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Sequences limits and cauchy sequences

  1. Dec 7, 2008 #1

    ibc

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    1. The problem statement, all variables and given/known data
    prove or refute:

    if lim(a(2n)-a(n)=o , then a(n) is a cauchy sequence


    2. Relevant equations



    3. The attempt at a solution
    I need to prove that for every m,n big enough a(m)-a(n)<epsilon
    so I know for all m and n I can say m=l*n, lim(a(m)-a(n))=lim(a(n*l)-a(n*l/2) +a(n*l/2) -a(n*l/4)..........+a(2n)-a(n)), which is the sum of alot of zeros, though if I take m to be 2^n or something like that, I get an inifinite amount of zeros, so I don't know what I can do with that.
    so I tried to find a sequence which contredicts it, though couldn't find any




    1. The problem statement, all variables and given/known data
    prove or refute:
    if |a(n+1)-a(n)|<9/10*|a(n)-a(n-1)|
    then a(n) is a cauchy sequence


    2. Relevant equations




    3. The attempt at a solution
    well fromt he equation I can get:
    (|a(n+1)-a(n)|)/(|a(n)-a(n-1))<9/10<1
    so what it gives me is that all that in the power of n is going to zero, which means is a cauchy sequence, thoguh I don't see how it helps me =\
     
  2. jcsd
  3. Dec 7, 2008 #2

    Dick

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    Yes, you want a counterexample for the first one. It's going to have to be a function that grows REALLY slowly. What kind of functions can you think of that do that? For the second one for n<m, a(n)-a(m)=(a(n)-a(n+1))+(a(n+1)-a(n+2))+(a(n+2)-a(n+3))+ ... (a(m-1)-a(m)). Think you can maybe bound that sum by a geometric series?
     
  4. Dec 8, 2008 #3

    ibc

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    I know I need something that grows really really slow, but I couldn't find any
    and we can't use functions yet in the course, so it'd have to be a sequence/series
    the slowest one I could think of is 1/1+1/2+1/3+1/4+...... which is good to refute the another question, which was "if lim(a(n+1)-a(n)=o , then a(n) is a cauchy sequence", though for this one it's not helpful.
     
  5. Dec 8, 2008 #4

    Dick

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    a(n)=log(n) grows pretty slowly, but not quite slowly enough. Try sqrt(log(n)).
     
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