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Proof of subsequence convergence

  1. Nov 5, 2013 #1
    prove if ## a_{2k} \rightarrow l ## and ## a_{2k-1} \rightarrow l ## then ## a_n \rightarrow l ## where ## a_{2k} ## and ## a_{2k-1} ## are subsequences of ## a_n ##

    my attempt:

    since: ## a_{2k} \rightarrow l ## then ## \forall \epsilon > 0 ## ##\exists N_1 \in \mathbb{R}## s.t. ##2k > N_1 \Rightarrow |a_{2k} - l|< \epsilon##

    also, since ## a_{2k-1} \rightarrow l ## then ## \forall \epsilon > 0 ## ##\exists N_2 \in \mathbb{R} ## s.t. ## 2k -1 > N_2 \Rightarrow |a_{2k-1} - l | < \epsilon ##

    from the above we have ## k > \dfrac{N_1}{2} ## and ## k > \dfrac{N_2 + 1}{2} ## then let ## N = max \{ \dfrac{N_1}{2}, \dfrac{N_2 + 1}{2} \} ## then for ## n >N ## ## |a_n - l | < \epsilon ##

    does this proof make sense, if not where could I improve?
     
  2. jcsd
  3. Nov 5, 2013 #2

    LCKurtz

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    Your proof looks good. Personally, I prefer to begin such a proof with "Suppose ##\epsilon > 0##" and then dispense with the ##\forall## statements by just using that ##\epsilon## in the argument.
     
  4. Nov 5, 2013 #3
    thanks,

    in my book they have done a different method: http://gyazo.com/6043e5a4f42915dac0f29b629c56f14c

    going by their method wouldn't my method be incorrect and I could simply say let N be the maximum of N_1 and N_2?
     
  5. Nov 5, 2013 #4

    LCKurtz

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    I think your proof is OK as it is. You have that if either ##2k## or ##2k-1## is greater than ##N## then ##a_{2k}## or ##a_{2k-1}## is within ##\epsilon## of the limit. If ##n>N## it fits one of those two cases.
     
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