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    Hello everyone,

    I am so confused by this proof question. It's kinda long. Any help will be greatly appreciated, like some hints etc.

    The Question is

    "Show that if a(n) converges to a not equal to 0 then a(n+1)/a(n) tends to 1 as n tends to infinity.. Show that the opposite statement is not true. By giving an example of a divergent sequence a(n) such that a(n+1)/a(n) tends to 1 as n tends to infinity"

    I just don't know where to start. I thought maybe I could prove monotonicity for convergence but then I have no idea what to do next.
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    (Original post by Mr XcX)
    Hello everyone,
    "Show that if a(n) converges to a not equal to 0 then a(n+1)/a(n) tends to 1 as n tends to infinity.. Show that the opposite statement is not true. By giving an example of a divergent sequence a(n) such that a(n+1)/a(n) tends to 1 as n tends to infinity"
    Remember that if a(n) converges to a !=0, then given any \epsilon >0 then there is an N such that n > N implies a(n) is within \epsilon of a. So if n>N, what are the bounds on a(n+1)/a(n)?

    The second part is very very easy!
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    (Original post by Gregorius)
    Remember that if a(n) converges to a !=0, then given any \epsilon >0 then there is an N such that n > N implies a(n) is within \epsilon of a. So if n>N, what are the bounds on a(n+1)/a(n)?

    The second part is very very easy!
    Okay, thank you.
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    (Original post by Mr XcX)
    Hello everyone,

    I am so confused by this proof question. It's kinda long. Any help will be greatly appreciated, like some hints etc.

    The Question is

    "Show that if a(n) converges to a not equal to 0 then a(n+1)/a(n) tends to 1 as n tends to infinity.. Show that the opposite statement is not true. By giving an example of a divergent sequence a(n) such that a(n+1)/a(n) tends to 1 as n tends to infinity"

    I just don't know where to start. I thought maybe I could prove monotonicity for convergence but then I have no idea what to do next.
    I think that you can approach this by considering that:

    1. a_n converges implies that a_n is Cauchy
    2. A Cauchy sequence is bounded.

    Then bear in mind that you are going to be considering the quantity |\frac{a_{n+1}}{a_n}-1|
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    (Original post by atsruser)
    I think that you can approach this by considering that:

    1. a_n converges implies that a_n is Cauchy
    2. A Cauchy sequence is bounded.

    Then bear in mind that you are going to be considering the quantity |\frac{a_{n+1}}{a_n}-1|
    Thank you
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    (Original post by atsruser)
    I think that you can approach this by considering that:

    1. a_n converges implies that a_n is Cauchy
    2. A Cauchy sequence is bounded.

    Then bear in mind that you are going to be considering the quantity |\frac{a_{n+1}}{a_n}-1|
    I have to say I'm not really seeing how this will work (for sure, you can make it work, but it won't be any easier than working from 1st principles I think).

    Note in particular that you will have to use the fact that a_n \not \to 0, since without this condition the desired result need not hold.
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    I seem to remember a theorem which maybe you have seen as well it says if a sequence of real numbers converges to some real number a then all subsequences of a_n will converge to that same number a. Hence you know that a_n and a_(n+1) both converge to a (since a_(n+1) is a subsequence of a_n) so you can use the algebra of limits for quotients since in this case a!=0 and the result follows that the limit is one. Alternatively you could argue from the definitions.

    For the second part just have a think and you can come up with some sequences easily.
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    (Original post by poorform)
    I seem to remember a theorem which maybe you have seen as well it says if a sequence of real numbers converges to some real number a then all subsequences of a_n will converge to that same number a. Hence you know that a_n and a_(n+1) both converge to a (since a_(n+1) is a subsequence of a_n) so you can use the algebra of limits for quotients since in this case a!=0 and the result follows that the limit is one.
    That's quite clever (in that I think it does manage to use known results while avoiding doing anything too obviously like assuming the answer).

    At the same time, the proof of the a_n/b_n = a/b result is so close in approach to the result here that I am certain quoting it is not what is intended.
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    (Original post by DFranklin)
    That's quite clever (in that I think it does manage to use known results while avoiding doing anything too obviously like assuming the answer).

    At the same time, the proof of the a_n/b_n = a/b result is so close in approach to the result here that I am certain quoting it is not what is intended.
    Yeah I suspect he is meant to argue using the definitions as well. But it depends on the context really he may not have even seen what I was quoting but I thought I'd post it as it came to mind that if you are allowed to use results like AOL etc. The result comes without much effort at all.
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    (Original post by DFranklin)
    I have to say I'm not really seeing how this will work (for sure, you can make it work, but it won't be any easier than working from 1st principles I think).
    I must admit that I did the working in my head, and the inequality that went the right way in my head went the wrong way on paper. But it seems easy to make it work, AFAICS:

    \epsilon > |a_{n+1}-a_n| = |a_n||\frac{a_{n+1}}{a_n}-1| > K |\frac{a_{n+1}}{a_n}-1|

    is what I had in mind.

