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!

Convergence Proof need help

  1. Oct 23, 2014 #1
    • OP has been warned that posts need to show what was attempted
    1. The problem statement, all variables and given/known data

    an = [sin(n)]/n

    Prove that this sequence converges using Cauchy theorem

    2. Relevant equations

    Cauchy theorem states that:

    A sequence is called a Cauchy theorem if for all ε > 0, there exists N , for all n > N s.t. |xn+1 - xn| < ε


    I do not know how to approach this proof.

    I would appreciate some help.
     
  2. jcsd
  3. Oct 23, 2014 #2

    RUber

    User Avatar
    Homework Helper

    Can you think of an inequality that would apply to ##\left|\frac {\sin(n)}{n}-\frac {\sin(n+1)}{n+1}\right|##?
     
  4. Oct 23, 2014 #3

    RUber

    User Avatar
    Homework Helper

    You should end up with a statement that says, given ##\epsilon>0##
    ##\left|\frac {\sin(n)}{n}-\frac {\sin(n+1)}{n+1}\right| < \epsilon ## for any ##n \geq N \geq f(\epsilon)##
     
  5. Oct 23, 2014 #4

    pasmith

    User Avatar
    Homework Helper

    That is not the correct definition. You have stated the condition for [itex]x_{n+1} - x_n \to 0[/itex], which is a necessary but not a sufficient condition for [itex]x_n[/itex] to converge.

    The correct definition is:

    A sequence [itex](a_n)[/itex] is Cauchy if and only if for every [itex]\epsilon > 0[/itex] there exists an [itex]N \in \mathbb{N}[/itex] such that for all [itex]n \in \mathbb{N}[/itex] and all [itex]m \in \mathbb{N}[/itex], if [itex]n \geq N[/itex] and [itex]m \geq N[/itex] then [itex]|a_n - a_m| < \epsilon[/itex].

    There is a theorem which states that a real sequence converges if and only if it is Cauchy.

    You may find it helpful show that [itex]|m \sin n - n \sin m| < n + m[/itex].
     
  6. Oct 23, 2014 #5
    I used ##0 \le |\sin x| \le 1##
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Convergence Proof need help
Loading...