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Homework Help: 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


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    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


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    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


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    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##
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