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Thread: sequence has a limit

  1. #1
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    sequence has a limit

    I have the sequence from the picture and I have to demonstrate that this sequence has a limit.
    I always get stuck at this kind of exercises.How to approach an exercise like this?
    Attached Thumbnails Attached Thumbnails sequence has a limit-1.png  
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    Re: sequence has a limit

    The usual approach to a problem like this is to see if you can show the sequence is bounded and monotone. Have you tried that?
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    Re: sequence has a limit

    Yes, I tried to solve x_{n+1}-x_{n} but I didn;t get too far.
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    Re: sequence has a limit

    Well, it's hard to help you when you don't show what you have tried or tell us what you are thinking. You didn't give us the exact statement of the problem so I don't know whether you are given that the sequence is convergent and asked to prove it, or whether you are to determine whether or not it is convergent. If you suspect (or know) it to be convergent, then I would try to decide whether it is monotone as a first or second step. If it might not be convergent I would suppose the limit exists and see what that tells you. Actually, in the second case I would try that first. So tell us more about what you are doing.
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    Re: sequence has a limit

    Quote Originally Posted by Vali View Post
    I have the sequence from the picture and I have to demonstrate that this sequence has a limit.
    I always get stuck at this kind of exercises.How to approach an exercise like this?
    Vali, do you ever post any of your work so that we can actually help you?
    We want to help, but we think simply supplying an answer is not helping.
    That said, note that if $t\in(0,1)$ then is it true that $\dfrac{1}{t}>0$.
    Is it true that if $\varepsilon>0$ then $t<t+\varepsilon~?$
    What can you say about the monotonic nature of the given sequence?
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    Re: sequence has a limit

    If the sequence has a limit, X, then we can take the limit on both sides of the recursion equation, $\displaystyle x_{n+1}= x_n+ \frac{1}{n}$. That gives $\displaystyle X= X+ \frac{1}{X}$ which is the same as $\displaystyle \frac{1}{X}= 0$. That is impossible.
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    Re: sequence has a limit

    Which shows that if you wait long enough, someone will work it for you.
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    Re: sequence has a limit

    show by induction

    $\displaystyle x_n>\sqrt{n}$
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    Re: sequence has a limit

    I tried to calculate x_(n+1) - x_(n) which is 1/x_(n) which is positive so x_(n) increases.Then, I assumed that x_(n)>0 and demonstrate that x_(n+1) > 0 but x_(n+1)=x_(n)+1/x_(n) so x_(n+1)>0 and now I don't know how to continue.
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  10. #10
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    Re: sequence has a limit

    If $\displaystyle x_0> 0$ and the sequence is increasing, of course $\displaystyle x_n> 0$ for all n! But why are you trying to prove that?
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    Re: sequence has a limit

    If xn > 0 that means that the sequence has a limit.I don't need to find the limit, just to prove that it's there.
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    Re: sequence has a limit

    Quote Originally Posted by Vali View Post
    If xn > 0 that means that the sequence has a limit.
    Why?

    -Dan
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  13. #13
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    Re: sequence has a limit

    Because from school I know that any increasing sequence has a limit.In this case it can be infinity if the sequence is unbounded or it can be a number if the sequence is bounded.If I assume that the sequence is bounded I get L-L=1/L, contradiction.
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    Re: sequence has a limit

    What about this increasing sequence: 1, 2, 3, 4, 5, 6, 7, 8, ...? Does that have a limit?
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    Re: sequence has a limit

    Oops, I misread. But "infinity" is not a limit.
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