn < 23 Let us prove that if ∑ sn converges, then so does ∑ sp

[PDF] Homework 8

7 Let (sn) be a convergent sequence and suppose that lim sn > a Prove that there exists a number N such that n>N 

[PDF] Solutions for Homework  Math 451(Section 3, Fall 2014)

8 5a) Claim: Suppose that (an), (bn) and (sn) are three sequences and that Now, suppose that (sn) converges to 0 Let ϵ > 0 Since limsn = 0, there exists N 

[PDF] HOMEWORK 4 - UCLA Math

Let (sn) be a bounded decreasing sequence Then (−sn) is a bounded increasing sequence, so −sn → L for some limit L Hence (sn) is convergent with sn 

[PDF] 2 Sequences

Theorem 2 3 If (sn) converges, then its limit is unique Proof Suppose s and t are two limits Take c = s−t 2 in the definition of limit Then ∃N1, N2 such that

[PDF] be a sequence with positive terms such that lim n→∞ an = L > 0 Let

For the inductive step, suppose we have defined b1, ,bn and bn = rl = ak Let { an} be a bounded sequence such that every convergent subsequence of {an} n + 1 This is an example of a telescoping series Since ∞ lim n=1 sn = ∞ lim

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2 Sequences

2.7 Limits of Sequences & 2.8 A Discussion about Proofs

Sequences of numbers are the fundamental tool of our approach to analysis. Definition 2.1.Asequenceis a functionswith domainfn2Z:nmgfor some integerm.

Alternatively, a sequence is an ordered set:

(sn)¥n=m= (sm,sm+1,sm+2,...) This is strictly the definition of an infinitesequence. We won"t consider finite sequences. Most commonly m=0 or 1 so that the initial term of the sequence iss0ors1. If the domain is understood or not r elevant,we might simply r eferto the sequence (sn). The codomain of a sequence can be any set. In elementary analysis, ty picallyevery snis a real number: in such a case we will say that "(sn)is a sequence of real numbers." Towards the end of the course, se shall consider sequences of functions (e.g. examples 2 & 3 below).

Examples

1.

For each n2N, letsn=1+1n

n. Then s

1=2,s2=94

,s3=6427 2. For each n2N0, letsnbe the functionsn:[0,1]!Rdefined by s n(x) =nxn(1x) 3.

For each n2N, define the functionsn:R!Rby

s

01,sn+1=1+Z

x

0sn(t)dt

so that s

1(x) =1+x,s2(x) =1+x+12

x2,s3(x) =1+x+12 x2+16 x3, ...

Limits

We want to describe what it means for the terms of a sequence to approach arbitrarily close to some value. In a calculus class you should have become used to writing expressions such as lim n!¥2n2+3n13n22=23 and limn!¥pn

2+4n=limn!¥4pn

2+4+n=0

Our first order of business is to make this logically watertight. Definition 2.2.Let(sn)be a sequence of real numbers and lets2R.

We say that(sn)converges to s, if

8e>0,9Nsuch thatn>N=)jsnsj We callsthelimitof(sn)and write limsn=sor simplysn!s(readsnapproaches s). We say that(sn)convergesif it has a limit, and that itdivergesotherwise. A limit must be finite! We shall discuss sequences whichdiverge to infinitylater. It is your choicewhether to insist thatNbe an integer or to allow it to be a (general) real number; the definitions are equivalent.

1Unless stated otherwise, we"ll assumeN2R. You should

certainly stateN2Nif something in your answer requires it! It is common but unnecessary to see n!¥written: e.g. limn!¥sn=sorsn!n!¥s. Feel free to do so if you feel it useful.

Below is a clickable version

2of the limit definition for the sequence withnthterm

s n=1+32 en/20cosn4 You should believe without proof thats=limsn=1. Try viewing the definition as a game: Givene>0, wechoose Nso that all termssncomingafter Nare closer tosthane. A proof amounts to a strategy that shows you will always win the game! We"ll not give an explicit proof here (try it later once you"ve seen more examples...). Instead use the animation to help you understand that, asegets smaller, we"re forced to chooseNlarger in order to satisfy the definition.1

This follows from the Archimidean principle: if9N2Rsatisfying the definition, then9˜N2Nsuch that˜NN.

Certainlyn>˜N=)n>N...

2If you want to the picture to move, you"ll need to open these notes in a full-function pdf reader such as Acrobat. A

lightweight pdf viewer or a web-broswer will likely only show a single still frame. 2

A Fully Worked Example

Weprovethat the sequence defined bysn=21pn

converges tos=2.

The definition requires us to show that a 'for all" statement is true. Our proof should therefore have

the following structure:

Start by supposing that e>0 has been given to us.

Describe how to choosea numberN(dependent one).

Check (usually a dir ectpr oofwith simple algebra) that if n>Nthenjsnsj...guarantees thatsnlivesher enbeing at leastthis lar ge...Scratch work.It is usually difficult to choose a suitableN, so it is a good idea to start with what you

want to be true and let it inspire you.

W ewant n>N=)

21pn
2This r equires

1pn 1e 2.

Choosing N=1e

2should be enough to complete the proof!

Warning!

W edo not yet have a pr oof!If your ar gumentfinishes " ...=)N=1e

2" then your con-

clusion is incorrect. Rearrange your scratch work to make it clear that you"ve satisfied the definition!

