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Math Help - Convergence of a sequence example

  1. #1
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    Convergence of a sequence example

    Starting to look at Convergence of sequences and given an example without a solution. The solution would really help my understanding of what's going on here:

    x_{n} = \frac{9n^{3}+3}{2n^{3}+5}

    Define what is meant by saying x_{n} converges to the limit l as n goes to infinity. Show that x_{n} converges to \frac{9}{2} as n tends to infinity.

    Now I know you can pull out the fraction and show that the rest tends to 1 (right?), simple enough you can already see the \frac{9}{2} right there but how do you show this? Basically that l = \frac{9}{2}
    Last edited by Shizaru; December 30th 2011 at 06:06 PM.
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  2. #2
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    Re: Convergence of a sequence example

    Perhaps you got typo, as written, the sequence diverges.
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    Re: Convergence of a sequence example

    Quote Originally Posted by Shizaru View Post
    Starting to look at Convergence of sequences and given an example without a solution. The solution would really help my understanding of what's going on here:
    x_{n} = \frac{9n^{3}+3}{2n^{2}+5}
    Define what is meant by saying x_{n} converges to the limit l as n goes to infinity. Show that x_{n} converges to \frac{9}{2} as n tends to infinity.
    There is a real problem with your post:
    \lim _{n \to \infty } \frac{9n^{3}+3}{2n^{2}+5} = \infty

    If it were \lim _{n \to \infty } \frac{9n^{2}+3}{2n^{2}+5} = \frac{9}{2} that is correct.

    Which is it?
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    Re: Convergence of a sequence example

    Yep, sorry, both powers are 3. Corrected it.
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    Re: Convergence of a sequence example

    Quote Originally Posted by Shizaru View Post
    Starting to look at Convergence of sequences and given an example without a solution. The solution would really help my understanding of what's going on here:

    x_{n} = \frac{9n^{3}+3}{2n^{3}+5}

    Define what is meant by saying x_{n} converges to the limit l as n goes to infinity. Show that x_{n} converges to \frac{9}{2} as n tends to infinity.

    Now I know you can pull out the fraction and show that the rest tends to 1 (right?), simple enough you can already see the \frac{9}{2} right there but how do you show this? Basically that l = \frac{9}{2}
    The standard way to deal with a problem like this is to divide top and bottom of the fraction by the highest power of n in sight, and then use some theorems about limits of sums, products and quotients. In this case, the first step is to divide through by n^3:

     \frac{9n^{3}+3}{2n^{3}+5} = \frac{9+\frac3{n^3}}{2+\frac5{n^3}}.

    In the numerator, \tfrac3{n^3}\to0 as n\to\infty. There is a theorem saying that the limit of a sum is the sum of the limits. That theorem tells you that the limit of the numerator is 9+0=9. Similarly, the limit of the denominator is 2+0=2. Finally, another theorem says that if the numerator and denominator of a fraction both have limits (and the denominator has a nonzero limit) then the limit of the quotient is the quotient of the limits. Apply that theorem to see that the limit of \frac{9+\frac3{n^3}}{2+\frac5{n^3}} is 9/2.
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    Re: Convergence of a sequence example

    Another way (using the definition of limit):

    \left |\frac{9n^3+3}{2n^3+5}-\frac{9}{2}\right |<\epsilon \Leftrightarrow \frac{39}{4n^3+10}<\epsilon \Leftrightarrow \sqrt[3]{\frac{39-10\epsilon}{4\epsilon}}<n

    This condition is satisfied for n\geq n_0=\left\lfloor \sqrt[3]{\frac{39-10\epsilon}{4\epsilon}}\right\rfloor +1
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  7. #7
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    Re: Convergence of a sequence example

    Thanks - how would I write up something like this. I know the theorems, and that basically each part can be taken on its own, like as you put it sum of limits = limit of sum and likewise for quotient. I know these rules but I don't know how to express it... to me it just seems like proving 1 + 1 = 2, its intuitive but I need a way of expressing it... I hope you get what I mean,,

    Plus what I was trying to say is that I would rearrange x_{n} into ( \frac{9}{2})( \frac{n^{3} + \frac{1}{3}}{n^{3}+\frac{5}{2}}) show that the second bracket tends to 1 .. is this the same as what you were saying? (just the opposite?) this was my initial working.
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