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Math Help - Need help on showing convergence of sqrt(n+1)-sqrt(n)

  1. #1
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    Need help on showing convergence of sqrt(n+1)-sqrt(n)

    y_n := sqrt(n+1) - sqrt(n)

    the question asks us to show that y_n and sqrt(n)*y_n converge and to find their limits

    i can see that y_n would converge to 0, but i can't show how. i don't even know what the second converges to.

    my strategy was to use the x_(n+1)/x_n theorem, but I was left with a messy equation whose comparison to 1 i can't illustrate.

    at the very least i ask what theorem i should use. thanks.
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  2. #2
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    Quote Originally Posted by Skerven View Post
    y_n := sqrt(n+1) - sqrt(n)

    the question asks us to show that y_n and sqrt(n)*y_n converge and to find their limits

    i can see that y_n would converge to 0, but i can't show how. i don't even know what the second converges to.

    my strategy was to use the x_(n+1)/x_n theorem, but I was left with a messy equation whose comparison to 1 i can't illustrate.

    at the very least i ask what theorem i should use. thanks.

    Multiply the expression by 1:

    \left(\sqrt{n+1}-\sqrt{n}\right)\left(\frac{\sqrt{n+1}+\sqrt{n}}{\s  qrt{n+1}+\sqrt{n}}\right)

    Develop the above and get that the limit indeed is zero.

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    Multiply the expression by 1:

    \left(\sqrt{n+1}-\sqrt{n}\right)\left(\frac{\sqrt{n+1}+\sqrt{n}}{\s  qrt{n+1}+\sqrt{n}}\right)

    Develop the above and get that the limit indeed is zero.

    Tonio
    Ok I tried the conjugate technique but I got a messy equation. Ok I'll do it again thanks..
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  4. #4
    MHF Contributor chisigma's Avatar
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    The following identity may be useful for You...

    \sqrt{n+1} - \sqrt{n} = \frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}= \frac{1}{\sqrt{n+1}+\sqrt{n}}

    Kind regards

    \chi \sigma
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  5. #5
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    Quote Originally Posted by chisigma View Post
    The following identity may be useful for You...

    \sqrt{n+1} - \sqrt{n} = \frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}= \frac{1}{\sqrt{n+1}+\sqrt{n}}

    Kind regards

    \chi \sigma
    how will this help me in finding the limit of sqrt(n)*y_n?

    *edit* sorry guys i solved it myself sqrt(n)*y_n --> 1/2. You just reduce the num. and the denom. to convergent sequences through alg. manipulation and apply a theorem or two.
    Last edited by Skerven; October 20th 2009 at 08:26 AM.
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  6. #6
    MHF Contributor chisigma's Avatar
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    Is...

    \lim_{n \rightarrow \infty} \frac{\sqrt{n}}{\sqrt{n+1}+\sqrt{n}} = \lim_{n \rightarrow \infty} \frac{1}{\sqrt{1+\frac{1}{n}}+1} = \frac{1}{2}

    Kind regards

    \chi \sigma
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  7. #7
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    Quote Originally Posted by Skerven View Post
    how will this help me in finding the limit of sqrt(n)*y_n?

    *edit* sorry guys i solved it myself sqrt(n)*y_n --> 1/2. You just reduce the num. and the denom. to convergent sequences through alg. manipulation and apply a theorem or two.

    Yes, of course...but you could NOT do that before you simplified the expression, right? We already knew the limit is 1/2, you didn't.

    Tonio
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