Proof.Lete>0 be given. LetN=1e

2. Then

n>N=)n>1e

2=)1pn

Thussn!2 as required.With practice, you might be able to produce a correct argument immediately for such a simple exam-

ple. However, in most cases even experts expect to first need some scratch work. 3 Uniqueness of LimitAs suggested by the definite article (...calls thelimit...) in Definition 2.2... Theorem 2.3.If(sn)converges, then its limit is unique.

Proof.Supposesandtare two limits. Takee=jstj2

in the definition of limit. Then9N1,N2such that n>N1=)jsnsjN2=)jsntjLetn>maxfN1,N2g. Then, j stj=jssn+sntjjsnsj+jsntj(4-inequality) jstj2 +jstj2 jstj

Contradiction.The idea of the proof is very simple: there exists atailof the sequence (all termssncomingaftersome

N) all of whose terms are close toboth limits: this is complete nonsense!st s+etes+t2

For alln>N,snmust lie bothher eand her e!Further ExamplesWe give several more examples of using the limit definition. In all cases,onlythe

formal argument needs to be presented. The challenge is figuring out what to write, so we present varying amounts of scratch work first. 1. Generalizing our pr eviousexample, we show that, for any k2R+the sequence defined by s n=1n khassn!0 Again a little scratch work, but faster this time. Givene>0, we want to chooseNsuch that n>N=)1n kThis amounts to havingn>1e

1/k. We can now write a formal argument:

Proof.Lete>0 be given. LetN=1e

1/k. Then

n>N=)n>1e

1/k=)1n

We conclude that 1n k!0.4

2.W epr ovethat sn=2n+13n7converges to23

First some scratch work: we must conclude

2n+13n723

3n7>17e ()n>73 +179e

We now have enough for a proof:

Proof 1.Lete>0 be given and letN=73

+179e. Then
n>N=)n>73 +179e=)0<173(3n7) =)2n+13n723 =3(2n+1)2(3n7)3(3n7) =173(3n7)73 . We conclude that2n+13n7!23 .Here is an alternative proof where we use a simpler expression forN.

Proof 2.Lete>0 be given and letN=max

7,3e . Then n>N=)2n+13n723 =173(3n7) <176n (sincen>7=)3n7>2n) 3n <3N e(sinceN3e

Hence result.The plot illustrates the two choices ofNas functions ofe. Note that the second is always larger

than the first! This is fine: if a particular choice ofN=N1(e)works in a proof, so will any other N

2(e)which is larger thanN1! Use this to your advantage to produce simpler arguments.020406080100

N 0 0.2 0.4 0.6 0.8 1 eN 1=73 +179e
N 2=max 7,3e 5

3.W epr ovethat sn=4n45n+13n4+2n2+3converges to43

We want to conclude that

4n45n+13n4+2n2+343

=8n215n93(3n4+2n2+3) 8n215n93(3n4+2n2+3) <9n29n4 =1n 2

Indeed it is enough to have

9n28n2+15n+9

which, by solving the quadratic, holds whenn15+p261 2 . Round this up to 16 and we have enough for the proof.

Proof 1.Lete>0 be given and letN=maxf16,1pe

g. Then n>N=)sn43 =8n215n93(3n4+2n2+3) <9n29n4 (sincen>16) 1n 2<1N

2eIf this were a formal answer, it would be wise to give a little scratch work to justify whyn>16

is sufficient for the inequality. Here is an alternative: this time we include a little of the scratch work in the answer, as you might do for a homework submission.

Proof 2.Suppose thatn24, then

n15+9n =)n215n+9=)9n28n2+15n+9

Lete>0 be given, and letN=maxf24,1pe

g. Then n>N=)sn43 =4n45n+13n4+2n2+343 =8n215n93(3n4+2n2+3) <9n29n4 1n 2<1N

2e(sinceN1pe

)Thereforesn!43 , as required. 6

Divergent sequences

Definition 2.4.Negating Definition 2.2 says that a sequence(sn)does not converge to sif,

9e>0 such that8N,9n>Nwithjsnsje

Furthermore, we say that(sn)isdivergentif it does not converge to any limits2R. Otherwise said,

8s2R,9e>0 such that8N,9n>Nwithjsnsje

It can be helpful when proving divergence to assumeN2Nso that you can quickly definenin terms ofN. At the bottom of the page we"ll explain what happens if you don"t...

Examples

1.

W epr ovethat the sequence with sn=7n

does not converge tos=1. We need to show that

9e>0 such that8N,9n>Nwith7n

1e() This is easy tovisualize: we know thatsn!0, so the sequence must eventually be nearly a distance 1 froms=1. Any value ofesmaller that 1 should satisfy(). We prove twice: once using the Definition directly, and once by contradiction.0234567s n01020s+e se Ns=1 nEvery tail of the sequence after her e ...contains some elements s nthat donotlieher e oDirect Proof.Lete=12 . We need to force7n 112
. Since we are only concerned withlarge values ofn, the term in the absolute value isnegative,7n 112
()17n 12 ()7n 12 ()n14

GivenN2N, letn=maxf14,N+1gto see that7n

1e. We conclude thatsn91.If we had only assumed thatNwerereal,then the definition ofnfails to be aninteger.This can

be fixed in a couple of ways:

Define n=maxf14,dNe+1gusing the ceiling function.

Appeal to the Ar chimideanpr opertyto show that 9n2Nsuch thatn>maxf13,Ng. Restricting toN2Nmakes the argument easier to follow: just remember to state it to help the reader! 7

Contradiction Proof.Supposesn!1. Then

8e>0,9N2Nsuch thatn>N=)7n

1In particular, this should hold fore=12 . But then, for all largen, we would require 7n 1<12quotesdbs_dbs17.pdfusesText_23