Page 1 of 3 123 LastLast
Results 1 to 15 of 31

Math Help - Limit of a Sequence

  1. #1
    Junior Member
    Joined
    Mar 2010
    Posts
    62

    Limit of a Sequence

    Is there a sequence (a sub n) which is bounded and is not convergent such that a sub n+1 - a sub n tens to 0 as n tends to infinity?

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member Black's Avatar
    Joined
    Nov 2009
    Posts
    105
    1,0,1,0,1,0,...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2010
    Posts
    62
    Quote Originally Posted by Black View Post
    1,0,1,0,1,0,...
    You mean (a sub n)=(1,0,1,0,1,0,...) and (a sub n+1)=(1,0,1,0,1,0,...)?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Oct 2009
    Posts
    195
    Quote Originally Posted by Markeur View Post
    You mean (a sub n)=(1,0,1,0,1,0,...) and (a sub n+1)=(1,0,1,0,1,0,...)?
    That doesn't even make sense. If (a_n) = (1,0,1,0,1,0,...) how could (a_n+1) equal the very same thing?

    Also, lim (a_n) - lim (a_n+1) is not the same as lim(a_n - a_n+1).

    Edit: That wasn't right, because it's minus.

    Well, if the lim (a_n - a_n+1) = 0, then the series is Cauchy, and therefore convergent. So the answer is no.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Mar 2010
    Posts
    62
    Quote Originally Posted by davismj View Post
    That doesn't even make sense. If (a_n) = (1,0,1,0,1,0,...) how could (a_n+1) equal the very same thing?

    Also, lim (a_n) - lim (a_n+1) is not the same as lim(a_n - a_n+1).

    Edit: That wasn't right, because it's minus.

    Well, if the lim (a_n - a_n+1) = 0, then the series is Cauchy, and therefore convergent. So the answer is no.
    Actually,the question that I posed was in Cambridge tripos 2006,Q11.
    http://www.math.cam.ac.uk/teaching/p.../PaperIA_1.pdf

    Since it was asked in the question,there must be some sequences that have this property.Just that I still can't think of any.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member mabruka's Avatar
    Joined
    Jan 2010
    From
    Mexico City
    Posts
    150
    Quote Originally Posted by davismj View Post
    That doesn't even make sense. If (a_n) = (1,0,1,0,1,0,...) how could (a_n+1) equal the very same thing?

    Also, lim (a_n) - lim (a_n+1) is not the same as lim(a_n - a_n+1).

    Edit: That wasn't right, because it's minus.

    Well, if the lim (a_n - a_n+1) = 0, then the series is Cauchy, and therefore convergent. So the answer is no.

    That is not right. A cauchy sequence does have that property but being cauchy is so much stronger.


    For example:

    x_n=\sum_{1}^{n} \frac{1}{i}


    Clearly it diverges so is not Cauchy but |x_{n+1}-x_n|=\frac{1}{n+1} \to 0


    However this is not the sequence Markeur needs because it is not bounded.
    Black idea is to define

    Let  x_n=   1 if n is odd or 0 if n is even

    then
    x_{n+1}-x_n= 0 for all n thus x_{n+1}-x_n \longrightarrow 0 , it is bounded but divergent.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member Black's Avatar
    Joined
    Nov 2009
    Posts
    105
    Actually, now that I looked at the link, my example will fail. I think the sequence

    a_n=\text{sin}\left(\sum_{k=1}^n\frac{1}{k}\right)

    will work.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member mabruka's Avatar
    Joined
    Jan 2010
    From
    Mexico City
    Posts
    150
    Why would it fail? I cant see the reason
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member Black's Avatar
    Joined
    Nov 2009
    Posts
    105
    Because, if you have the sequence 1,0,1,0,... , then |a_{n+1}-a_n|=1, \, \forall n \in \mathbb{N}.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member mabruka's Avatar
    Joined
    Jan 2010
    From
    Mexico City
    Posts
    150
    Ouch LOL

    you are right i totally missed that


    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member
    Joined
    Oct 2009
    Posts
    195
    Quote Originally Posted by mabruka View Post
    That is not right. A cauchy sequence does have that property but being cauchy is so much stronger.


