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showing that a sequence converges - help Watch

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    Are you sure you can just square it? Wouldn't it be better to just keep it as sqrt( n/(n+1) )?

    Were you not writing a delta-epsilon, I would mention that n=(n+1)-(1).

    But I really have no idea how to do these formal proof thingies
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    (Original post by Kaya_01)
    I need to show that this sequence Attachment 203242 converges as n tends to infinity.

    I know I have to use the fact that abs(a_n - L) < e (e being epsilon)

    I've started off by removing the square root by squaring the expression. I now have n/(n+1). What is the limit for this?

    My proof is starting with stating with the fact that e > 0


    Any help is hugely hugely appreciated! thank you!
    \displaystyle \frac{\sqrt{n}}{\sqrt{n+1}} = \frac{\sqrt{n+1-1}}{\sqrt{n+1}}=
    \displaystyle \sqrt{1-\frac{1}{n+1}}
    As \displaystyle n \rightarrow \infty so \displaystyle \frac{1}{n+1}\rightarrow 0 that is it will be smaller and smaller
    and
    \displaystyle \frac{\sqrt{n}}{\sqrt{n+1}} \rightarrow \sqrt{1-0}=1
    that is convergent
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    (Original post by ztibor)
    \displaystyle \frac{\sqrt{n}}{\sqrt{n+1}} = \frac{\sqrt{n+1-1}}{\sqrt{n+1}}=
    \displaystyle \sqrt{1-\frac{1}{n+1}}
    As \displaystyle n \rightarrow \infty so \displaystyle \frac{1}{n+1}\rightarrow 0
    and
    \displaystyle \frac{\sqrt{n}}{\sqrt{n+1}} \rightarrow \sqrt{1-0}=1
    that is convergent
    hey there - thank you so so much!!

    can I ask what the purpose of rationalizing this expression is?
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    As someone else has said, you can't simply square it without knowing that  x \mapsto x^2 is continuous for all of \mathbb{R}, because that's the only way you can square it and conclude the convergence implies the convergence of \dfrac{\sqrt{n}}{\sqrt{n+1}}
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    (Original post by ztibor)
    \displaystyle \frac{\sqrt{n}}{\sqrt{n+1}} = \frac{\sqrt{n+1-1}}{\sqrt{n+1}}=
    \displaystyle \sqrt{1-\frac{1}{n+1}}
    As \displaystyle n \rightarrow \infty so \displaystyle \frac{1}{n+1}\rightarrow 0 that is it will be smaller and smaller
    and
    \displaystyle \frac{\sqrt{n}}{\sqrt{n+1}} \rightarrow \sqrt{1-0}=1
    that is convergent
    Good explanation!

    (i`d used that the sequence equals: (1+\frac{1}{n})^{-1/2}
 
 
 
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