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Math Help - Convergence Sequence (Analysis)

  1. #1
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    Convergence Sequence (Analysis)

    Hi, im not sure if this is in calculus or number theory section...

    Using the definition of a convergent sequence show that <br />
\sqrt{n+1} - \sqrt{n}<br />
converges to zero.

    Thanks in advance!!
    -ynot-
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ynotidas View Post
    Hi, im not sure if this is in calculus or number theory section...

    Using the definition of a convergent sequence show that <br />
\sqrt{n+1} - \sqrt{n}<br />
converges to zero.

    Thanks in advance!!
    -ynot-
    Big Hint: multiply by the conjugate over itself and simplify. of course, you have to do this within the framework of the definition
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  3. #3
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    Thanks!!
    i got to

    1/(\sqrt{n+1} + \sqrt{n})

    i have no idea what to do next??
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ynotidas View Post
    Thanks!!
    i got to

    1/(\sqrt{n+1} + \sqrt{n})

    i have no idea what to do next??
    ok, another hint:

    \Bigg| \frac 1{\sqrt{n + 1} + \sqrt{n} }\Bigg| \le \Bigg| \frac 1{\sqrt{n} + \sqrt{n} } \Bigg| = \Bigg| \frac 1{2 \sqrt{n} }\Bigg|

    now see if you can finish up. look at the definition of convergence to see what you have to show
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  5. #5
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    thanks, do i do this?

    \frac 1{2\sqrt{n}}<epsilon

    then n>1/(4epsilon^2)

    thus |An - 0| < epsilon if n>1 & n>1/(4epsilon^2)

    n*> max{1,1/(4epsilon^2)}

    n>=n*, we have |An-0|<epsilon

    [SORRY it's taking me forever to type it in the [math} mode]
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ynotidas View Post
    thanks, do i do this?

    \frac 1{2\sqrt{n}}<epsilon

    then n>1/(4epsilon^2)

    thus |An - 0| < epsilon if n>1 & n>1/(4epsilon^2)

    n*> max{1,1/(4epsilon^2)}

    n>=n*, we have |An-0|<epsilon

    [SORRY it's taking me forever to type it in the [math} mode]
    you want to show that

    For every \epsilon > 0, there exists an N \in \mathbb{N}, such that n > N implies

    |\sqrt{n + 1} - \sqrt{n} - 0| < \epsilon

    So let \epsilon > 0 and choose N such that N > \frac 1{4 \epsilon ^2}

    Then blah blah blah < \epsilon

    QED
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