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A sequence {an}is called a Cauchy sequence if for any given ϵ > 0, there exists N ∈ N Proof Since {an}forms a Cauchy sequence, for ϵ = 1 there exists N ∈ N such that (i) lima1/n = 1, if a > 0 (ii) limnαxn = 0, if x < 1 and α ∈ IR Solution: 



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Lecture 5

1 Cauchy sequences

Denition 1.0.1.A sequencefangis called a Cauchy sequence if for any given >0, there existsN2Nsuch thatn;mN=) janamj< : Example 1.0.2.Letfangbe a sequence such thatfangconverges toL(say). Let >0be given.

Then there existsN2Nsuch that

janLj<2 8nN:

Thus ifn;mN, we have

janamj janLj+jamLj<2 +2

Thusfangis Cauchy.

Lemma 1.0.3.Iffangis a Cauchy sequence, thenfangis bounded. Proof.Sincefangforms a Cauchy sequence, for= 1 there existsN2Nsuch that janamj<1;8n;mN:

In particular,

janaNj<1;8nN:

Hence ifnN, then

janj janaNj+jaNj<1 +jaNj;8nN: LetM= maxfja1j;ja2j;:::;jaN1j;1 +jaNjg. Thenjanj Mfor alln2N. Hencefangis bounded./// Theorem 1.0.4.Iffangis a Cauchy sequence, thenfangis convergent. Proof.Letankbe a monotone subsequence of the Cauchy sequencefang. Thenankis a bounded, monotone subsequence. Hencefankgconverges toL(say). Now we claim that the sequencefang itself converges toL. Let >0. ChooseN1;N2such that n;n kN1=) janankj< =2 n kN2=) jankaj< =2: Then n;n kmaxfN;N1g=) janaj janankj+kankaj< :

Hence the claim.///

1

Therefore, we have the following Criterion:

Cauchy's Criterion for convergence:A sequencefangconverges if and only if for every >0, there existsNsuch that janamj< 8m;nN:

Example 1.0.5.Letfangbe dened asa1= 1;an+1= 1 +1a

n. The show thatfangis Cauchy.

Note thatan>1andanan1=an1+ 1>2. Then

jan+1anj=jan1ana nan1j 12 janan1j 12 n1ja2a1j;8n2: Hence jamanj jamam1j+jam1am2j+::::+jan+1anj ja2a1jn11;=12

So given, >0, we can chooseNsuch that12

N1<2

Indeed the following holds,

Theorem 1.0.6.Letfangbe a sequence such thatjan+1anj< janan1jfor allnNfor someNand0< <1. Thenfangis a Cauchy sequence.

Proof.Proof follows as in the previous example.

In the above theorem if= 1, then we cannot say if the sequence is Cauchy or Not. For example

Example 1.0.7.Letan=nX

k=11k . Then it is easy to see that jan+1anjjanan1=nn+ 1<1:

But the sequencefangdiverges. Indeed,a2n>1 +n2

, by induction.

On the other hand if we choosean=nX

k=11k

2. Then

jan+1anjjanan1=n2(n+ 1)2<1: But the sequencefangconverges (This will be proved in the next section while studying innite series). 2quotesdbs_dbs19.pdfusesText_25