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Math Help - Sequence Proof

  1. #1
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    Sequence Proof

    Prove that if x_n --> infinity, then the sequence given by

    x_n / x_n+1 is convergent.


    I am supposed to use the definition of divergence, so I know there is a number n for which all x_n > M for any M chosen.
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  2. #2
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    Re: Sequence Proof

    Quote Originally Posted by veronicak5678 View Post
    Prove that if x_n --> infinity, then the sequence given by

    x_n / x_n+1 is convergent.


    I am supposed to use the definition of divergence, so I know there is a number n for which all x_n > M for any M chosen. i think you are using an incorrect definition of divergence
    to support what i have written in red:
    define x_n as x_{2k}=0, x_{2k+1}=1 \,  \forall k \in \mathbb{Z}^+.
    then x_n is divergent since its not convergent. But if we take M=5(say) we can't find any n which would satisfy x_n>M.

    Also can you please explicitly write what definition of x_n \rightarrow \infty you are using??
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  3. #3
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    Re: Sequence Proof

    Quote Originally Posted by veronicak5678 View Post
    Prove that if x_n --> infinity, then the sequence given by

    x_n / x_n+1 is convergent.
    Consider the sequence x_n  = \left\{ \begin{gathered}  n,\text{ n odd} \hfill \\  2^n ,\text{ n even} \hfill \\ \end{gathered}  \right..

    Is it true that \left( {x_n } \right) \to \infty ~?

    What can you say about \frac{{x_n }}{{x_{n + 1} }}~?
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    Re: Sequence Proof

    Thanks for the answers. The definition of a limit at infinity from my book is as follows:

    sn → ∞ as n → ∞
    provided that for every number M there is an integer N so that
    sn ≥ M
    whenever n ≥ N.
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  5. #5
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    Re: Sequence Proof

    Quote Originally Posted by veronicak5678 View Post
    The definition of a limit at infinity from my book is as follows:
    sn → ∞ as n → ∞
    provided that for every number M there is an integer N so that
    sn ≥ M whenever n ≥ N.
    That definition is standard. Now apply it to the example I gave you.
    It will show that (x_n)\to\infty.

    BUT does \frac{{x_n }}{{x_{n + 1} }} converge as claimed.
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    Re: Sequence Proof

    Not sure I follow...
    We could use the same definition to show that
    x_n-> inf implies x_n+1-> inf, but how does that prove that x_n / x_n+1 converges?
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  7. #7
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    Re: Sequence Proof

    Quote Originally Posted by veronicak5678 View Post
    Not sure I follow...
    We could use the same definition to show that
    x_n-> inf implies x_n+1-> inf, but how does that prove that x_n / x_n+1 converges?
    Which of these two do you mean:
    A~\left( {\frac{{x_n }}{{x_{n + 1} }}} \right)\text{  or  }B~\left( {\frac{{x_n }}{{x_n  + 1}}} \right)

    If it is A then the statement is false.
    If it is B then the proof is trivial.
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  8. #8
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    Re: Sequence Proof

    Sorry for being unclear. I mean B.
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  9. #9
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    Re: Sequence Proof

    Quote Originally Posted by veronicak5678 View Post
    Sorry for being unclear. I mean B.
    If (x_n)\to\infty then \left( {\frac{1}{{x_n+1 }}} \right) \to 0.

    Note that \left( {\frac{{x_n }}{{x_n  + 1}}} \right) = \left( {1 - \frac{1}{{x_n  + 1}}} \right)
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  10. #10
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    Re: Sequence Proof

    I understand that, but how can I prove this using just the definition of divergence, and no theorems about limits?
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  11. #11
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    Re: Sequence Proof

    Quote Originally Posted by veronicak5678 View Post
    I understand that, but how can I prove this using just the definition of divergence, and no theorems about limits?
    If \varepsilon  > 0 use the divergence definition to make x_n  + 1 > \frac{1}{{1 + \varepsilon }}.

    Having to do it this is just busy work.
    So that is as far as I am willing to take it.
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  12. #12
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    Re: Sequence Proof

    OK. Thanks for helping me.
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