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Math Help - Proof of a limit of an infinite sequence

  1. #1
    Junior Member Kanwar245's Avatar
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    Proof of a limit of an infinite sequence

    Find and prove

    \lim_{n\rightarrow\infty}(\sqrt{n+3} - \sqrt{n})
    Last edited by Kanwar245; March 15th 2013 at 07:34 PM.
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    Junior Member Kanwar245's Avatar
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    Re: Proof of a limit of an infinite sequence

    I'm thinking of rationalizing it first, also I can see that the limit approaches 0
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    Re: Proof of a limit of an infinite sequence

    That sounds good- multiply by \frac{\sqrt{n+3}+ \sqrt{n}}{\sqrt{n+3}+ \sqrt{n}}
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    Junior Member Kanwar245's Avatar
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    Re: Proof of a limit of an infinite sequence

    yes and it becomes
    \frac{3}{(\sqrt{n+3}+\sqrt{n})}
    Last edited by Kanwar245; March 15th 2013 at 07:35 PM.
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    Re: Proof of a limit of an infinite sequence

    From
    a_n=\frac{3}{\sqrt{n+3}+\sqrt{n}}

    You can show that this is always less than
    b_n=\frac{3}{\sqrt{n}+\sqrt{n}}

    and always greater than
    c_n=0

    Using the sandwich theorem show that an converges to zero.
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    Re: Proof of a limit of an infinite sequence

    Quote Originally Posted by Shakarri View Post
    From
    a_n=\frac{3}{\sqrt{n+3}+\sqrt{n}}

    You can show that this is always less than
    b_n=\frac{3}{\sqrt{n}+\sqrt{n}}

    and always greater than
    c_n=0

    Using the sandwich theorem show that an converges to zero.
    Isn't the sandwich theorem overkill here? The denominator clearly goes to infinity, so the whole fraction goes to 0.
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    Re: Proof of a limit of an infinite sequence

    Oh yes you could do it that way
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    Junior Member Kanwar245's Avatar
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    Re: Proof of a limit of an infinite sequence

    right but when proving, for all epsilon greater than 0, what N would I choose so that |\frac{3}{2\sqrt{n}}| < \epsilon
    If i try to square it, epsilon gets squared too...
    Last edited by Kanwar245; March 15th 2013 at 07:36 PM.
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    Re: Proof of a limit of an infinite sequence

    Quote Originally Posted by Kanwar245 View Post
    right but when proving, for all epsilon greater than 0, what N would I choose so that |3/2sqrt(n)| < epsilon
    If i try to square it, epsilon gets squared too...
    Why bother squaring anything? Since this is a sequence, we know n is positive, and so the stuff inside the absolute value has to be positive. This means its equal to its absolute value (or the absolute value sign can be neglected).
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    Re: Proof of a limit of an infinite sequence

    Right so we have this:
    |\frac{3}{\sqrt{n} + \sqrt{n}}| < \epsilon
    \Rightarrow |\frac{3}{2\sqrt{n}}| < \epsilon
    \Rightarrow \frac{3}{2\sqrt{n}} < \epsilon
    Now from this point on, if I try to isolate n, at some point I will have to square \sqrt{n} which will yield {\epsilon}^2 ...
    Last edited by Kanwar245; March 15th 2013 at 07:37 PM.
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    Re: Proof of a limit of an infinite sequence

    Quote Originally Posted by Kanwar245 View Post
    Right so we have this:
    |\frac{3}{\sqrt{n} + \sqrt{n}}| < \epsilon
    \Rightarrow |\frac{3}{2\sqrt{n}}| < \epsilon
    \Rightarrow \frac{3}{2\sqrt{n}} < \epsilon
    Now from this point on, if I try to isolate n, at some point I will have to square \sqrt{n} which will yield {\epsilon}^2 ...
    What's wrong with that?
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    Re: Proof of a limit of an infinite sequence

    Quote Originally Posted by Prove It View Post
    What's wrong with that?
    Oh so we can do this?
    \sqrt{n} > \frac{3}{2\epsilon}

    n > \frac{9}{4\epsilon^2}

    So when proving, I can say
    Given \epsilon > 0
    Choose N = \frac{9}{4\epsilon^2}?
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    Re: Proof of a limit of an infinite sequence

    Quote Originally Posted by Kanwar245 View Post
    Oh so we can do this?
    \sqrt{n} > \frac{3}{2\epsilon}

    n > \frac{9}{4\epsilon^2}

    So when proving, I can say
    Given \epsilon > 0
    Choose N = \frac{9}{4\epsilon^2}?
    You certainly can
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    Re: Proof of a limit of an infinite sequence

    Okay Perfect!
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