    For example:

    x_n=\sum_{1}^{n} \frac{1}{i}


    Clearly it diverges so is not Cauchy but |x_{n+1}-x_n|=\frac{1}{n+1} \to 0


    However this is not the sequence Markeur needs because it is not bounded.
    Black idea is to define

    Let  x_n=   1 if n is odd or 0 if n is even

    then
    x_{n+1}-x_n= 0 for all n thus x_{n+1}-x_n \longrightarrow 0 , it is bounded but divergent.
    That's true, about what I said. My mistake. About what you said. (1,0,1,0,...) - (0,1,0,1,...) is (1,-1,1,-1,...), which doesn't converge, let alone to zero.

    Regarding black's other idea...

    I don't think so. Any way you split up sin(x), the n+1th term will never come arbitrarily close to the nth term without being convergent.

    In general, (a_n+1) - (a_n) would require the limit of (a_n+1) to converge to the limit of (a_n) and therefore would require (a_n) to have a limit. Since this is clearly not the case, the only other option is that (a_n+1) = (a_n) for every n, which (since this is pretty much induction) means that the sequence is constant. Here again, the sequence converges.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Member
    Joined
    Aug 2009
    Posts
    130
    3, 3.1, 3.14, 3.141, 3.1415... is a bounded sequence, where  (a_{n+1} - a_n) \rightarrow 0 but isn't convergent in  Q

    But I guess you're talking about sequences in R?
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by JG89 View Post
    3, 3.1, 3.14, 3.141, 3.1415... is a bounded sequence, where  (a_{n+1} - a_n) \rightarrow 0 but isn't convergent in  Q

    But I guess you're talking about sequences in R?
    That's a good example, but I think if he was looking for something like that he'd say "Cauchy but not convergent".
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Member mabruka's Avatar
    Joined
    Jan 2010
    From
    Mexico City
    Posts
    150
    What is wrong with a_n=\sin\left(\sum_{k=1}^n\frac{1}{k}\right) ?


    We have it is not convergent.
    It is bounded.
    And since sin is continuous, for every \epsilon >  0 there is \delta >  0 such that

    if \frac{1}{n+1}=\left | \sum_{k=1}^{n+1}\frac{1}{k}   -\sum_{k=1}^n\frac{1}{k} \right | < \delta then

    \left | \sin\left(\sum_{k=1}^{n+1}\frac{1}{k}\right)   -\sin\left(\sum_{k=1}^n\frac{1}{k}\right) \right | < \epsilon

    So for every \epsilon there is an N \in \mathbb N such that if n+1>N    , then

    \frac{1}{n+1} < \delta and

    |a_{n+1}-a_n|=\left | \sin\left(\sum_{k=1}^{n+1}\frac{1}{k}\right)  -\sin\left(\sum_{k=1}^n\frac{1}{k}\right) \right | < \epsilon
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Junior Member
    Joined
    Mar 2010
    Posts
    62
    I wonder whether there is a sequence x_n such that \displaystyle\lim_{n\to\infty}x_n^{2} = L and x_{n+1}-x_n \longrightarrow 0 but x_n diverges?
    Follow Math Help Forum on Facebook and Google+

Page 1 of 3 123 LastLast

Similar Math Help Forum Discussions

  1. Limit of Sequence
    Posted in the Calculus Forum
    Replies: 13
    Last Post: August 6th 2011, 04:22 PM
  2. limit of sequence
    Posted in the Calculus Forum
    Replies: 3
    Last Post: June 28th 2011, 05:09 PM
  3. Replies: 2
    Last Post: October 26th 2010, 10:23 AM
  4. Sequence and limit
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: June 4th 2009, 02:04 PM
  5. Limit of a sequence
    Posted in the Calculus Forum
    Replies: 5
    Last Post: May 25th 2008, 09:07 AM

Search Tags


/mathhelpforum @mathhelpforum