# Cauchy sequences

by garyljc
Tags: cauchy, sequences
 P: 105 By definition, a sequence a(n) has the Cauchy sequence if for eery E>0 ,there exist a natural number N such that Abs(a(n) - a(m) ) < E for all n, m > N Could anyone tell me what is a(m) ? is it a subsequence of a(n) , or could it be any other non related sequence ?
 Emeritus Sci Advisor PF Gold P: 4,500 a(m) is the same sequence as a(n)
 HW Helper P: 2,264 a(m) and a(n) are not sequences they are elements of a sequence Pehaps the difficulty will be eased by restating the definition differently a sequence is Cauchy if for any E>0 there exist a natural number N such that the difference between any two terms beyond N cannot exceed N or a sequence is Cauchy if for any E>0 there exist a natural number N such that Abs(a(N+n) - a(N+m) ) < E for all n,m that are natural numbers
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,682 Cauchy sequences Neither a_m nor a_n in that is a sequence. They are, rather, any two numbers from the original sequence {a_i}, with, of course, m and n larger than N.
 P: 105 OK thanks One more question what's the difference between Lim sup a(n) and sup A(n) does the limit tells me something else ?
P: 6,104
 Quote by garyljc OK thanks One more question what's the difference between Lim sup a(n) and sup A(n) does the limit tells me something else ?
It is easier to explain by example. Consider the sequence 1, 1/2, 1/3, 1/4,...

The sup is 1, while the lim sup is 0.
 P: 95 what basically is the change that lim produced in sup why it changes sup=1 to lim sup=0 and do this thing hold in all cases that lim sup is not the part of the sequence
 P: 587 The limit superior of a sequence $(a_n)_{n\geq 0}$ is the largest accumulation (or cluster) point of this sequence. An accumulation point is a number c such that in any neighbourhood of c there are infintely many members of the sequence. Analogously, the limit inferior is the least such accumulation point. If $(a_n)_{n\geq 0}$ is convergent, say with limit a, then $$\lim_{n\to\infty} {a_n} = \limsup_{n\to\infty}{a_n} = \liminf_{n\to\infty}{a_n} = a$$

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