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Math Help - subsequences and convergence.

  1. #1
    Super Member Showcase_22's Avatar
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    subsequences and convergence.

    This is a question that seems obvious except my method doesn't look quite right.

    Suppose we have a sequence a_n and are trying to prove that it converges. Assume that we have shown that the subsequences (a_{2n}) and (a_{2n+1}) both converge to the same limit a. Prove that (a_n) \rightarrow a converges.
    Here's what i've done so far:

    a_{2n} \rightarrow a
    a_{2n+1} \rightarrow a

    N.T.S (need to show) a_n converges.

    (a_{2n})^\infty_{n=1}:

    a_2<a_4<a_6<........<a_{2n}<a

    (a_{2n+1})^ \infty_{n=1}

    a_3<a_5<a_7<.........<a_{2n+1}<a

    Adding these two expressions gives:

    a_2+a_3<a_4+a_5<a_6+a_7<.......<a_{2n}+a_{2n+1}<2a

    \frac{a_2+a_3}{2}<\frac{a_4+a_5}{2}<\frac{a_6+a_7}  {2}<.....<\frac{a_{2n}+a_{2n+1}}{2}<a

    From here i'm not sure where to proceed. I'm not even sure if this is the right way of doing it!

    I have noticed that I can get all the terms except for a_1. However, I can't see a way of connecting the two subsequences apart from adding them together.

    I was thinking of making a_n=\frac{a_{2n}+a_{2n+1}}{2} and then I would have a nice increasing sequence that has a limit a. Otherwise i'm having some difficulty in seeing what to do next.

    Could someone please give me a nudge in the right direction? I would be grateful if they could!

    EDIT: I've just noticed that you can get a_1 if youmake n=0 for the second subsequence. I can't actually see how this helps unless I can somehow see the property that a_{2n+1} is less than a_n. Even then I would have trouble justifying the order a_1<a_2_a_3......etc
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    This is a question that seems obvious except my method doesn't look quite right.



    Here's what i've done so far:

    a_{2n} \rightarrow a
    a_{2n+1} \rightarrow a

    N.T.S (need to show) a_n converges.

    (a_{2n})^\infty_{n=1}:

    a_2<a_4<a_6<........<a_{2n}<a

    (a_{2n+1})^ \infty_{n=1}

    a_3<a_5<a_7<.........<a_{2n+1}<a

    Adding these two expressions gives:

    a_2+a_3<a_4+a_5<a_6+a_7<.......<a_{2n}+a_{2n+1}<2a

    \frac{a_2+a_3}{2}<\frac{a_4+a_5}{2}<\frac{a_6+a_7}  {2}<.....<\frac{a_{2n}+a_{2n+1}}{2}<a

    From here i'm not sure where to proceed. I'm not even sure if this is the right way of doing it!

    I have noticed that I can get all the terms except for a_1. However, I can't see a way of connecting the two subsequences apart from adding them together.

    I was thinking of making a_n=\frac{a_{2n}+a_{2n+1}}{2} and then I would have a nice increasing sequence that has a limit a. Otherwise i'm having some difficulty in seeing what to do next.

    Could someone please give me a nudge in the right direction? I would be grateful if they could!

    EDIT: I've just noticed that you can get a_1 if youmake n=0 for the second subsequence. I can't actually see how this helps unless I can somehow see the property that a_{2n+1} is less than a_n. Even then I would have trouble justifying the order a_1<a_2_a_3......etc
    i am probably oversimplifying this, but it seems to me you can do this

    Since a_{2k} \to a, we have that given \epsilon > 0, \exists N_1 \in \mathbb{N} such that 2k > N_1 implies |a_{2k} - a| < \epsilon

    similarly, we have for \{ a_{2k + 1} \}, there is an N_2 \in \mathbb{N} such that 2k + 1 > N_2 implies |a_{2k + 1} - a| < \epsilon.

    Choose N = \text{max} \{ N_1, N_2 \}. then, n > N implies |a_n - a| < \epsilon. (since n is either even or odd, and in either case, |a_n - a| < \epsilon by the two previous convergences)
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  3. #3
    Super Member Showcase_22's Avatar
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    When I was in a lecture I noticed there was a fundamental flaw with what I had written. I had only acknowledged the situation where a is an upper bound!

    I have an alternative method but it is still longer than yours.

    Have you made n=2k and then used the relationship n>N to generate <br />
|a_{2k} - a| < \epsilon<br />
? The same goes for the second inequality.

    If so, that's a really good way of thinking about it. I'll remember that for some other horrible question!
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    When I was in a lecture I noticed there was a fundamental flaw with what I had written. I had only acknowledged the situation where a is an upper bound!

    I have an alternative method but it is still longer than yours.

    Have you made n=2k and then used the relationship n>N to generate <br />
|a_{2k} - a| < \epsilon<br />
? The same goes for the second inequality.

    If so, that's a really good way of thinking about it. I'll remember that for some other horrible question!
    yes, n can be odd or even, so i took turns making n = 2k and then n = 2k + 1. and we know about the subsequences with subscripts like that. so we use that information
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