Results 1 to 4 of 4

Math Help - seemingly simple limit proof

  1. #1
    Newbie
    Joined
    Sep 2010
    Posts
    24

    seemingly simple limit proof

    Problem Statement:
    suppose that { s_n} and { t_n} are sequences such that s_n \leq t_n for all n. Show that if s_n \rightarrow s and t_n \rightarrow t, then s \leq t. Hint: an " \varepsilon-argument" works well here.

    Relevant Theorems:
    Thm 3.3-a (Baby-Rudin) - \lim_{n \rightarrow \inf}(s_n+t_n)=s+t
    My Work:
    For some N_1, when n \geq N_1 , then |s-s_n| < \varepsilon given \varepsilon > 0.
    For some N_2, when n \geq N_2 , then |t- t_n| < \varepsilon given \varepsilon > 0.
    now, \lim_{n \rightarrow \inf}s_n=s and \lim_{n \rightarrow \inf}t_n=t.
    From thm 3.3, we have that \lim_{n \rightarrow \inf}(t_n-s_n)=t-s and since (t_n-s_n) \geq 0 for all n, then t-s \geq 0
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    Quote Originally Posted by DontKnoMaff View Post
    Problem Statement:
    suppose that { s_n} and { t_n} are sequences such that s_n \leq t_n for all n. Show that if s_n \rightarrow s and t_n \rightarrow t, then s \leq t. Hint: an " \varepsilon-argument" works well here.

    Relevant Theorems:
    Thm 3.3-a (Baby-Rudin) - \lim_{n \rightarrow \inf}(s_n+t_n)=s+t
    My Work:
    For some N_1, when n \geq N_1 , then |s-s_n| < \varepsilon given \varepsilon > 0.
    For some N_2, when n \geq N_2 , then |t- t_n| < \varepsilon given \varepsilon > 0.
    now, \lim_{n \rightarrow \inf}s_n=s and \lim_{n \rightarrow \inf}t_n=t.
    (*) From thm 3.3, we have that \lim_{n \rightarrow \inf}(t_n-s_n)=t-s and since (t_n-s_n) \geq 0 for all n, then t-s \geq 0
    How does (*) follow from any of the above? What you're saying is true (that if a convergent sequence is always non-negative, then its limit is non-negative), but it's equivalent to what you're trying to show. And I don't see how it follows from what you've written. (What have you done with \epsilon, N_1, \mbox{and }N_2, apart from summoning them into existence?)

    A proof by contradiction works well. Suppose that a_n \geq 0 \ \forall n and that a_n \to a. Suppose that a<0. Take \epsilon>0 small enough that a+\epsilon < 0. What do you get if you find N large enough that |a_n-a|<\epsilon for n\geq N? (Break up the absolute value into two inequalities; one of them will contradict the non-negativity of a_n).

    Then take a_n to be the sequence t_n-s_n, as you did.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2010
    Posts
    24
    thanks alot. I am horrible at writing these proofs. OK here is what I've got now:

    take { a_n} where a_n=t_n-s_n for all n and { a_n} \rightarrow a=t-s.
    We have assumed that s_n \leq t_n which yields t_n-s_n \geq 0, so a_n \geq 0 for all n.
    We now want a>0.
    Suppose a<0, take \epsilon > 0 s.t. a+\epsilon < 0.
    Now take N s.t. when n \geq N, then , |a_n-a| < \epsilon.
    Now since |a| > \epsilon and a <0 and a_n >0, this is a contradiction and so a \geq 0 which gives t-s \geq 0

    I think that works now.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    Everything is fine until the last line.
    What you should have written is that since |a_n-a| < \epsilon, we have -\epsilon < a_n-a < \epsilon (this is by definition of the absolute value). But a_n-a < \epsilon \Rightarrow a_n < a+\epsilon < 0, which contradicts a_n \geq 0.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. A seemingly simple proof
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: July 8th 2010, 10:09 PM
  2. Replies: 2
    Last Post: May 16th 2010, 09:42 AM
  3. seemingly simple derivative
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 17th 2009, 07:05 PM
  4. Help with a (seemingly?) simple proof.
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: May 27th 2009, 10:38 PM
  5. Another seemingly simple integral
    Posted in the Calculus Forum
    Replies: 5
    Last Post: February 25th 2009, 02:44 PM

Search Tags


/mathhelpforum @mathhelpforum