    Note in particular that you will have to use the fact that a_n \not \to 0, since without this condition the desired result need not hold.
    That was explicitly stated in the question, so I can't see any objection to relying on it.
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    (Original post by atsruser)
    I must admit that I did the working in my head, and the inequality that went the right way in my head went the wrong way on paper. But it seems easy to make it work, AFAICS:

    \epsilon > |a_{n+1}-a_n| = |a_n||\frac{a_{n+1}}{a_n}-1| > K |\frac{a_{n+1}}{a_n}-1|
    What is K here?

    Spoiler:
    Show
    To my mind the "key point" you're going to need to show in a proof of the original question is that you can bound a_n away from 0. (i.e. that we can find m > 0 and N s.t. |a_n| > m for n > N). The rest is just routine manipulation.

    My problem with all the responses other than Gregorius is that no-one really seems to be showing this, or even acknowledging that it needs to be shown. Without explaining what K is, it seems to me you are falling into the same trap.


    That was explicitly stated in the question, so I can't see any objection to relying on it.
    You misunderstand. You absolutely have to rely on it. And so an argument that says "you can solve this by using X, Y and Z" where none of X, Y or Z have anything to do with the key fact that a_n doesn't tend to 0 seems to me to be missing the point.
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    (Original post by DFranklin)
    What is K here?
    Spoiler:
    Show
    Well, if you want a specific value, take K = |a|/2 > 0 and then |a_n|>K for sufficiently large n


    Spoiler:
    Show
    To my mind the "key point" you're going to need to show in a proof of the original question is that you can bound a_n away from 0. (i.e. that we can find m > 0 and N s.t. |a_n| > m for n > N). The rest is just routine manipulation.

    My problem with all the responses other than Gregorius is that no-one really seems to be showing this, or even acknowledging that it needs to be shown. Without explaining what K is, it seems to me you are falling into the same trap.
    Well, this is someone's homework, and we don't want to do it all, do we?

    You misunderstand. You absolutely have to rely on it. And so an argument that says "you can solve this by using X, Y and Z" where none of X, Y or Z have anything to do with the key fact that a_n doesn't tend to 0 seems to me to be missing the point.
    Maybe I've got my misunderstanding head on today, but I think what I wrote above only works if a !=0, so this is implicit in the argument (and it's easy to see a counter-example if a_n tends to 0 anyway).
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    I can do the second part easily enough now using examples. It's just the proof, and they want clear written explanations. Anyone know any good websites or videos of an example like this cause I still am just not understanding it.
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    Should I prove that a(n+1) converges to a and then do the subsequence argument that a(n) also converges to a, then a(n+1)/a(n) tends to 1 as n tends to infinity.
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    (Original post by Mr XcX)
    Should I prove that a(n+1) converges to a and then do the subsequence argument that a(n) also converges to a, then a(n+1)/a(n) tends to 1 as n tends to infinity.
    No. You might get away with it if you just want to get the question done, but you won't learn anything.

    Gregorius gave the right suggestion in the first response to your post. Have you acted on it? Explicitly, if

    |a_n - a| < \epsilon, |a_{n+1} -a | < \epsilon

    then what are lower/upper bounds for a_{n+1} / a_n?

    (Edit: you may find it helpful to assume \epsilon is small relative to a as well - you'll need to quantify how small).
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    (Original post by DFranklin)
    No. You might get away with it if you just want to get the question done, but you won't learn anything.

    Gregorius gave the right suggestion in the first response to your post. Have you acted on it? Explicitly, if

    |a_n - a| < \epsilon, |a_{n+1} -a | < \epsilon

    then what are lower/upper bounds for a_{n+1} / a_n?

    (Edit: you may find it helpful to assume \epsilon is small relative to a as well - you'll need to quantify how small).
    I get 1 is less than 1 which is clearly incorrect for the upper or lower bounds. I am so horrific at sequences, its just so tragic.
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    (Original post by Mr XcX)
    I get 1 is less than 1 which is clearly incorrect for the upper or lower bounds. I am so horrific at sequences, its just so tragic.
    It's got nothing to do with sequences.

    Write A = a_n, B = a_{n+1}. Assume (for now) that the limit, a is positive.

    If a - \epsilon < A < a + \epsilon and a - \epsilon < B < a + \epsilon, and epsilon is small enough that a -\epsilon > 0, then:

    What is a lower bound for A / B? (Hint: what's a lower bound for A, and what's an upper bound for B?)

    What is an upper bound for A / B?

    You are going to have to post some actual maths, no-one is just going to post a proof for you.

    [META: Note for other posters - I am aware this is not the shortest way of going about things. I think the OP needs to do this from first principles].
 
 
 
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