Results 1 to 5 of 5

Thread: Convergent sequence

  1. #1
    Member
    Joined
    Jan 2009
    Posts
    145

    Convergent sequence

    Note: This is a homework problem.

    Suppose that {$\displaystyle {a_n}$} is a bounded sequence of reals such that for any n, $\displaystyle a_n \le \frac{a_{n-1} + a_{n+1}}{2} $.

    Show that $\displaystyle b_n = a_{n+1} - a_n$ is an increasing sequence and, $\displaystyle {a_n}$ converges.



    I've already done the first part, that $\displaystyle {b_n}$ is an increasing sequence. For the second part, my attempt so far has been that looking at $\displaystyle a_2 \le \frac{a_1 + a_3}{2} $, there are two possibilities:

    Case 1: It's possible that $\displaystyle a_3 \ge a_2 $. In that case, if for some integer k, $\displaystyle a_{k+1} \ge a_k $, then we can use that $\displaystyle a_{k+1} \le \frac{a_k + a_{k+2}}{2}$ to show that $\displaystyle 2*a_{k+1} \le a_k + a_{k+2} \le a_{k+1} + a_{k+2} $, and so $\displaystyle a_{k+1} \le a_{k+2} $, establishing by induction that {$\displaystyle a_n$} is also an increasing bounded sequence and then convergent.

    Case 2: It's also possible that $\displaystyle a_1 \ge a_2 $. This is where I'm having trouble. The only way I can see to continue with this is to show that the sequence {$\displaystyle a_n$} is decreasing in this case. After several failed attempts, I'm starting to become skeptical that it actually is decreasing. Is this not going to work, and should I abandon this attempt? If so, does anyone have a suggestion for what direction to go in?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Jan 2009
    Posts
    145
    In the second case, I assume that $\displaystyle a_k \ge a_{k+1} $ for some positive k. Then, $\displaystyle a_{k+1} \le \frac{a_{k+2} + a_k}{2} $, implying that $\displaystyle 2a_{k+1} \le a_{k+2} + a_k $.

    I have no idea where to go after this point. It doesn't seem like I can finish this in this way by showing that the sequence is either monotonically increasing (case 1) or decreasing (case 2). Is there a better way to attempt this?

    Thanks for any help!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,782
    Thanks
    2824
    Awards
    1
    Can you show that $\displaystyle (b_n)$ is bounded above?
    Increasing bounded sequences converge.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jan 2009
    Posts
    145
    Showing that $\displaystyle (b_n) $ is bounded above follows from the fact that $\displaystyle (a_n) $ is bounded, right? I can just pick an N that's larger than $\displaystyle 2*a_n$ for any n, and then $\displaystyle N > b_n = a_{n+1} - a_n $? So I can conclude that $\displaystyle (b_n} \rightarrow b $, for some b. Then I'd use this to show that the $\displaystyle (a_n) $ sequence converges by taking some $\displaystyle \epsilon > 0 $ and noting that for some N, if n > N then $\displaystyle |b + a_{n+1} - a_n| = |b - b_n| < \epsilon $.

    I don't really see how the left hand side of that last equality can be simplified to show the convergence of the $\displaystyle (a_n) $ sequence though :/

    Edit: Sorry, I'm not trying to fish for answers. I had tried looking at the limit of the $\displaystyle (b_n) $ sequence already and didn't have any luck simplifying the expression. I'm really frustrated trying to see how to evaluate limits in this way, so even a nudge in the right direction would be greatly appreciated.
    Last edited by Math Major; Oct 15th 2010 at 04:16 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Sep 2010
    Posts
    24
    b_n converges to 0 and a_n is a decreasing sequence. when a_n reaches its bound, b_n maxes out at 0
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. is this sequence convergent?
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: Feb 14th 2011, 08:39 AM
  2. convergent sequence
    Posted in the Differential Geometry Forum
    Replies: 7
    Last Post: Feb 27th 2010, 11:59 AM
  3. Convergent sequence
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Oct 26th 2009, 01:18 AM
  4. Convergent sequence
    Posted in the Differential Geometry Forum
    Replies: 6
    Last Post: Jul 7th 2009, 04:11 PM
  5. convergent sequence
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Oct 9th 2007, 03:34 PM

Search Tags


/mathhelpforum @mathhelpforum