Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Limits of Subsequences

  1. Feb 26, 2008 #1
    If two subsequences of a sequence {an} have different limits, does {an} converge? and Why?Could you prove it?
  2. jcsd
  3. Feb 26, 2008 #2


    User Avatar
    Science Advisor
    Homework Helper

    By definition, we say that a (sub)sequence converges if its limit exists (and is finite).
    By definition, if the limit exists, it is unique. You may prove this for yourself.

    For example, the sequence {-1, 1, -1, 1, -1, 1, ....} has two converging subsequences {1, 1, 1, 1, ...} and {-1, -1, -1, -1, ...} (actually, there are infinitely many, as long as it ends in just 1's or just -1's) which converge to 1 and -1 respectively. The sequence itself has no limit though.
  4. Feb 27, 2008 #3
    there is a theorem i guess, which states that if {an} converges then every subsequence of it converges and that to the same nr as {an}. Hence if we can find at least two subsequences of a sequence {an} that converge do different nrs, that is have different limits, then the sequence {an} does not converge!
  5. Feb 28, 2008 #4


    User Avatar
    Science Advisor

    Specifically, you can do this: suppose the subsequence of {an} has subsequence {an}i which converges to P and subsequence {an}j which converges to Q.

    Assume that {an} converges to L, and take [itex]\epsilon[/itex]= (1/2)|P- Q|. For any N, there will be n1> N such that an1, in the first subsequence, is arbitratily close to P and n2> N such that an2, in the second subsequence is arbitrarily close to Q. If they are not within [itex]2\epsilon[/itex] of each other, they cannot both be within [itex]\epsilon[/itex] of L, a contradiction.
  6. Oct 14, 2010 #5
    Hello everyone,

    I have tried to write a proof based on HallsofIvy's response, posted below. However, I am not able to derive a contradiction from what I have at the moment.

    Could someone please assist me with the conclusion of this proof?

    Thank you very much.


    [PLAIN]http://img222.imageshack.us/img222/7317/ps24proof.jpg [Broken]
    Last edited by a moderator: May 5, 2017
  7. Oct 14, 2010 #6
    Last edited by a moderator: May 5, 2017
  8. Oct 17, 2010 #7
    Suppose [tex]\lim a_n=L[/tex]. Let [tex]a_{n_i}[/tex] be an arbitrary subsequence. We wish to prove that [tex]\lim a_{n_i}=L[/tex].

    We know that [tex]\forall \epsilon>0[/tex] [tex] \exists N[/tex] such that [tex]n\geq N[/tex] implies [tex]|a_n-L|<\epsilon[/tex].

    We need to show that [tex]\forall \epsilon>0[/tex] [tex] \exists I[/tex] such that [tex]i\geq I[/tex] implies [tex]|a_{n_i}-L|<\epsilon[/tex]. But if we choose [tex]I[/tex] such that [tex]i\geq I[/tex] implies that [tex]n_i\geq N[/tex], then [tex]i\geq I[/tex] implies [tex]|a_{n_i}-L|<\epsilon[/tex]. The existence of such an [tex]I[/tex] is guaranteed by the definition of a subsequence (something to check).

    So what this proof says is that if a sequence is convergent, then all it's subsequences converge to the same limit. Now take the contra-positive of this statement and compare it with your question.

    I hope that helped.

    [edit] Just noticed the date of the OP's post... oh well.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook