1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Problem with convergent sequences

  1. Jul 3, 2012 #1
    Hi, I have the following problem and have done the first two questions, but I don't know how to solve the last two. Thanks for any help you can give me!

    1. The problem statement, all variables and given/known data
    Let [itex]a_{n}\rightarrow a[/itex], [itex]b_{n}\rightarrow b[/itex] be convergent sequences in [itex]\Re[/itex]. Prove, or give a counterexample to, the following statements:

    A) [itex]a_{n}[/itex] is a monotone sequence;
    B) if [itex]a_{n}>b_{n}+1/(n^3+4)[/itex], then [itex]a>b[/itex];
    C) if [itex]a_{n}>((n^3+1)/(2n^3+1))b_{n}[/itex], then a>b;
    D) if [itex]s_{n}=(1/n)(a_1+...+a_n)[/itex], then [itex]s_n \rightarrow a[/itex].


    2. Relevant equations



    3. The attempt at a solution

    I have solved the first two. For A I have given the counterexample [itex]a_n=sin(n)/n[/itex] and for B I have used the fact that as n goes to infinity, [itex]1/(n^3+4)[/itex] approaches 0, which would give [itex]a_n > b_n[/itex], which is a>b when n goes to infinity.

    I have tried the same thing with C, but it gives me [itex]a>(1/2)b[/itex], which doesn't lead me anywhere, I think. And for D, I think that as n goes to infinity, [itex]s_n[/itex] will be close to [itex]a_n[/itex] because [itex]s_n ≈ (1/n)*n*a_n[/itex], which is the same as saying [itex]s_n \rightarrow a[/itex]. However, I don't know if this is correct, and if it is, how am I supposed to express it?

    Thanks a lot!
     
  2. jcsd
  3. Jul 3, 2012 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Think about that last statement if ##a_n = \frac 2 {n^3+4}## and ##b_n=0##.
     
  4. Jul 4, 2012 #3
    Thanks!

    So that would give me [itex]2/(n^3+4)>1/(n^3+4)[/itex], which holds when n goes to infinity.

    So could I use something similar for part C then, something like [itex]b_n=0[/itex] or [itex]b_n=1[/itex]? With [itex]b_n=1[/itex], I could have [itex]a_n=(2n^3+1)/(2n^3+1)[/itex]?
     
  5. Jul 4, 2012 #4
    However, they would not hold for negative values of n? I am confused :/
     
  6. Jul 4, 2012 #5

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I'm not sure you understand my example and I don't know what you mean when you say it "holds when n goes to infinity". The ##a_n## and ##b_n## in my example go ##a=0## and ##b=0## respectively. The ##a_n>b_n## does not hold in the limit.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Problem with convergent sequences
  1. Sequence Convergence (Replies: 2)

Loading...