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

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

    N.T.S (need to show) $\displaystyle a_n$ converges.

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

    $\displaystyle a_2<a_4<a_6<........<a_{2n}<a$

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

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

    Adding these two expressions gives:

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

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

    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 $\displaystyle a_1$. However, I can't see a way of connecting the two subsequences apart from adding them together.

    I was thinking of making $\displaystyle a_n=\frac{a_{2n}+a_{2n+1}}{2}$ 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:

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

    N.T.S (need to show) $\displaystyle a_n$ converges.

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

    $\displaystyle a_2<a_4<a_6<........<a_{2n}<a$

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

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

    Adding these two expressions gives:

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

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

    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 $\displaystyle a_1$. However, I can't see a way of connecting the two subsequences apart from adding them together.

    I was thinking of making $\displaystyle a_n=\frac{a_{2n}+a_{2n+1}}{2}$ 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 $\displaystyle a_{2k} \to a$, we have that given $\displaystyle \epsilon > 0$, $\displaystyle \exists N_1 \in \mathbb{N}$ such that $\displaystyle 2k > N_1$ implies $\displaystyle |a_{2k} - a| < \epsilon$

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

    Choose $\displaystyle N = \text{max} \{ N_1, N_2 \}$. then, $\displaystyle n > N$ implies $\displaystyle |a_n - a| < \epsilon$. (since $\displaystyle n$ is either even or odd, and in either case, $\displaystyle |a_n - a| < \epsilon$ 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 $\displaystyle
    |a_{2k} - a| < \epsilon
    $? 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 $\displaystyle
    |a_{2k} - a| < \epsilon
    $? 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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